Chapter 4: Determinants

For a $$2 \times 2$$ determinant $$\begin{vmatrix} a & b \\ c & d \end{vmatrix}$$, the value is $$ad - bc$$.

Here $$a = 2$$, $$b = 4$$, $$c = -1$$, $$d = 2$$. Therefore

$$\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = (2)(2) - (4)(-1) = 4 + 4 = 8.$$

Expand the $$2 \times 2$$ determinant as $$ad - bc$$.

$$\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix} = (x)(x) - (x+1)(x-1)$$

Using $$(x+1)(x-1) = x^2 - 1$$,

$$= x^2 - (x^2 - 1) = x^2 - x^2 + 1 = 1.$$

Two entries of the third column are $$0$$, so expanding along the third column $$C_3$$ is the shortest route. The only nonzero term comes from the $$(1,3)$$ entry.

$$\Delta = 4\,(-1)^{1+3}\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} + 0 + 0$$

$$= 4(1)\big[(-1)(1) - (3)(4)\big] = 4(-1 - 12)$$

$$= 4(-13) = -52.$$

Expand along the first row.

$$\Delta = 0\begin{vmatrix} 0 & \sin\beta \\ -\sin\beta & 0 \end{vmatrix} - \sin\alpha\begin{vmatrix} -\sin\alpha & \sin\beta \\ \cos\alpha & 0 \end{vmatrix} + (-\cos\alpha)\begin{vmatrix} -\sin\alpha & 0 \\ \cos\alpha & -\sin\beta \end{vmatrix}$$

The first term vanishes. For the second $$2 \times 2$$ determinant,

$$\begin{vmatrix} -\sin\alpha & \sin\beta \\ \cos\alpha & 0 \end{vmatrix} = (-\sin\alpha)(0) - (\sin\beta)(\cos\alpha) = -\sin\beta\cos\alpha$$

For the third,

$$\begin{vmatrix} -\sin\alpha & 0 \\ \cos\alpha & -\sin\beta \end{vmatrix} = (-\sin\alpha)(-\sin\beta) - (0)(\cos\alpha) = \sin\alpha\sin\beta$$

Substituting back,

$$\Delta = -\sin\alpha(-\sin\beta\cos\alpha) + (-\cos\alpha)(\sin\alpha\sin\beta)$$

$$= \sin\alpha\sin\beta\cos\alpha - \sin\alpha\sin\beta\cos\alpha = 0.$$

Evaluate each determinant separately.

$$\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = 3(1) - x(x) = 3 - x^2$$

$$\begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix} = 3(1) - 2(4) = 3 - 8 = -5$$

Equating the two values,

$$3 - x^2 = -5 \implies x^2 = 8 \implies x = \pm 2\sqrt{2}.$$

The area of a triangle with vertices $$(x_1, y_1)$$, $$(x_2, y_2)$$, $$(x_3, y_3)$$ is

$$\text{Area} = \frac{1}{2}\,\big|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\,\big|$$

Take $$(x_1,y_1)=(3,8)$$, $$(x_2,y_2)=(-4,2)$$, $$(x_3,y_3)=(5,1)$$:

$$\text{Area} = \frac{1}{2}\,\big|\,3(2 - 1) + (-4)(1 - 8) + 5(8 - 2)\,\big|$$

$$= \frac{1}{2}\,\big|\,3(1) + (-4)(-7) + 5(6)\,\big| = \frac{1}{2}\,|3 + 28 + 30|$$

$$= \frac{1}{2}(61) = \frac{61}{2} \text{ square units.}$$

Given points:

• A(1, 3)
• B(0, 0)
• D(k, 0) (to be found)

1. Equation of the line through A and B (determinant form)

For any point P(x, y) on AB,

$$\begin{vmatrix} x & y & 1\\ 1 & 3 & 1\\ 0 & 0 & 1 \end{vmatrix}=0$$

Expanding along the first row:

$$x(3\cdot1-1\cdot0)-y(1\cdot1-0\cdot0)+1(1\cdot0-3\cdot0)=0$$

$$3x-y=0$$

Hence the required equation is

$$y=3x$$

2. Using the given area to find k

The area of a triangle with vertices $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$ is

$$\text{Area}=\dfrac12\left|\,x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$$

Substitute A(1, 3), B(0, 0), D(k, 0):

$$\text{Area}=\dfrac12\left|1\,(0-0)+0\,(0-3)+k\,(3-0)\right|$$

$$\text{Area}=\dfrac12\left|3k\right|=\dfrac32\,|k|.$$

Given area = 3 sq units, so

$$\dfrac32\,|k|=3\;\;\Longrightarrow\;\;|k|=2.$$

Therefore

$$k=2 \quad\text{or}\quad k=-2.$$

Conclusion

• Equation of AB: $$y=3x.$$
• Possible values of k: $$k=\pm2.$$

Step 1 – Identify the position of the element 6
In the given determinant $$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix},$$ 6 lies in the 2^nd row and the 3^rd column. Hence its minor will be denoted by $$M_{23}$$.

Step 2 – Form the sub-determinant
To obtain $$M_{23}$$ we delete the 2^nd row and 3^rd column that contain the element 6. The remaining 2 × 2 determinant is $$\begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix}.$$

Step 3 – Evaluate the 2 × 2 determinant
For a 2 × 2 determinant $$\begin{vmatrix} a & b \\ c & d \end{vmatrix}$$ the value is $$ad - bc$$. Therefore

$$M_{23} = (1)(8) - (2)(7) = 8 - 14 = -6.$$

Conclusion
The minor of the element 6 in the given determinant is $$-6$$.

The given determinant is

$$\begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}$$

Label the entries as $$a_{11}=1,\;a_{12}=-2,\;a_{21}=4,\;a_{22}=3.$$

1. Minor and cofactor of $$a_{11}$$

Delete row 1 and column 1:

Sub-matrix $$\bigl[3\bigr] \;\Rightarrow\; M_{11}=3.$$

Cofactor $$A_{11}=(-1)^{1+1}M_{11}= (+1)(3)=3.$$

2. Minor and cofactor of $$a_{12}$$

Delete row 1 and column 2:

Sub-matrix $$\bigl[4\bigr] \;\Rightarrow\; M_{12}=4.$$

Cofactor $$A_{12}=(-1)^{1+2}M_{12}=(-1)(4)=-4.$$

3. Minor and cofactor of $$a_{21}$$

Delete row 2 and column 1:

Sub-matrix $$\bigl[-2\bigr] \;\Rightarrow\; M_{21}=-2.$$

Cofactor $$A_{21}=(-1)^{2+1}M_{21}=(-1)(-2)=2.$$

4. Minor and cofactor of $$a_{22}$$

Delete row 2 and column 2:

Sub-matrix $$\bigl[1\bigr] \;\Rightarrow\; M_{22}=1.$$

Cofactor $$A_{22}=(-1)^{2+2}M_{22}=(+1)(1)=1.$$

Summary

Let the given determinant be

$$\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}.$$

For any element $$a_{ij}$$ of a determinant,

• its minor $$M_{ij}$$ is the determinant obtained after deleting the i-th row and the j-th column,
• its cofactor $$A_{ij}$$ is $$A_{ij}=(-1)^{i+j}M_{ij}.$$

We need $$M_{11},\,A_{11},\,M_{21},\,A_{21}$$.

1. Minor and cofactor of $$a_{11}$$

Delete row 1 and column 1:

$$M_{11}=\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}=a_{22}a_{33}-a_{23}a_{32}.$$

Now

$$A_{11}=(-1)^{1+1}M_{11}=(+1)\bigl(a_{22}a_{33}-a_{23}a_{32}\bigr)$$

so

$$A_{11}=a_{22}a_{33}-a_{23}a_{32}.$$

2. Minor and cofactor of $$a_{21}$$

Delete row 2 and column 1:

$$M_{21}=\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix}=a_{12}a_{33}-a_{13}a_{32}.$$

The cofactor is

$$A_{21}=(-1)^{2+1}M_{21}=(-1)^3M_{21}=-M_{21}.$$

Therefore

$$A_{21}=-(a_{12}a_{33}-a_{13}a_{32})=a_{13}a_{32}-a_{12}a_{33}.$$

Hence

• $$M_{11}=a_{22}a_{33}-a_{23}a_{32},\qquad A_{11}=a_{22}a_{33}-a_{23}a_{32},$$
• $$M_{21}=a_{12}a_{33}-a_{13}a_{32},\qquad A_{21}=a_{13}a_{32}-a_{12}a_{33}.$$

Given determinant

$$\Delta = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$$

For each element $$a_{ij}$$ of the determinant:

• The minor $$M_{ij}$$ is the determinant obtained after deleting its i-th row and j-th column.
• The cofactor $$A_{ij}$$ is $$A_{ij}=(-1)^{i+j}M_{ij}.$$

Below, every intermediate 2 × 2 determinant is expanded step-by-step.

Row 1

1. $$M_{11}=\begin{vmatrix}0 & 4\\5 & -7\end{vmatrix}=0(-7)-4\cdot5=-20\;\;\Rightarrow\;\;A_{11}=(-1)^{1+1}M_{11}=-20$$
2. $$M_{12}=\begin{vmatrix}6 & 4\\1 & -7\end{vmatrix}=6(-7)-4\cdot1=-42-4=-46\;\;\Rightarrow\;\;A_{12}=(-1)^{1+2}M_{12}=46$$
3. $$M_{13}=\begin{vmatrix}6 & 0\\1 & 5\end{vmatrix}=6\cdot5-0\cdot1=30\;\;\Rightarrow\;\;A_{13}=(-1)^{1+3}M_{13}=30$$

Row 2

4. $$M_{21}=\begin{vmatrix}-3 & 5\\5 & -7\end{vmatrix}=(-3)(-7)-5\cdot5=21-25=-4\;\;\Rightarrow\;\;A_{21}=(-1)^{2+1}M_{21}=4$$
5. $$M_{22}=\begin{vmatrix}2 & 5\\1 & -7\end{vmatrix}=2(-7)-5\cdot1=-14-5=-19\;\;\Rightarrow\;\;A_{22}=(-1)^{2+2}M_{22}=-19$$
6. $$M_{23}=\begin{vmatrix}2 & -3\\1 & 5\end{vmatrix}=2\cdot5-(-3)\cdot1=10+3=13\;\;\Rightarrow\;\;A_{23}=(-1)^{2+3}M_{23}=-13$$

Row 3

7. $$M_{31}=\begin{vmatrix}-3 & 5\\0 & 4\end{vmatrix}=(-3)\cdot4-5\cdot0=-12\;\;\Rightarrow\;\;A_{31}=(-1)^{3+1}M_{31}=-12$$
8. $$M_{32}=\begin{vmatrix}2 & 5\\6 & 4\end{vmatrix}=2\cdot4-5\cdot6=8-30=-22\;\;\Rightarrow\;\;A_{32}=(-1)^{3+2}M_{32}=22$$
9. $$M_{33}=\begin{vmatrix}2 & -3\\6 & 0\end{vmatrix}=2\cdot0-(-3)\cdot6=18\;\;\Rightarrow\;\;A_{33}=(-1)^{3+3}M_{33}=18$$

Summary

Verification of the relation

$$a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33} =2(-12)+(-3)(22)+5(18) =-24-66+90=0.$$

The given relation is therefore proved.

Step 1 — Write the matrix.

$$A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$$

Step 2 — Find each minor by deleting the row and column of the element. For a $$2 \times 2$$ matrix, the minor of an entry is just the single remaining entry.

• $$M_{11} = 4$$ (delete row 1 and column 1; entry left is $$4$$)
• $$M_{12} = 1$$ (delete row 1 and column 2; entry left is $$1$$)
• $$M_{21} = 3$$ (delete row 2 and column 1; entry left is $$3$$)
• $$M_{22} = 2$$ (delete row 2 and column 2; entry left is $$2$$)

Step 3 — Convert each minor to a cofactor using $$A_{ij}=(-1)^{i+j}M_{ij}$$.

• $$A_{11}=(-1)^{1+1}M_{11}=4$$
• $$A_{12}=(-1)^{1+2}M_{12}=-1$$
• $$A_{21}=(-1)^{2+1}M_{21}=-3$$
• $$A_{22}=(-1)^{2+2}M_{22}=2$$

These cofactors form the cofactor matrix

$$\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}=\begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}.$$

Step 4 — The adjoint is the transpose of the cofactor matrix.

$$adj\,A = \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}^{T}=\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$$

Given matrix

$$A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$$

1. Determinant of $$A$$
   Expand along the first row:
   $$|A| = 1\,(4\cdot4-3\cdot3)-3\,(1\cdot4-3\cdot1)+3\,(1\cdot3-4\cdot1)$$
   $$|A| = 1\,(16-9)-3\,(4-3)+3\,(3-4)=7-3-3=1$$
2. Cofactor matrix $$C=[C_{ij}]$$
   

   $$C_{11}=\bigl|\begin{smallmatrix}4&3\\3&4\end{smallmatrix}\bigr|=16-9=7$$	$$C_{12}=-\bigl|\begin{smallmatrix}1&3\\1&4\end{smallmatrix}\bigr|=-(4-3)=-1$$	$$C_{13}=\bigl|\begin{smallmatrix}1&4\\1&3\end{smallmatrix}\bigr|=3-4=-1$$
   $$C_{21}=-\bigl|\begin{smallmatrix}3&3\\3&4\end{smallmatrix}\bigr|=-(12-9)=-3$$	$$C_{22}=\bigl|\begin{smallmatrix}1&3\\1&4\end{smallmatrix}\bigr|=4-3=1$$	$$C_{23}=-\bigl|\begin{smallmatrix}1&3\\1&3\end{smallmatrix}\bigr|=-(3-3)=0$$
   $$C_{31}=\bigl|\begin{smallmatrix}3&3\\4&3\end{smallmatrix}\bigr|=9-12=-3$$	$$C_{32}=-\bigl|\begin{smallmatrix}1&3\\1&3\end{smallmatrix}\bigr|=-(3-3)=0$$	$$C_{33}=\bigl|\begin{smallmatrix}1&3\\1&4\end{smallmatrix}\bigr|=4-3=1$$

   
   Thus
   $$C = \begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}$$
3. Adjugate of $$A$$
   $$adj\,A = C^{T}=\begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$$
4. Verification of $$A\,adj\,A = |A|\,I$$
   Multiply:

Row 1: $$[1\;3\;3]\begin{bmatrix}7\\-1\\-1\end{bmatrix}=7-3-3=1$$, etc.
Complete multiplication gives
$$A\,adj\,A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=|A|\,I$$

5. Inverse of $$A$$
   Because $$|A|=1$$,
   $$A^{-1}=\dfrac{1}{|A|}\,adj\,A = adj\,A=\begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$$

Given matrices

$$A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$$

We have to verify that $$(AB)^{-1} = B^{-1}A^{-1}.$$

1. Find $$A^{-1}$$

For a $$2 \times 2$$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$, $$\det = ad-bc$$ and $$\bigl(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\bigr)^{-1} = \dfrac1{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$

Determinant of $$A$$:
$$\det A = (2)(-4) - (3)(1) = -8 - 3 = -11.$$

Therefore

$$A^{-1} = \dfrac1{-11}\begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} \dfrac{4}{11} & \dfrac{3}{11} \\ \dfrac{1}{11} & -\dfrac{2}{11} \end{bmatrix}.$$

2. Find $$B^{-1}$$

Determinant of $$B$$:
$$\det B = (1)(3) - (-2)(-1) = 3 - 2 = 1.$$

Hence

$$B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}.$$

3. Compute $$AB$$

$$AB = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 2(1)+3(-1) & 2(-2)+3(3) \\ 1(1)+(-4)(-1) & 1(-2)+(-4)(3) \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix}.$$

4. Find $$(AB)^{-1}$$

Determinant of $$AB$$:
$$\det(AB) = (-1)(-14) - (5)(5) = 14 - 25 = -11.$$

Thus

$$(AB)^{-1} = \dfrac1{-11}\begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix} = \begin{bmatrix} \dfrac{14}{11} & \dfrac{5}{11} \\ \dfrac{5}{11} & \dfrac{1}{11} \end{bmatrix}.$$

5. Compute $$B^{-1}A^{-1}$$

$$B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \dfrac{4}{11} & \dfrac{3}{11} \\ \dfrac{1}{11} & -\dfrac{2}{11} \end{bmatrix}.$$

Carry out the multiplication element-wise:

Therefore

$$B^{-1}A^{-1} = \begin{bmatrix} \dfrac{14}{11} & \dfrac{5}{11} \\ \dfrac{5}{11} & \dfrac{1}{11} \end{bmatrix}.$$

6. Comparison

$$(AB)^{-1} = \begin{bmatrix} \dfrac{14}{11} & \dfrac{5}{11} \\ \dfrac{5}{11} & \dfrac{1}{11} \end{bmatrix} = B^{-1}A^{-1}.$$

Hence verified.

The given matrix is $$A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$$.

Step 1: Compute $$A^2$$.

$$A^2 = A\,A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2\cdot2+3\cdot1 & 2\cdot3+3\cdot2 \\ 1\cdot2+2\cdot1 & 1\cdot3+2\cdot2 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}.$$

Step 2: Compute $$-4A$$.

$$4A = 4\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix}, \qquad -4A = \begin{bmatrix} -8 & -12 \\ -4 & -8 \end{bmatrix}.$$

Step 3: Add the three matrices.

The identity matrix is $$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$ Therefore

$$A^2 - 4A + I = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} + \begin{bmatrix} -8 & -12 \\ -4 & -8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O.$$

Hence $$A^2 - 4A + I = O$$ is verified.

Step 4: Derive $$A^{-1}$$ from the equation.

From $$A^2 - 4A + I = O$$ we write

$$I = 4A - A^2 = A(4I - A).$$

Premultiplying by $$A^{-1}$$ gives

$$A^{-1} = 4I - A.$$

Step 5: Evaluate $$4I - A$$.

$$4I - A = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}.$$

Therefore $$A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}.$$

(A quick check shows $$AA^{-1} = I$$.)

Step 1  Write the system in matrix form

The linear equations $$2x+5y=1, \; 3x+2y=7$$ can be written as $$\begin{bmatrix}2 & 5\\[2pt] 3 & 2\end{bmatrix}\begin{bmatrix}x\\[2pt]y\end{bmatrix}=\begin{bmatrix}1\\[2pt]7\end{bmatrix}.$$

Step 2  Compute the determinant $$\Delta$$ of the coefficient matrix

$$\Delta=\begin{vmatrix}2 & 5\\ 3 & 2\end{vmatrix}=2\times2-5\times3=4-15=-11.$$

Because $$\Delta\neq0$$ (it is $$-11$$), the system has a unique solution and Cramer’s rule applies.

Step 3  Find the numerator determinant $$\Delta_x$$ for $$x$$

Replace the first column of the coefficient matrix by the constants:

$$\Delta_x=\begin{vmatrix}1 & 5\\ 7 & 2\end{vmatrix}=1\times2-5\times7=2-35=-33.$$

Step 4  Calculate $$x$$

$$x=\dfrac{\Delta_x}{\Delta}=\dfrac{-33}{-11}=3.$$

Step 5  Find the numerator determinant $$\Delta_y$$ for $$y$$

Replace the second column of the coefficient matrix by the constants:

$$\Delta_y=\begin{vmatrix}2 & 1\\ 3 & 7\end{vmatrix}=2\times7-1\times3=14-3=11.$$

Step 6  Calculate $$y$$

$$y=\dfrac{\Delta_y}{\Delta}=\dfrac{11}{-11}=-1.$$

Step 7  Verification

• First equation: $$2(3)+5(-1)=6-5=1\, \checkmark$$
• Second equation: $$3(3)+2(-1)=9-2=7\, \checkmark$$

The solution satisfies both equations, so it is correct.

We write the system in matrix form $$A\,X=B$$ where

$$A=\begin{bmatrix}3&-2&3\\2&1&-1\\4&-3&2\end{bmatrix},\;X=\begin{bmatrix}x\\y\\z\end{bmatrix},\;B=\begin{bmatrix}8\\1\\4\end{bmatrix}.$$

1. Determinant of $$A$$

Expanding along the first row,

$$|A|=3\begin{vmatrix}1&-1\\-3&2\end{vmatrix}-(-2)\begin{vmatrix}2&-1\\4&2\end{vmatrix}+3\begin{vmatrix}2&1\\4&-3\end{vmatrix}.$$

Calculate each minor:

• $$\begin{vmatrix}1&-1\\-3&2\end{vmatrix}=1\cdot2-(-1)(-3)=2-3=-1.$$
• $$\begin{vmatrix}2&-1\\4&2\end{vmatrix}=2\cdot2-(-1)\cdot4=4+4=8.$$
• $$\begin{vmatrix}2&1\\4&-3\end{vmatrix}=2(-3)-1\cdot4=-6-4=-10.$$

Hence $$|A|=3(-1)+2\cdot8+3(-10)=-3+16-30=-17.$$

Because $$|A|\neq0$$, $$A^{-1}$$ exists and the system has a unique solution.

2. Cofactor matrix of $$A$$

Compute every cofactor $$C_{ij}=(-1)^{i+j}M_{ij}$$ (each $$M_{ij}$$ is the corresponding minor):

Thus the cofactor matrix is

$$C=\begin{bmatrix}-1&-8&-10\\-5&-6&1\\-1&9&7\end{bmatrix}.$$

3. Adjoint and inverse of $$A$$

The adjugate is the transpose of the cofactor matrix:

$$\operatorname{adj}A=C^{\mathrm T}=\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}.$$

Hence $$A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A=-\dfrac{1}{17}\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}.$$

4. Solution vector $$X=A^{-1}B$$

First compute $$\operatorname{adj}A\,B$$:

$$\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}\begin{bmatrix}8\\1\\4\end{bmatrix}=\begin{bmatrix}-1\cdot8+(-5)\cdot1+(-1)\cdot4\\-8\cdot8+(-6)\cdot1+9\cdot4\\-10\cdot8+1\cdot1+7\cdot4\end{bmatrix}=\begin{bmatrix}-17\\-34\\-51\end{bmatrix}.$$

Now multiply by $$1/|A|=-1/17$$:

$$X=A^{-1}B=-\dfrac{1}{17}\begin{bmatrix}-17\\-34\\-51\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}.$$

5. Therefore $$x=1,\;y=2,\;z=3.$$

Step 1 – Introduce variables
Let the three required numbers be $$x$$, $$y$$ and $$z$$ (in the given order).

Step 2 – Translate each statement into an equation

• Their sum is $$6$$  ⇒  $$x + y + z = 6$$.
• “Multiply the third number by $$3$$ and add the second number to it, we get $$11$$”  ⇒  $$y + 3z = 11$$.
• “Adding the first and third numbers gives double the second number”  ⇒  $$x + z = 2y$$.

Thus we have the system

$$\begin{aligned} (1)&\;& x + y + z &=&\; 6,\\[2pt] (2)&\;& y + 3z &=&\; 11,\\[2pt] (3)&\;& x + z &=&\; 2y. \end{aligned}$$

Step 3 – Write the matrix equation

$$\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 3\\ 1 & -2 & 1\end{bmatrix} \begin{bmatrix}x\\ y\\ z\end{bmatrix} = \begin{bmatrix}6\\ 11\\ 0\end{bmatrix}.$$

Denote the coefficient matrix by $$A$$, the column of variables by $$X$$ and the constant column by $$B$$, so that $$AX=B$$.

Step 4 – Check that $$\det A \neq 0$$ (so the system is solvable)

$$\det A = \begin{vmatrix}1&1&1\\0&1&3\\1&-2&1\end{vmatrix} = 1(1\cdot1-3\cdot(-2)) - 1(0\cdot1-3\cdot1) + 1(0\cdot(-2)-1\cdot1) = 7 + 3 - 1 = 9 \neq 0.$$

Step 5 – Use Cramer’s rule

Let $$\Delta = \det A = 9$$.

(i) Replace the first column of $$A$$ by $$B$$ to get

$$\Delta_1 = \begin{vmatrix}6&1&1\\11&1&3\\0&-2&1\end{vmatrix} = 6(1\cdot1-3\cdot(-2)) - 1(11\cdot1-3\cdot0) + 1(11\cdot(-2)-1\cdot0) = 42 - 11 - 22 = 9.$$

Hence $$x = \dfrac{\Delta_1}{\Delta} = \dfrac{9}{9} = 1.$$

(ii) Replace the second column of $$A$$ by $$B$$:

$$\Delta_2 = \begin{vmatrix}1&6&1\\0&11&3\\1&0&1\end{vmatrix} = 1(11\cdot1-3\cdot0) - 6(0\cdot1-3\cdot1) + 1(0\cdot0-11\cdot1) = 11 + 18 - 11 = 18.$$

Hence $$y = \dfrac{\Delta_2}{\Delta} = \dfrac{18}{9} = 2.$$

(iii) Replace the third column of $$A$$ by $$B$$:

$$\Delta_3 = \begin{vmatrix}1&1&6\\0&1&11\\1&-2&0\end{vmatrix} = 1(1\cdot0-11\cdot(-2)) - 1(0\cdot0-11\cdot1) + 6(0\cdot(-2)-1\cdot1) = 22 + 11 - 6 = 27.$$

Hence $$z = \dfrac{\Delta_3}{\Delta} = \dfrac{27}{9} = 3.$$

Step 6 – State the result
The three numbers are $$1$$, $$2$$ and $$3$$.

For a 2 × 2 matrix $$\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$, the determinant equals $$ad - bc$$.

Here, $$a = 2$$, $$b = 4$$, $$c = -5$$ and $$d = -1$$.

Substituting:

$$\det = (2)(-1) - (4)(-5)$$

$$= -2 - (-20)$$

$$= -2 + 20$$

$$= 18$$

Hence $$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} = 18$$.

For the given matrix

$$A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$$

first compute its determinant.

$$|A| = (1)(2) - (2)(4) = 2 - 8 = -6$$

Now find the matrix $$2A$$ by multiplying every element of $$A$$ by 2:

$$2A = 2 \times \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$$

Calculate its determinant.

$$|2A| = (2)(4) - (4)(8) = 8 - 32 = -24$$

Next, evaluate $$4|A|$$ (because the order of the matrix is 2, we expect a factor of $$2^2 = 4$$):

$$4|A| = 4(-6) = -24$$

Since $$|2A| = -24$$ and $$4|A| = -24$$, we have proved

$$|2A| = 4|A|.$$

Given matrix

$$A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$$

Let us verify that $$|3A| = 27|A|$$ step by step.

1. Order of matrix

$$A$$ is a $$3\times3$$ matrix. Hence, whenever every row (or every column) of the determinant is multiplied by the same scalar $$k$$, the value of the determinant gets multiplied by $$k^3$$ (three rows ⇒ factor $$k$$ appears thrice).

2. Determinant of $$A$$

The matrix $$A$$ is upper-triangular, so its determinant equals the product of its diagonal entries.

$$|A| = 1 \times 1 \times 4 = 4$$

3. The matrix $$3A$$

Multiply every entry of $$A$$ by $$3$$:

$$3A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}$$

4. Determinant of $$3A$$

Again, $$3A$$ is upper-triangular, so

$$|3A| = 3 \times 3 \times 12 = 108$$

5. Compare $$|3A|$$ with $$27|A|$$

$$27|A| = 27 \times 4 = 108$$

Thus,

$$|3A| = 108 = 27|A|$$

Hence proved.

The given matrix is

$$A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$$

For a 3 × 3 matrix $$A = \bigl[a_{ij}\bigr]$$, the determinant $$|A|$$ can be expanded along the first row:

$$|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$$

where $$C_{1j} = (-1)^{1+j} M_{1j}$$ and $$M_{1j}$$ is the minor obtained by deleting the first row and the j-th column.

1. First cofactor $$C_{11}$$

Minor $$M_{11}$$ (delete row 1, column 1):

$$M_{11}=\begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix}= (1)(-9) - (-3)(4)= -9 + 12 = 3$$

Since $$(-1)^{1+1}=1$$, we get $$C_{11}=3$$.

2. Second cofactor $$C_{12}$$

Minor $$M_{12}$$ (delete row 1, column 2):

$$M_{12}=\begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix}= (2)(-9) - (-3)(5)= -18 + 15 = -3$$

Because $$(-1)^{1+2} = -1$$,

$$C_{12}= -M_{12}= -(-3)= 3$$

3. Third cofactor $$C_{13}$$

Minor $$M_{13}$$ (delete row 1, column 3):

$$M_{13}=\begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}= (2)(4) - (1)(5)= 8 - 5 = 3$$

Here $$(-1)^{1+3}=1$$, so $$C_{13}=3$$.

4. Assemble all terms

First row elements are $$a_{11}=1$$, $$a_{12}=1$$ and $$a_{13}=-2$$; therefore

$$\begin{aligned}|A| & = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}\\ & = (1)(3) + (1)(3) + (-2)(3)\\ & = 3 + 3 - 6\\ & = 0\end{aligned}$$

Hence, the determinant of $$A$$ vanishes.

The equality of the two determinants is

$$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}.$$

Step 1  Evaluate the determinant containing $$x$$:

$$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=x\cdot x-2\cdot18=x^{2}-36.$$

Step 2  Evaluate the constant determinant:

$$\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}=6\cdot6-2\cdot18=36-36=0.$$

Step 3  Equate the two results and solve for $$x$$:

$$x^{2}-36=0\;\;\Longrightarrow\;\;x^{2}=36\;\;\Longrightarrow\;\;x=\pm6.$$

Hence both $$6$$ and $$-6$$ satisfy the given equation, so the correct option is (B) $$\pm6$$.

For any three points $$P(x_1,y_1)$$, $$Q(x_2,y_2)$$, $$R(x_3,y_3)$$ the area of the triangle they form is given by

$$\dfrac12\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}.$$

If this determinant is $$0$$, the area is $$0$$ and the points are collinear.

Here

• $$A(a,\; b+c)$$  so  $$x_1=a,\;y_1=b+c,$$
• $$B(b,\; c+a)$$  so  $$x_2=b,\;y_2=c+a,$$
• $$C(c,\; a+b)$$  so  $$x_3=c,\;y_3=a+b.$$

Form the determinant:

$$\Delta=\begin{vmatrix}a & b+c & 1\\ b & c+a & 1\\ c & a+b & 1\end{vmatrix}.$$

Expand along the first row:

$$\begin{aligned}\Delta&=a\bigl((c+a)(1)-1(a+b)\bigr) \\[-3pt]&\quad-(b+c)\bigl(b(1)-1\,c\bigr) \\[-3pt]&\quad+1\bigl(b(a+b)-(c+a)c\bigr).\end{aligned}$$

Simplify each bracket:

• $$(c+a)(1)-1(a+b)=c+a-a-b=c-b,$$
• $$b(1)-1\,c=b-c,$$
• $$b(a+b)-(c+a)c=ab+b^2-c^2-ac.$$

Substitute back:

$$\begin{aligned}\Delta&=a(c-b)-(b+c)(b-c)+ab+b^2-c^2-ac.\end{aligned}$$

Next evaluate $$(b+c)(b-c)=b^2-c^2$$ and collect like terms:

$$\begin{aligned}\Delta&=ac-ab-(b^2-c^2)+ab+b^2-c^2-ac\\&=(ac-ac)+(\!-ab+ab)+(\!-b^2+b^2)+(c^2-c^2)=0.\end{aligned}$$

Because $$\Delta=0$$, the area of $$\triangle ABC$$ is zero. Hence points $$A$$, $$B$$ and $$C$$ are collinear.

For three vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ the area of the triangle is given by

$$\text{Area}=\frac12\left|\begin{vmatrix}x_1&y;_1&1\\x_2&y;_2&1\\x_3&y;_3&1\end{vmatrix}\right|.$$

Substitute $$(x_1,y_1)=(2,-6),\,(x_2,y_2)=(5,4),\,(x_3,y_3)=(k,4):$$

$$\text{Area}=\frac12\left|\begin{vmatrix}2&-6&1\\5&4&1\\k&4&1\end{vmatrix}\right|=35.$$

Evaluate the determinant by expanding along the first row:

$$\begin{aligned} \begin{vmatrix}2&-6&1\\5&4&1\\k&4&1\end{vmatrix}&=2\bigl(4\cdot1-1\cdot4\bigr)-(-6)\bigl(5\cdot1-1\cdot k\bigr)+1\bigl(5\cdot4-4\cdot k\bigr)\\[2mm] &=2(0)+6(5-k)+(20-4k)\\[2mm] &=30-6k+20-4k\\[2mm] &=50-10k. \end{aligned}$$

Hence

$$\frac12\left|50-10k\right|=35\;\Longrightarrow\;\left|50-10k\right|=70.$$

Divide by $$10$$:

$$\left|5-k\right|=7.$$

This gives two linear equations:

• $$5-k=7\;\Longrightarrow\;k=-2,$$
• $$5-k=-7\;\Longrightarrow\;k=12.$$

Therefore $$k=12$$ or $$k=-2$$, so the correct option is (D).

Given determinant

$$\Delta = \begin{vmatrix} 5 & 3 & 8\\ 2 & 0 & 1\\ 1 & 2 & 3 \end{vmatrix}$$

We are asked to expand it by cofactors of the elements of the second row (row 2).

The general cofactor attached to an element $$a_{ij}$$ is

$$A_{ij}=(-1)^{i+j}M_{ij},$$

where $$M_{ij}$$ is the minor obtained after deleting the i-th row and j-th column.

Step 1: Find minors of second-row elements

• Element $$a_{21}=2$$ (row 2, column 1)
  Minor $$M_{21}=\begin{vmatrix} 3 & 8\\ 2 & 3 \end{vmatrix}=3\times3-8\times2=9-16=-7.$$

• Element $$a_{22}=0$$ (row 2, column 2)
  Minor $$M_{22}=\begin{vmatrix} 5 & 8\\ 1 & 3 \end{vmatrix}=5\times3-8\times1=15-8=7.$$

• Element $$a_{23}=1$$ (row 2, column 3)
  Minor $$M_{23}=\begin{vmatrix} 5 & 3\\ 1 & 2 \end{vmatrix}=5\times2-3\times1=10-3=7.$$

Step 2: Find cofactors of those elements

• $$A_{21}=(-1)^{2+1}\,M_{21}=(-1)^3(-7)=7.$$
• $$A_{22}=(-1)^{2+2}\,M_{22}=(-1)^4(7)=7.$$
• $$A_{23}=(-1)^{2+3}\,M_{23}=(-1)^5(7)=-7.$$

Step 3: Expand the determinant along the second row

$$\Delta=a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}$$

Substituting each value:

$$\Delta = 2\times7 + 0\times7 + 1\times(-7).$$

Step 4: Evaluate

$$\Delta = 14 + 0 - 7 = 7.$$

Hence, the value of the determinant is

$$\boxed{7}$$

Given determinant

$$\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$$

We are asked to expand it by the cofactors of the third column.

Step 1 : Compute the cofactors of the third-column elements

Step 2 : Expansion along the third column

\[\begin{aligned} \Delta &= yz\,C_{13}+zx\,C_{23}+xy\,C_{33}\\ &= yz\,(y-x)+zx\,(x-z)+xy\,(y-x)\\ &= (y-x)(yz+xy)+xz(x-z).\tag{1} \end{aligned}\]

Step 3 : Initial simplification

Since $$yz+xy=y(z+x)$$, relation (1) becomes

\[\Delta = y(y-x)(z+x)+xz(x-z).\tag{2}\]

Step 4 : Identify the factors

• Put $$x=y$$ in the original determinant; the first two rows become identical, so $$\Delta=0$$. Therefore $$(x-y)$$ is a factor.
• Put $$y=z$$; rows 2 and 3 coincide ⇒ $$\Delta=0$$, hence $$(y-z)$$ is a factor.
• Put $$z=x$$; rows 1 and 3 coincide ⇒ $$\Delta=0$$, hence $$(z-x)$$ is a factor.

Thus $$\Delta$$ contains $$(x-y)(y-z)(z-x)$$ as a factor. Expression (1) (or (2)) is a cubic polynomial in $$x,y,z$$, and the product $$ (x-y)(y-z)(z-x) $$ is already of degree 3; hence

$$\Delta = k\,(x-y)(y-z)(z-x)$$ for some constant $$k$$ independent of $$x,y,z$$.

Step 5 : Find the constant $$k$$

Choose convenient values, say $$x=1,\;y=2,\;z=3$$.

Evaluate $$\Delta$$ with these numbers:

$$\Delta = \begin{vmatrix}1&1&6\\1&2&3\\1&3&2\end{vmatrix} = 2.$$

Compute the product:

$$(x-y)(y-z)(z-x) = (1-2)(2-3)(3-1)=(-1)(-1)(2)=2.$$

Thus $$k=1$$ and

$$\boxed{\;\Delta=(x-y)(y-z)(z-x)\;}$$

The determinant equals the Vandermonde product of the three variables.

The determinant of a square matrix can be expanded along any one row or any one column.

For the given determinant

$$\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$

let $$A_{ij}$$ denote the cofactor of the element $$a_{ij}$$.

• Expansion along the first column (column 1) is

$$\Delta = a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$$

because each term is the element from column 1 multiplied by its own cofactor, with alternating signs already built into the cofactors.

Comparing this with the options provided, we see that it matches option (D) exactly:

$$a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$$

Hence, the value of $$\Delta$$ is given by option (D).

Step 1 – Identify each element of the matrix

Let $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ so that $$a = 1,\; b = 2,\; c = 3,\; d = 4$$.

Step 2 – Find the minors and cofactors

• For $$a_{11}=1$$: its minor is $$M_{11}=4$$.
  Cofactor $$C_{11}=(-1)^{1+1}M_{11}=4$$.
• For $$a_{12}=2$$: its minor is $$M_{12}=3$$.
  Cofactor $$C_{12}=(-1)^{1+2}M_{12}=-3$$.
• For $$a_{21}=3$$: its minor is $$M_{21}=2$$.
  Cofactor $$C_{21}=(-1)^{2+1}M_{21}=-2$$.
• For $$a_{22}=4$$: its minor is $$M_{22}=1$$.
  Cofactor $$C_{22}=(-1)^{2+2}M_{22}=1$$.

Step 3 – Form the cofactor matrix

$$C = \begin{bmatrix} C_{11} & C_{12} \\ C_{21} & C_{22} \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$$

Step 4 – Transpose the cofactor matrix to get the adjoint

$$\operatorname{adj} A = C^{\mathrm T} = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$$

Step 1 : Write the matrix

$$A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$$

Step 2 : Find all the 2 × 2 minors and their cofactors

Step 3 : Form the cofactor matrix

$$C = \begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}$$

Step 4 : Transpose the cofactor matrix to get the adjoint

$$\operatorname{adj}A = C^{\mathrm T} = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$$

Therefore, the adjoint of the given matrix has been found.

Let

$$A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}.$$

1. Determinant of $$A$$

$$|A| = (2)(-6) - (3)(-4) = -12 + 12 = 0.$$

2. Adjoint of $$A$$

First find the cofactors.

Cofactor matrix:

$$C = \begin{bmatrix} -6 & 4 \\ -3 & 2 \end{bmatrix}.$$

Adjoint is the transpose of $$C$$:

$$\operatorname{adj} A = C^{T} = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}.$$

3. Product $$A (\operatorname{adj} A)$$

$$ A(\operatorname{adj} A)=\begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix} \begin{bmatrix}-6 & -3\\4 & 2\end{bmatrix} =\begin{bmatrix} 2(-6)+3(4) & 2(-3)+3(2)\\ -4(-6)+(-6)(4) & -4(-3)+(-6)(2) \end{bmatrix} =\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}. $$

4. Product $$(\operatorname{adj} A) A$$

$$ (\operatorname{adj} A)A=\begin{bmatrix}-6 & -3\\4 & 2\end{bmatrix} \begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix} =\begin{bmatrix} (-6)(2)+(-3)(-4) & (-6)(3)+(-3)(-6)\\ 4(2)+2(-4) & 4(3)+2(-6) \end{bmatrix} =\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}. $$

5. Right-hand side $$|A| I$$

Since $$|A| = 0$$,

$$|A| I_2 = 0 \times \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}.$$

Both products obtained in Steps 3 and 4 are equal to $$|A| I_2$$. Hence

$$A(\operatorname{adj} A) = (\operatorname{adj} A)A = |A| I_2.$$

The relation is verified for the given matrix.

Given matrix

$$A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$$

We have to show that $$A(\operatorname{adj}A)=(\operatorname{adj}A)A = |A|\,I$$.

1. Determinant of $$A$$

$$|A| = 1\,\begin{vmatrix} 0 & -2 \\ 0 & 3 \end{vmatrix} -(-1)\,\begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} + 2\,\begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix}$$ $$\;\;\; = 1(0-0)+1\bigl(3\cdot3-(-2)\cdot1\bigr)+2(3\cdot0-0\cdot1)$$ $$\;\;\; = 0 + 1(9+2)+0 = 11.$$

Thus $$|A|=11.$$

2. Cofactors

3. Adjugate of $$A$$

The cofactor matrix is $$\begin{bmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{bmatrix}.$$ Transposing, we obtain $$\operatorname{adj}A=\begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}.$$

4. Product $$A(\operatorname{adj}A)$$

$$A(\operatorname{adj}A)=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} =\begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} =11I.$$

5. Product $$(\operatorname{adj}A)A$$

$$(\operatorname{adj}A)A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} =\begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} =11I.$$

6. Conclusion

Since $$|A|=11$$ and both products equal $$11I$$, we have verified that

$$A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|\,I.$$

Step 1 — Write the matrix

Let $$A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}.$$

Step 2 — Find the determinant |A|

For a $$2 \times 2$$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ the determinant is $$ad - bc$$. Thus

$$|A| = (2)(3) - (-2)(4) = 6 + 8 = 14.$$

Because $$|A| \neq 0$$, the inverse exists.

Step 3 — Find the adjugate \(\operatorname{adj} A\)

For a $$2 \times 2$$ matrix the adjugate is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:

$$\operatorname{adj} A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}.$$

Step 4 — Compute the inverse

$$A^{-1} = \dfrac{1}{|A|}\,\operatorname{adj} A = \dfrac{1}{14}\begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}.$$

Simplify each entry:

$$A^{-1} = \begin{bmatrix} \dfrac{3}{14} & \dfrac{1}{7} \\[4pt] -\dfrac{2}{7} & \dfrac{1}{7} \end{bmatrix}.$$

Let $$A = \begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$$.

Step 1 : Determinant.
$$|A| = (-1)(2) - (5)(-3) = -2 + 15 = 13$$
Since $$|A| \neq 0$$, the inverse exists.

Step 2 : Cofactors.

Hence the cofactor matrix is $$\begin{bmatrix} 2 & 3 \\ -5 & -1 \end{bmatrix}$$.

Step 3 : Adjoint.
$$\operatorname{adj}(A)=\begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$$

Step 4 : Inverse.
$$A^{-1}=\dfrac{1}{13}\begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$$

Given matrix $$A=\begin{bmatrix}1&2&3\\0&2&4\\0&0&5\end{bmatrix}$$.

Step 1 : Check whether $$A$$ is invertible.

Because $$A$$ is upper-triangular,

$$\det(A)=1\times2\times5=10\neq0,$$

so $$A$$ is non-singular and the inverse exists.

Step 2 : Use the Gauss–Jordan method.

(i) Divide the second and third rows by their pivots:

(ii) $$R_2\to R_2-2R_3$$

(iii) $$R_1\to R_1-3R_3$$

(iv) $$R_1\to R_1-2R_2$$

The left half is now the identity matrix, so the right half is $$A^{-1}$$.

Inverse of $$A$$:

$$A^{-1}=\begin{bmatrix}1&-1&\dfrac15\\0&\dfrac12&-\dfrac25\\0&0&\dfrac15\end{bmatrix}$$

(Multiplying $A$ and $A^{-1}$ indeed gives the identity matrix.)

Let $$A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$$.

1. Determinant of $$A$$

Using the first row expansion,

$$|A| = 1\,\Bigl|\begin{smallmatrix} 3 & 0 \\ 2 & -1 \end{smallmatrix}\Bigr| - 0 + 0 = 1\,(3\,(-1) - 0\cdot 2) = -3.$$

Since $$|A| = -3 \neq 0$$, the inverse exists.

2. Cofactor matrix

• Row 1:
  $$C_{11}=(-1)^{1+1}\,\Bigl|\begin{smallmatrix} 3 & 0 \\ 2 & -1 \end{smallmatrix}\Bigr| = -3,\; C_{12}=(-1)^{1+2}\,\Bigl|\begin{smallmatrix} 3 & 0 \\ 5 & -1 \end{smallmatrix}\Bigr| = +3,\; C_{13}=(-1)^{1+3}\,\Bigl|\begin{smallmatrix} 3 & 3 \\ 5 & 2 \end{smallmatrix}\Bigr| = -9.$$
• Row 2:
  $$C_{21}=(-1)^{2+1}\,\Bigl|\begin{smallmatrix} 0 & 0 \\ 2 & -1 \end{smallmatrix}\Bigr| = 0,\; C_{22}=(-1)^{2+2}\,\Bigl|\begin{smallmatrix} 1 & 0 \\ 5 & -1 \end{smallmatrix}\Bigr| = -1,\; C_{23}=(-1)^{2+3}\,\Bigl|\begin{smallmatrix} 1 & 0 \\ 5 & 2 \end{smallmatrix}\Bigr| = -2.$$
• Row 3:
  $$C_{31}=(-1)^{3+1}\,\Bigl|\begin{smallmatrix} 0 & 0 \\ 3 & 0 \end{smallmatrix}\Bigr| = 0,\; C_{32}=(-1)^{3+2}\,\Bigl|\begin{smallmatrix} 1 & 0 \\ 3 & 0 \end{smallmatrix}\Bigr| = 0,\; C_{33}=(-1)^{3+3}\,\Bigl|\begin{smallmatrix} 1 & 0 \\ 3 & 3 \end{smallmatrix}\Bigr| = 3.$$

Thus the cofactor matrix is

$$\begin{bmatrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{bmatrix}.$$

3. Adjugate of $$A$$

Adj $$A$$ is the transpose of the cofactor matrix:

$$\operatorname{Adj}\,A = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}.$$

4. Inverse of $$A$$

$$A^{-1} = \dfrac{1}{|A|}\,\operatorname{Adj}\,A = \frac{1}{-3}\,\begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & \dfrac{1}{3} & 0 \\ 3 & \dfrac{2}{3} & -1 \end{bmatrix}.$$

Hence the required inverse has been obtained.

Let $$A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$$.

1. Determinant of $$A$$

Using the first row expansion,

$$|A| = 2\big((-1)(1)-0\cdot 2\big) - 1\big(4\cdot 1-0\cdot(-7)\big) + 3\big(4\cdot 2-(-1)(-7)\big).$$

Compute every bracket:

• $$(-1)(1)-0\cdot 2 = -1$$
• $$4\cdot 1-0\cdot(-7)=4$$
• $$4\cdot 2-(-1)(-7)=8-7=1$$

Therefore

$$|A| = 2(-1)-1(4)+3(1) = -2-4+3 = -3.$$

Since $$|A|\neq 0$$, the inverse of $$A$$ exists.

2. Cofactor matrix

Each cofactor $$C_{ij}=(-1)^{i+j}M_{ij}$$ where $$M_{ij}$$ is the minor obtained by deleting the i-th row and j-th column.

Hence the cofactor matrix is

$$C=\begin{bmatrix}-1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6\end{bmatrix}.$$

3. Adjoint of $$A$$

The adjoint is the transpose of the cofactor matrix:

$$\operatorname{adj}A = C^{\mathsf T} = \begin{bmatrix}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{bmatrix}.$$

4. Inverse of $$A$$

$$A^{-1}=\dfrac{1}{|A|}\,\operatorname{adj}A = -\dfrac{1}{3}\begin{bmatrix}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{bmatrix}.$$

Multiplying each entry by $$-\dfrac{1}{3}$$, we get

$$A^{-1}=\begin{bmatrix}\dfrac{1}{3} & -\dfrac{5}{3} & -1 \\ \dfrac{4}{3} & -\dfrac{23}{3} & -4 \\ -\dfrac{1}{3} & \dfrac{11}{3} & 2\end{bmatrix}.$$

Thus the inverse exists and is as obtained above.

Let $$A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$$.

Step 1 : Determinant of  $$A$$
Expand along the first row:
$$|A| = 1\left|\begin{matrix} 2 & -3 \\ -2 & 4 \end{matrix}\right| - (-1)\left|\begin{matrix} 0 & -3 \\ 3 & 4 \end{matrix}\right| + 2\left|\begin{matrix} 0 & 2 \\ 3 & -2 \end{matrix}\right|$$
$$|A| = 1(2\cdot4-(-3)(-2)) + 1(0\cdot4-(-3)\cdot3) + 2(0\cdot(-2)-2\cdot3)$$
$$|A| = 1(8-6)+1(0+9)+2(0-6)$$
$$|A| = 2 + 9 - 12 = -1$$
Since $$|A| \neq 0$$, the inverse exists.

Step 2 : Cofactor matrix of  $$A$$

Let $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$$.

Step 1 : Determinant of $$A$$

Expanding along the first row,

$$|A| = 1\begin{vmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} - 0 + 0.$$

But

$$\begin{vmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} = \cos\alpha(-\cos\alpha) - \sin\alpha\,\sin\alpha = -\cos^{2}\alpha - \sin^{2}\alpha = -1.$$

Hence $$|A| = -1 \neq 0$$, so the inverse of $$A$$ exists for every real $$\alpha$$.

Step 2 : Verify that $$A$$ is its own inverse

Compute $$A^{2}$$:

$$A^{2} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}.$$

• First row becomes $$[1,0,0]$$.
• Second row:
  • $(2,1)$ entry: $$0$$,
  • $(2,2)$ entry: $$\cos^{2}\alpha+\sin^{2}\alpha = 1$$,
  • $(2,3)$ entry: $$\cos\alpha\sin\alpha-\sin\alpha\cos\alpha = 0$$.
• Third row:
  • $(3,1)$ entry: $$0$$,
  • $(3,2)$ entry: $$\sin\alpha\cos\alpha-\cos\alpha\sin\alpha = 0$$,
  • $(3,3)$ entry: $$\sin^{2}\alpha+\cos^{2}\alpha = 1$$.

Thus $$A^{2}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=I_{3}.$$

Step 3 : Inverse

Since $$A^{2}=I_{3}$$, we have $$A^{-1}=A$$.

Therefore

$$A^{-1}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}.$$

Step 1 : Write the given matrices.
$$A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix},\qquad B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$$

Step 2 : Find $$A^{-1}$$.

• Determinant : $$|A| = 3\times5 - 7\times2 = 15 - 14 = 1$$ (non-zero, so the inverse exists).
• Adjugate : $$\operatorname{adj}A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$$
• Inverse : $$A^{-1} = \dfrac{1}{|A|}\operatorname{adj}A = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$$

Step 3 : Find $$B^{-1}$$.

• Determinant : $$|B| = 6\times9 - 8\times7 = 54 - 56 = -2$$.
• Adjugate : $$\operatorname{adj}B = \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$$
• Inverse : $$B^{-1} = \dfrac{1}{|B|}\operatorname{adj}B = -\dfrac12\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -\dfrac{9}{2} & 4 \\ \dfrac{7}{2} & -3 \end{bmatrix}$$

Step 4 : Form $$AB$$ and then $$(AB)^{-1}$$.

• Product : $$AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 3\times6+7\times7 & 3\times8+7\times9 \\[2pt] 2\times6+5\times7 & 2\times8+5\times9 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$$
• Determinant : $$|AB| = 67\times61 - 87\times47 = 4087 - 4089 = -2$$.
• Adjugate : $$\operatorname{adj}(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$$
• Inverse : $$ (AB)^{-1} = \dfrac{1}{|AB|}\operatorname{adj}(AB) = -\dfrac12\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -\dfrac{61}{2} & \dfrac{87}{2} \\ \dfrac{47}{2} & -\dfrac{67}{2} \end{bmatrix} $$

Step 5 : Compute $$B^{-1}A^{-1}$$.

Step 6 : Compare the two results.
$$(AB)^{-1} = \begin{bmatrix} -\dfrac{61}{2} & \dfrac{87}{2} \\ \dfrac{47}{2} & -\dfrac{67}{2} \end{bmatrix} \;\;\text{and}\;\; B^{-1}A^{-1} = \begin{bmatrix} -\dfrac{61}{2} & \dfrac{87}{2} \\ \dfrac{47}{2} & -\dfrac{67}{2} \end{bmatrix}$$

The two matrices are identical; hence $$ (AB)^{-1} = B^{-1}A^{-1} $$ is verified.

Given matrix
$$A = \begin{bmatrix} 3 & 1\\ -1 & 2 \end{bmatrix}$$

1. Calculate \(A^2\)

$$A^2 = A\,A = \begin{bmatrix} 3 & 1\\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1\\ -1 & 2 \end{bmatrix}$$

\[\begin{aligned} A^2 &= \begin{bmatrix} 3\times3 + 1\times(-1) & 3\times1 + 1\times2\\ -1\times3 + 2\times(-1) & -1\times1 + 2\times2 \end{bmatrix} \\[4pt] &= \begin{bmatrix} 9-1 & 3+2\\ -3-2 & -1+4 \end{bmatrix} \\[4pt] &= \begin{bmatrix} 8 & 5\\ -5 & 3 \end{bmatrix} \end{aligned}\]

2. Compute \(-5A\)

$$-5A = -5\begin{bmatrix} 3 & 1\\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -15 & -5\\ 5 & -10 \end{bmatrix}$$

3. Form the expression \(A^2 - 5A + 7I\)

\[\begin{aligned} A^2 - 5A &= \begin{bmatrix} 8 & 5\\ -5 & 3 \end{bmatrix} + \begin{bmatrix} -15 & -5\\ 5 & -10 \end{bmatrix} \\[4pt] &= \begin{bmatrix} 8-15 & 5-5\\ -5+5 & 3-10 \end{bmatrix} \\[4pt] &= \begin{bmatrix} -7 & 0\\ 0 & -7 \end{bmatrix} \\[4pt] &= -7I \end{aligned}\]

Adding \(7I\):

$$A^2 - 5A + 7I = -7I + 7I = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} = O$$

Hence the relation is proved.

4. Finding \(A^{-1}\) using the relation

From $$A^2 - 5A + 7I = O$$ we get $$A^2 - 5A = -7I$$.

Factorising $$A$$ gives $$A\,(A - 5I) = -7I$$.

Multiply both sides on the right by $$A^{-1}$$ (which exists because the determinant of $$A$$ is non-zero):

$$A\,(A - 5I)A^{-1} = -7I A^{-1}\;\;\Rightarrow\;\; (A-5I) = -7A^{-1}$$

Therefore

$$A^{-1} = -\tfrac{1}{7}(A - 5I).$$

Compute $$A - 5I$$:

$$A - 5I = \begin{bmatrix} 3 & 1\\ -1 & 2 \end{bmatrix} - \begin{bmatrix} 5 & 0\\ 0 & 5 \end{bmatrix} = \begin{bmatrix} -2 & 1\\ -1 & -3 \end{bmatrix}$$

Thus

$$A^{-1} = -\tfrac{1}{7}\begin{bmatrix} -2 & 1\\ -1 & -3 \end{bmatrix} = \begin{bmatrix} \dfrac{2}{7} & -\dfrac{1}{7}\\ \dfrac{1}{7} & \dfrac{3}{7} \end{bmatrix}$$

Verification (optional)
$$A\,A^{-1} = \begin{bmatrix} 3 & 1\\ -1 & 2 \end{bmatrix}\begin{bmatrix} 2/7 & -1/7\\ 1/7 & 3/7 \end{bmatrix} = I,$$ confirming the result.

Let $$A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$$. We require real numbers $$a$$ and $$b$$ such that

$$A^2 + aA + bI = O,$$ where $$I$$ is the identity matrix and $$O$$ is the null matrix.

1. Compute $$A^2$$

$$A^2 = A\,A = \begin{bmatrix}3 & 2 \\ 1 & 1\end{bmatrix}\begin{bmatrix}3 & 2 \\ 1 & 1\end{bmatrix} = \begin{bmatrix}3\cdot3 + 2\cdot1 & 3\cdot2 + 2\cdot1 \\ 1\cdot3 + 1\cdot1 & 1\cdot2 + 1\cdot1\end{bmatrix} = \begin{bmatrix}11 & 8 \\ 4 & 3\end{bmatrix}.$$

2. Write $$A^2 + aA + bI$$

$$aA = a\begin{bmatrix}3 & 2 \\ 1 & 1\end{bmatrix} = \begin{bmatrix}3a & 2a \\ a & a\end{bmatrix}, \qquad bI = b\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}b & 0 \\ 0 & b\end{bmatrix}.$$

Hence

$$A^2 + aA + bI = \begin{bmatrix}11 & 8 \\ 4 & 3\end{bmatrix} + \begin{bmatrix}3a & 2a \\ a & a\end{bmatrix} + \begin{bmatrix}b & 0 \\ 0 & b\end{bmatrix} = \begin{bmatrix}11 + 3a + b & 8 + 2a \\ 4 + a & 3 + a + b\end{bmatrix}.$$

3. Equate to the zero matrix

Setting each entry equal to zero gives the system

• $$11 + 3a + b = 0$$
• $$8 + 2a = 0$$
• $$4 + a = 0$$
• $$3 + a + b = 0$$

From $$4 + a = 0$$ we get $$a = -4$$. Substituting $$a = -4$$:

$$8 + 2(-4) = 0 \qquad\text{(satisfied)},$$

$$11 + 3(-4) + b = 0 \;\Rightarrow\; -1 + b = 0 \;\Rightarrow\; b = 1,$$

and

$$3 + (-4) + 1 = 0 \qquad\text{(satisfied)}.$$

4. Result

Therefore $$a = -4$$ and $$b = 1$$ satisfy $$A^2 + aA + bI = O$$.

Given matrix:

$$A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & -3\\ 2 & -1 & 3 \end{bmatrix}$$

Step 1 – Find $$A^2$$

$$A^2 = A\,A$$

Thus $$A^2 = \begin{bmatrix} 4 & 2 & 1\\ -3 & 8 & -14\\ 7 & -3 & 14 \end{bmatrix}$$

Step 2 – Find $$A^3$$

$$A^3 = A^2 A$$

Thus $$A^3 = \begin{bmatrix} 8 & 7 & 1\\ -23 & 27 & -69\\ 32 & -13 & 58 \end{bmatrix}$$

Step 3 – Form the expression $$A^3-6A^2+5A+11I$$

Compute each part:

• $$-6A^2 = \begin{bmatrix} -24 & -12 & -6\\ 18 & -48 & 84\\ -42 & 18 & -84 \end{bmatrix}$$
• $$5A = \begin{bmatrix} 5 & 5 & 5\\ 5 & 10 & -15\\ 10 & -5 & 15 \end{bmatrix}$$
• $$11I = \begin{bmatrix} 11 & 0 & 0\\ 0 & 11 & 0\\ 0 & 0 & 11 \end{bmatrix}$$

Add the four matrices entry-wise:

Every entry is 0; hence

$$A^3 - 6A^2 + 5A + 11I = O$$

Step 4 – Derive $$A^{-1}$$ from the identity

Because $$A$$ is invertible, post-multiplying by $$A^{-1}$$ gives

$$A^2 - 6A + 5I + 11A^{-1} = O$$

Hence

$$A^{-1} = -\dfrac{1}{11}\,(A^2 - 6A + 5I)$$

Compute $$A^2 - 6A + 5I$$

• $$A^2$$ found earlier.
• $$-6A = \begin{bmatrix} -6 & -6 & -6\\ -6 & -12 & 18\\ -12 & 6 & -18 \end{bmatrix}$$
• $$5I = \begin{bmatrix} 5 & 0 & 0\\ 0 & 5 & 0\\ 0 & 0 & 5 \end{bmatrix}$$

Adding them:

$$A^2 - 6A + 5I = \begin{bmatrix} 3 & -4 & -5\\ -9 & 1 & 4\\ -5 & 3 & 1 \end{bmatrix}$$

Therefore

$$A^{-1} = -\dfrac{1}{11}\,\begin{bmatrix} 3 & -4 & -5\\ -9 & 1 & 4\\ -5 & 3 & 1 \end{bmatrix} = \begin{bmatrix} -\dfrac{3}{11} & \dfrac{4}{11} & \dfrac{5}{11}\\[4pt] \dfrac{9}{11} & -\dfrac{1}{11} & -\dfrac{4}{11}\\[4pt] \dfrac{5}{11} & -\dfrac{3}{11} & -\dfrac{1}{11} \end{bmatrix}$$

The identity is proved and the inverse obtained.

Step 1 : Write the matrix.
$$A = \begin{bmatrix} 2 & -1 & 1\\ -1 & 2 & -1\\ 1 & -1 & 2 \end{bmatrix}$$

Step 2 : Compute $$A^{2}=A\,A$$.

Hence
$$A^{2}=\begin{bmatrix} 6 & -5 & 5\\ -5 & 6 & -5\\ 5 & -5 & 6 \end{bmatrix}$$

Step 3 : Compute $$A^{3}=A^{2}A$$.

Thus
$$A^{3}=\begin{bmatrix} 22 & -21 & 21\\ -21 & 22 & -21\\ 21 & -21 & 22 \end{bmatrix}$$

Step 4 : Form the expression $$A^{3}-6A^{2}+9A-4I$$.

First,

• $$6A^{2}=\begin{bmatrix} 36 & -30 & 30\\ -30 & 36 & -30\\ 30 & -30 & 36 \end{bmatrix}$$
• $$9A=\begin{bmatrix} 18 & -9 & 9\\ -9 & 18 & -9\\ 9 & -9 & 18 \end{bmatrix}$$
• $$4I=\begin{bmatrix} 4 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 4 \end{bmatrix}$$

Now compute step by step.

$$A^{3}-6A^{2}=\begin{bmatrix} 22-36 & -21-(-30) & 21-30\\ -21-(-30) & 22-36 & -21-(-30)\\ 21-30 & -21-(-30) & 22-36 \end{bmatrix}=\begin{bmatrix} -14 & 9 & -9\\ 9 & -14 & 9\\ -9 & 9 & -14 \end{bmatrix}$$

Add $$9A$$:

$$(A^{3}-6A^{2})+9A=\begin{bmatrix} -14+18 & 9-9 & -9+9\\ 9-9 & -14+18 & 9-9\\ -9+9 & 9-9 & -14+18 \end{bmatrix}=\begin{bmatrix} 4 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 4 \end{bmatrix}=4I$$

Finally, $$4I-4I=O$$, so
$$A^{3}-6A^{2}+9A-4I=O$$, as desired.

Step 5 : Deduce $$A^{-1}$$.

Post-multiplying the verified relation by $$A^{-1}$$ gives

$$A^{2}-6A+9I-4A^{-1}=O$$

Hence

$$4A^{-1}=A^{2}-6A+9I$$

Compute the right-hand matrix.

We already have $$A^{2}$$ and $$6A$$. Their difference is

$$A^{2}-6A=\begin{bmatrix} -6 & 1 & -1\\ 1 & -6 & 1\\ -1 & 1 & -6 \end{bmatrix}$$

Add $$9I$$:

$$A^{2}-6A+9I=\begin{bmatrix} 3 & 1 & -1\\ 1 & 3 & 1\\ -1 & 1 & 3 \end{bmatrix}$$

Therefore

$$A^{-1}=\dfrac14\begin{bmatrix} 3 & 1 & -1\\ 1 & 3 & 1\\ -1 & 1 & 3 \end{bmatrix}$$

Both parts of the question are proved.

Let the order of the non-singular square matrix $$A$$ be $$n=3$$.

We recall two standard facts that hold for every non-singular square matrix of order $$n$$:

1. The adjugate (also called the adjoint) of $$A$$ satisfies
   $$\operatorname{adj}A = |A|\,A^{-1}.$$

2. The determinant of a product equals the product of the determinants, i.e. for any conformable matrices $$B$$ and $$C$$,
   $$|BC| = |B|\,|C|.$$

Using (1) with (2) we obtain a direct formula for the determinant of the adjugate:

Take determinants on both sides of $$\operatorname{adj}A = |A|\,A^{-1}$$:

$$|\operatorname{adj}A| = |\,|A|\,A^{-1}|.$$

Because $$|A|$$ is an ordinary (non-zero) scalar, we can pull it out of the determinant. The matrix $$|A|\,A^{-1}$$ is the product of the scalar matrix $$|A|I$$ and $$A^{-1}$$, so

$$|\operatorname{adj}A| = |\,|A|I\,|\,\cdot|\,A^{-1}|.$$

The determinant of a scalar multiple of the identity of order $$n$$ is the scalar raised to the $$n^{\text{th}}$$ power:

$$|\,|A|I| = |A|^{n}.$$

Also, $$|A^{-1}| = \dfrac{1}{|A|}.$$ Hence

$$|\operatorname{adj}A| = |A|^{n}\,\cdot\,\dfrac{1}{|A|} = |A|^{n-1}.$$

For the present question we have $$n = 3$$, so

$$|\operatorname{adj}A| = |A|^{3-1} = |A|^{2}.$$

Therefore the correct option is (B).

Let $$A$$ be an invertible (non-singular) matrix of order 2.

We use the multiplicative property of determinants:

$$\det(AB)=\det(A)\,\det(B)$$ for any two square matrices $$A$$ and $$B$$ of the same order.

Choose $$B=A^{-1}$$. Then $$AB=AA^{-1}=I_2$$, where $$I_2$$ is the identity matrix of order 2.

Taking determinants,

$$\det(AB)=\det(I_2).$$

The determinant of the identity matrix is 1, so

$$\det(A)\,\det(A^{-1}) = 1.$$

Since $$A$$ is invertible, $$\det(A) \neq 0$$. Hence

$$\det(A^{-1}) = \dfrac{1}{\det(A)}.$$

Therefore the correct option is (B).

For a pair of simultaneous linear equations

$$a_1x + b_1y = c_1$$
$$a_2x + b_2y = c_2$$

write the coefficient (or augmented-matrix) determinants

• Main determinant (of coefficients)
  $$\Delta = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1$$
• Column-replacement determinants (needed only if $$\Delta \neq 0$$)
  $$\Delta_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1b_2 - c_2b_1$$
  $$\Delta_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1c_2 - a_2c_1$$

The system is

$$x + 2y = 2$$
$$2x + 3y = 3$$

so $$a_1 = 1, \; b_1 = 2, \; c_1 = 2$$ and $$a_2 = 2, \; b_2 = 3, \; c_2 = 3$$.

Step 1 – Determinant of the coefficient matrix

$$\Delta = (1)(3) - (2)(2) = 3 - 4 = -1 \neq 0$$

Because $$\Delta \neq 0$$, the system is consistent and possesses exactly one solution. (If $$\Delta = 0$$ we would have examined $$\Delta_x,\,\Delta_y$$ to decide further.)

Step 2 – (Optional) Compute the unique solution

$$\Delta_x = (2)(3) - (3)(2) = 6 - 6 = 0$$
$$\Delta_y = (1)(3) - (2)(2) = 3 - 4 = -1$$

Hence by Cramer’s rule
$$x = \dfrac{\Delta_x}{\Delta} = \dfrac{0}{-1} = 0,$$
$$y = \dfrac{\Delta_y}{\Delta} = \dfrac{-1}{-1} = 1.$$

Conclusion

The given pair of linear equations is consistent and has a unique solution $$x = 0,\, y = 1$$.

Step 1 : Write the system in matrix form

$$\begin{aligned} 2x-y &= 5,\\ x+y &= 4 \end{aligned}$$

Let $$A$$ be the coefficient matrix, $$X$$ the column of variables and $$B$$ the column of constants.

$$A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}, \; X = \begin{bmatrix} x \\ y \end{bmatrix}, \; B = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$$

Then $$AX = B$$ represents the given system.

Step 2 : Find the rank of the coefficient matrix $$A$$

The determinant of $$A$$ (which is the only $$2\times2$$ minor) is

$$|A| = \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (-1)(1) = 2 + 1 = 3$$

Since $$|A| \neq 0$$, the rank of $$A$$ is $$\operatorname{rank}(A)=2$$.

Step 3 : Find the rank of the augmented matrix $$[A\,|\,B]$$

The augmented matrix is $$[A\,|\,B] = \begin{bmatrix} 2 & -1 & 5 \\ 1 & 1 & 4 \end{bmatrix}$$.

Because $$|A| \neq 0$$, every $$2\times2$$ minor drawn from the first two columns is non-zero, and inserting the column $$B$$ cannot reduce the rank. Hence

$$\operatorname{rank}([A\,|\,B]) = 2$$.

Step 4 : Apply the consistency criterion

For a system of $$m$$ equations in $$n$$ unknowns:

• The system is consistent if and only if $$\operatorname{rank}(A)=\operatorname{rank}([A\,|\,B])$$.
• If this common rank equals $$n$$, there is a unique solution.

Here $$\operatorname{rank}(A)=\operatorname{rank}([A\,|\,B])=2$$ and $$n=2$$.

Therefore the system is consistent with a unique solution.

Step 5 : (Optional) Find the solution

Since $$|A|\neq0$$, use Cramer’s rule:

$$\begin{aligned} |A_x| &= \begin{vmatrix} 5 & -1 \\ 4 & 1 \end{vmatrix} = 5(1) - (-1)(4) = 5 + 4 = 9,\\ |A_y| &= \begin{vmatrix} 2 & 5 \\ 1 & 4 \end{vmatrix} = 2(4) - 5(1) = 8 - 5 = 3. \end{aligned}$$

Hence

$$x = \dfrac{|A_x|}{|A|} = \dfrac{9}{3} = 3, \qquad y = \dfrac{|A_y|}{|A|} = \dfrac{3}{3} = 1.$$

The ordered pair $$ (x,y) = (3,1) $$ indeed satisfies both equations.

Conclusion: The system is consistent and has the unique solution $$x = 3,\; y = 1$$.

The given system is

$$x+3y=5,\qquad 2x+6y=8.$$

Write it in matrix form $$AX=B$$, where

$$A=\begin{bmatrix}1&3\\2&6\end{bmatrix},\qquad X=\begin{bmatrix}x\\y\end{bmatrix},\qquad B=\begin{bmatrix}5\\8\end{bmatrix}.$$

Step 1 — Determinant of the coefficient matrix.

$$|A|=\begin{vmatrix}1&3\\2&6\end{vmatrix}=(1)(6)-(3)(2)=6-6=0.$$

Since $$|A|=0$$, the system does not have a unique solution; it is either inconsistent or has infinitely many solutions. To decide between the two, examine $$(\operatorname{adj}A)\,B$$.

Step 2 — Compute $$\operatorname{adj}A$$.

For $$A=\begin{bmatrix}1&3\\2&6\end{bmatrix}$$, interchange the diagonal entries and change the sign of the off-diagonal entries:

$$\operatorname{adj}A=\begin{bmatrix}6&-3\\-2&1\end{bmatrix}.$$

Step 3 — Compute $$(\operatorname{adj}A)\,B$$.

$$(\operatorname{adj}A)\,B=\begin{bmatrix}6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8\end{bmatrix}=\begin{bmatrix}(6)(5)+(-3)(8)\\(-2)(5)+(1)(8)\end{bmatrix}=\begin{bmatrix}30-24\\-10+8\end{bmatrix}=\begin{bmatrix}6\\-2\end{bmatrix}.$$

Step 4 — Conclusion.

Here $$|A|=0$$ while $$(\operatorname{adj}A)\,B=\begin{bmatrix}6\\-2\end{bmatrix}\neq O$$. When the determinant of the coefficient matrix is zero and $$(\operatorname{adj}A)\,B$$ is a non-zero matrix, the system has no solution.

This agrees with the geometric picture: $$\dfrac{a_1}{a_2}=\dfrac{1}{2}=\dfrac{b_1}{b_2}=\dfrac{3}{6}$$, but $$\dfrac{c_1}{c_2}=\dfrac{5}{8}$$ differs, so the two lines are parallel and distinct.

Hence the system is inconsistent; it has no solution.

Given system
$$\begin{cases} x+y+z=1 \\ 2x+3y+2z=2 \\ ax+ay+2az=4 \end{cases}$$

Step 1 : Coefficient (square) matrix and its determinant
$$A=\begin{bmatrix}1&1&1\\2&3&2\\a&a;&2a\end{bmatrix}$$

Compute $$|A|$$ by expansion along the first row:

$$|A|=1\left(3\cdot2a-2\cdot a\right)-1\left(2\cdot2a-2\cdot a\right)+1\left(2\cdot a-3\cdot a\right)$$ $$\;=1(6a-2a)-1(4a-2a)+(2a-3a)$$ $$\;=4a-2a-a=a$$

Step 2 : Discuss the two possible cases

• Case I : $$a\neq0$$
  $$|A|\neq0,$$ so a unique solution exists by Cramer’s rule. Hence the system is consistent.
• Case II : $$a=0$$
  With $$a=0$$ the third equation becomes $$0x+0y+0z=4\;\Rightarrow\;0=4,$$ an impossibility. Thus no solution exists and the system is inconsistent.

Conclusion
The given system is consistent for every real $$a\neq0$$ and inconsistent when $$a=0$$.

Given equations

$$3x-y-2z=2,\;2y-z=-1,\;3x-5y=3$$

Step 1 • Matrix form

$$A=\begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix},\;X=\begin{bmatrix}x\\y\\z\end{bmatrix},\;B=\begin{bmatrix}2\\-1\\3\end{bmatrix}$$ so that $$AX=B$$.

Step 2 • Augmented matrix

$$[A|B]=\begin{bmatrix}3&-1&-2&|&2\\0&2&-1&|&-1\\3&-5&0&|&3\end{bmatrix}$$

Step 3 • Row–reduction (elementary operations)

Make the element below the first pivot zero:

$$R_3\leftarrow R_3-R_1\;\Rightarrow\;[A|B]\to\begin{bmatrix}3&-1&-2&|&2\\0&2&-1&|&-1\\0&-4&2&|&1\end{bmatrix}$$

Use the second row to eliminate the entry in the third row:

$$R_3\leftarrow R_3+2R_2\;\Rightarrow\;[A|B]\to\begin{bmatrix}3&-1&-2&|&2\\0&2&-1&|&-1\\0&0&0&|&-1\end{bmatrix}$$

Step 4 • Ranks

• The last row of the coefficient part is $$[0\;0\;0]$$, so $$\operatorname{rank}(A)=2$$.
• In the augmented matrix the same row is $$[0\;0\;0\;|\;-1]$$, a non-zero row, therefore $$\operatorname{rank}([A|B])=3$$.

Step 5 • Consistency test

Since $$\operatorname{rank}(A)\neq\operatorname{rank}([A|B])$$, the system is inconsistent; no solution satisfies all three equations simultaneously.

Step 1 : Write the system in matrix form
Let $$A = \begin{bmatrix}5 & -1 & 4\\[2pt] 2 & 3 & 5\\[2pt] 5 & -2 & 6\end{bmatrix},\;\; X = \begin{bmatrix}x\\ y\\ z\end{bmatrix},\;\; B = \begin{bmatrix}5\\ 2\\ -1\end{bmatrix}$$ Then $$AX = B$$.

Step 2 : Compute the determinant of $$A$$
$$\begin{aligned} |A| &= 5\,(3\cdot6-5\cdot(-2)) -(-1)\,(2\cdot6-5\cdot5) + 4\,(2\cdot(-2)-3\cdot5)\\[4pt] &= 5\,(18+10) + (12-25) + 4\,(-4-15)\\[4pt] &= 5\times28 -13 + 4\times(-19)\\[4pt] &= 140 - 13 - 76\\[4pt] &= 51\neq0. \end{aligned}$$ Since $$|A|\ne0$$, $$\operatorname{rank}(A)=3$$ (full rank).

Step 3 : Rank of the augmented matrix
Because $$|A|\ne0$$, none of the elementary operations that create the augmented matrix can reduce the rank, so
$$\operatorname{rank}([A\,|\,B]) = \operatorname{rank}(A) = 3.$$

Conclusion on consistency
Ranks are equal and equal to the number of unknowns; therefore the system is consistent and possesses a unique solution.

Step 4 : Obtain the unique solution (Cramer’s rule)

Final result
The system is consistent and its solution is $$x=3,\;y=2,\;z=-2$$.

Rewrite the equations in matrix form $$AX = B$$:

$$A = \begin{bmatrix} 5 & 2\\ 7 & 3 \end{bmatrix},\; X = \begin{bmatrix} x\\ y \end{bmatrix},\; B = \begin{bmatrix} 4\\ 5 \end{bmatrix}.$$

To solve, find $$A^{-1}$$ (because $$X = A^{-1}B$$).

1. Determinant of A
$$|A| = 5\times3 - 7\times2 = 15 - 14 = 1.$$
Since $$|A| \neq 0$$, $$A^{-1}$$ exists.

2. Adjoint of A
For a 2 × 2 matrix $$\begin{bmatrix}a & b\\ c & d\end{bmatrix}$$, $$\operatorname{adj} A = \begin{bmatrix}d & -b\\ -c & a\end{bmatrix}$$. Thus

$$\operatorname{adj} A = \begin{bmatrix} 3 & -2\\ -7 & 5 \end{bmatrix}.$$

3. Inverse of A
$$A^{-1} = \dfrac{1}{|A|}\,\operatorname{adj} A = 1\times \begin{bmatrix} 3 & -2\\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2\\ -7 & 5 \end{bmatrix}.$$

4. Multiply by B

$$X = A^{-1}B = \begin{bmatrix} 3 & -2\\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4\\ 5 \end{bmatrix} = \begin{bmatrix} 3(4) + (-2)(5)\\ -7(4) + 5(5) \end{bmatrix} = \begin{bmatrix} 12 - 10\\ -28 + 25 \end{bmatrix} = \begin{bmatrix} 2\\ -3 \end{bmatrix}.$$

Hence $$x = 2$$ and $$y = -3$$.

Step 1 · Write the system in matrix form

$$\begin{aligned}2x-y&=-2\\[2pt]3x+4y&=3\end{aligned}$$

Let

$$A=\begin{bmatrix}2&-1\\3&4\end{bmatrix},\;X=\begin{bmatrix}x\\y\end{bmatrix},\;B=\begin{bmatrix}-2\\3\end{bmatrix}$$

Then $$AX=B$$.

Step 2 · Compute the determinant of A

$$|A|=\begin{vmatrix}2&-1\\3&4\end{vmatrix}=2\times4-(-1)\times3=8+3=11$$

Because $$|A|\neq0$$, matrix $$A$$ is invertible and a unique solution exists.

Step 3 · Find the adjugate and inverse of A

• For a $$2\times2$$ matrix $$\begin{bmatrix}a&b;\\c&d;\end{bmatrix}$$, the adjugate is $$\begin{bmatrix}d&-b\\-c&a;\end{bmatrix}$$.

Thus

$$\operatorname{adj}A=\begin{bmatrix}4&1\\-3&2\end{bmatrix}$$

Hence

$$A^{-1}=\frac1{|A|}\,\operatorname{adj}A=\frac1{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}$$

Step 4 · Solve for X

$$X=A^{-1}B=\frac1{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix}$$

First multiply the matrices:

$$\begin{bmatrix}4&1\\-3&2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix}=\begin{bmatrix}4(-2)+1(3)\\-3(-2)+2(3)\end{bmatrix}=\begin{bmatrix}-8+3\\6+6\end{bmatrix}=\begin{bmatrix}-5\\12\end{bmatrix}$$

Now divide by $$11$$:

$$X=\begin{bmatrix}x\\y\end{bmatrix}=\frac1{11}\begin{bmatrix}-5\\12\end{bmatrix}=\begin{bmatrix}-\dfrac{5}{11}\\\dfrac{12}{11}\end{bmatrix}$$

Step 5 · Write the solution

$$x=-\dfrac{5}{11},\quad y=\dfrac{12}{11}$$

Step 1 : Write the system in matrix form

The given equations are

$$4x-3y=3,\;3x-5y=7$$

Coefficient matrix $$A$$, column-vector of variables $$X$$ and constant column $$B$$ are

$$A=\begin{bmatrix}4&-3\\3&-5\end{bmatrix},\;X=\begin{bmatrix}x\\y\end{bmatrix},\;B=\begin{bmatrix}3\\7\end{bmatrix}$$

Thus $$AX=B$$.

Step 2 : Find $$|A|$$ (determinant of $$A$$)

$$|A|=4(-5)-(-3)(3)=-20+9=-11$$

Since $$|A|\neq0$$, the inverse $$A^{-1}$$ exists and the system has a unique solution.

Step 3 : Compute adjoint of $$A$$

For a $$2\times2$$ matrix $$\begin{bmatrix}a_{11}&a;_{12}\\a_{21}&a;_{22}\end{bmatrix}$$,

$$\operatorname{adj}A=\begin{bmatrix}a_{22}&-a_{12}\\-a_{21}&a;_{11}\end{bmatrix}$$

Hence

$$\operatorname{adj}A=\begin{bmatrix}-5&3\\-3&4\end{bmatrix}$$

Step 4 : Find $$A^{-1}$$

$$A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A=\dfrac{1}{-11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix} =\begin{bmatrix}\dfrac{5}{11}&-\dfrac{3}{11}\\\dfrac{3}{11}&-\dfrac{4}{11}\end{bmatrix}$$

Step 5 : Compute $$X=A^{-1}B$$

$$X=\begin{bmatrix}\dfrac{5}{11}&-\dfrac{3}{11}\\\dfrac{3}{11}&-\dfrac{4}{11}\end{bmatrix} \begin{bmatrix}3\\7\end{bmatrix}$$

Taking components one by one,

• First component (value of $$x$$): $$x=\dfrac{5}{11}\cdot3-\dfrac{3}{11}\cdot7=\dfrac{15-21}{11}=-\dfrac{6}{11}$$
• Second component (value of $$y$$): $$y=\dfrac{3}{11}\cdot3-\dfrac{4}{11}\cdot7=\dfrac{9-28}{11}=-\dfrac{19}{11}$$

Step 6 : Write the solution set

$$x=-\dfrac{6}{11},\;y=-\dfrac{19}{11}$$

Step 1 : Write the system in matrix form

The given equations are

$$5x + 2y = 3, \; 3x + 2y = 5$$

Put $$A = \begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}, \; X = \begin{bmatrix} x \\ y \end{bmatrix}, \; B = \begin{bmatrix} 3 \\ 5 \end{bmatrix}$$

Then $$AX = B$$.

Step 2 : Find the determinant of A

$$|A| = 5 \times 2 - 2 \times 3 = 10 - 6 = 4 \neq 0$$

Because $$|A| \neq 0$$, the inverse $$A^{-1}$$ exists and the system has a unique solution.

Step 3 : Compute the adjugate and inverse of A

For a $$2 \times 2$$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$, $$\operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$.

Hence

$$\operatorname{adj}(A) = \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}$$

Therefore

$$A^{-1} = \dfrac{1}{|A|}\,\operatorname{adj}(A) = \dfrac{1}{4}\begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}$$

Step 4 : Multiply $$A^{-1}$$ with B to get X

$$X = A^{-1}B = \dfrac{1}{4}\begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 5 \end{bmatrix}$$

Carry out the product:

First row: $$2 \times 3 + (-2) \times 5 = 6 - 10 = -4$$

Second row: $$-3 \times 3 + 5 \times 5 = -9 + 25 = 16$$

Thus

$$X = \dfrac{1}{4}\begin{bmatrix} -4 \\ 16 \end{bmatrix} = \begin{bmatrix} -1 \\ 4 \end{bmatrix}$$

Step 5 : Read the solution

$$x = -1, \; y = 4$$

The given system

$$2x+y+z=1, \; x-2y-z=\dfrac{3}{2}, \; 3y-5z=9$$

may be written in matrix form as $$A\,X=B$$ where

$$A=\begin{bmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{bmatrix}, \; X=\begin{bmatrix}x\\y\\z\end{bmatrix}, \; B=\begin{bmatrix}1\\\dfrac32\\9\end{bmatrix}.$$

1. Determinant of the coefficient matrix

$$|A|=\begin{vmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{vmatrix}=2\begin{vmatrix}-2&-1\\3&-5\end{vmatrix}-1\begin{vmatrix}1&-1\\0&-5\end{vmatrix}+1\begin{vmatrix}1&-2\\0&3\end{vmatrix}.$$

Calculate the minors:

\(\begin{vmatrix}-2&-1\\3&-5\end{vmatrix}=(-2)(-5)-(-1)(3)=10+3=13\)

\(\begin{vmatrix}1&-1\\0&-5\end{vmatrix}=1(-5)-(-1)\cdot0=-5\)

\(\begin{vmatrix}1&-2\\0&3\end{vmatrix}=1\cdot3-(-2)\cdot0=3\)

Hence

$$|A|=2(13)-1(-5)+1(3)=26+5+3=34.$$

Because $$|A|\neq0,$$ the system possesses a unique solution.

2. Adjoint of $$A$$

Compute each cofactor $$C_{ij}$$ (signs: $$+ - +; - + -; + - +$$):

\(C_{11}=13\)

\(C_{12}=-(-5)=5\)

\(C_{13}=3\)

\(C_{21}=-\bigl\{\begin{vmatrix}1&1\\3&-5\end{vmatrix}\bigr\}=-(-8)=8\)

\(C_{22}=\begin{vmatrix}2&1\\0&-5\end{vmatrix}=-10\)

\(C_{23}=-\bigl\{\begin{vmatrix}2&1\\0&3\end{vmatrix}\bigr\}=-6\)

\(C_{31}=\begin{vmatrix}1&1\\-2&-1\end{vmatrix}=1\)

\(C_{32}=-\bigl\{\begin{vmatrix}2&1\\1&-1\end{vmatrix}\bigr\}=3\)

\(C_{33}=\begin{vmatrix}2&1\\1&-2\end{vmatrix}=-5\)

The cofactor matrix is

$$C=\begin{bmatrix}13&5&3\\8&-10&-6\\1&3&-5\end{bmatrix}.$$

Transpose to get the adjoint:

$$\operatorname{Adj}(A)=C^{\mathrm T}=\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}.$$

3. Inverse of $$A$$

$$A^{-1}=\dfrac1{|A|}\operatorname{Adj}(A)=\dfrac1{34}\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}.$$

4. Product $$A^{-1}B$$

First compute $$P=\operatorname{Adj}(A)\,B$$:

Row 1: $$13(1)+8\left(\dfrac32\right)+1(9)=13+12+9=34$$

Row 2: $$5(1)+(-10)\left(\dfrac32\right)+3(9)=5-15+27=17$$

Row 3: $$3(1)+(-6)\left(\dfrac32\right)+(-5)(9)=3-9-45=-51$$

Thus $$P=\begin{bmatrix}34\\17\\-51\end{bmatrix}.$$

Now

$$X=A^{-1}B=\frac1{34}P=\begin{bmatrix}\dfrac{34}{34}\\\dfrac{17}{34}\\\dfrac{-51}{34}\end{bmatrix}=\begin{bmatrix}1\\\dfrac12\\-\dfrac32\end{bmatrix}.$$

5. Solution

$$x=1, \; y=\dfrac12, \; z=-\dfrac32.$$

Let the given system be written in matrix form $$A\,X = B$$, where

$$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, \; X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \; B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}.$$

To solve we need $$A^{-1}$$. First evaluate the determinant $$|A|$$ (using the first row):

$$|A| = 1\bigl(1\cdot 1 - (-3)\cdot 1\bigr) - (-1)\bigl(2\cdot 1 - (-3)\cdot 1\bigr) + 1\bigl(2\cdot 1 - 1\cdot 1\bigr) = 4 + 5 + 1 = 10.$$

Since $$|A| \neq 0$$, $$A$$ is non–singular and a unique solution exists.

Step 1 – Cofactors

Step 2 – Adjugate
The cofactor matrix is $$C = \begin{bmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{bmatrix}$$.
Hence $$\operatorname{adj}A = C^{T} = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}.$$

Step 3 – Inverse of $$A$$
$$A^{-1} = \dfrac{1}{|A|}\,\operatorname{adj}A = \dfrac{1}{10}\,\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}.$$

Step 4 – Solution vector

$$X = A^{-1}B = \dfrac{1}{10}\,\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}.$$

Compute the product inside:

First entry  $$= 4\cdot4 + 2\cdot0 + 2\cdot2 = 20$$
Second entry $$= -5\cdot4 + 0\cdot0 + 5\cdot2 = -10$$
Third entry  $$= 1\cdot4 + (-2)\cdot0 + 3\cdot2 = 10$$

Therefore $$X = \dfrac{1}{10}\,\begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}.$$

Hence $$x = 2,\; y = -1,\; z = 1.$$

Step 1  Write the system in matrix form

Let $$A=\begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix},\;X=\begin{bmatrix}x\\y\\z\end{bmatrix},\;B=\begin{bmatrix}5\\-4\\3\end{bmatrix}$$
Then $$AX=B$$.

Step 2  Find $$|A|$$

Expanding along the first row:

$$\begin{aligned}|A|&=2\Bigl((-2)(-2)-1(-1)\Bigr)-3\Bigl(1(-2)-1(3)\Bigr)+3\Bigl(1(-1)-(-2)(3)\Bigr)\\&=2(4+1)-3(-2-3)+3(-1+6)\\&=2\times5-3\times(-5)+3\times5\\&=10+15+15=40\neq0\end{aligned}$$

Since $$|A|\neq0$$, $$A^{-1}$$ exists and the system has a unique solution.

Step 3  Compute the adjoint of $$A$$

All cofactors $$C_{ij}$$:

Thus

$$\operatorname{adj}A=\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}$$

Step 4  Find $$A^{-1}$$

$$A^{-1}=\dfrac1{|A|}\,\operatorname{adj}A=\dfrac1{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}$$

Step 5  Compute $$X=A^{-1}B$$

First $$\operatorname{adj}A\,B=\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix}5\\-4\\3\end{bmatrix}=\begin{bmatrix}40\\80\\-40\end{bmatrix}$$

Hence

$$X=A^{-1}B=\dfrac1{40}\begin{bmatrix}40\\80\\-40\end{bmatrix}=\begin{bmatrix}1\\2\\-1\end{bmatrix}$$

Step 6  Write the solution

$$x=1,\;y=2,\;z=-1$$

Step 1 : Write the system in matrix form
$$A\,X=B$$
$$A=\begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix},\;X=\begin{bmatrix}x\\y\\z\end{bmatrix},\;B=\begin{bmatrix}7\\-5\\12\end{bmatrix}$$

Step 2 : Find |A|
Expand along the first row:
$$|A|=1\Bigl|\begin{smallmatrix}4&-5\\-1&3\end{smallmatrix}\Bigr|-(-1)\Bigl|\begin{smallmatrix}3&-5\\2&3\end{smallmatrix}\Bigr|+2\Bigl|\begin{smallmatrix}3&4\\2&-1\end{smallmatrix}\Bigr|$$
$$|A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4$$
Since $$|A|\neq0$$, the system has a unique solution and $$A^{-1}$$ exists.

Step 3 : Compute cofactors of A
\(C_{11}=7,\;C_{12}=-19,\;C_{13}=-11\)
\(C_{21}=1,\;C_{22}=-1,\;C_{23}=-1\)
\(C_{31}=-3,\;C_{32}=11,\;C_{33}=7\)

Step 4 : Adjugate and inverse of A
Adjugate is the transpose of the cofactor matrix:
$$\operatorname{adj}A=\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}$$
$$A^{-1}=\dfrac{1}{|A|}\,\operatorname{adj}A=\dfrac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}$$

Step 5 : Find X = A⁻¹B
First compute $$\operatorname{adj}A\,B$$:
$$\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\begin{bmatrix}7\\-5\\12\end{bmatrix}=\begin{bmatrix}7\times7+1(-5)+(-3)12\\-19\times7+(-1)(-5)+11\times12\\-11\times7+(-1)(-5)+7\times12\end{bmatrix}=\begin{bmatrix}8\\4\\12\end{bmatrix}$$
Multiply by $$\dfrac14$$:
$$X=A^{-1}B=\dfrac14\begin{bmatrix}8\\4\\12\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}$$

Step 6 : Solution of the system
$$x=2,\;y=1,\;z=3$$

Step 1 : Write the matrix
$$A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$$

Step 2 : Determinant of $$A$$
Expand along the first row:
$$|A| = 2\Bigl|(2,-4;1,-2)\Bigr| -(-3)\Bigl|(3,-4;1,-2)\Bigr| + 5\Bigl|(3,2;1,1)\Bigr|$$
$$|A| = 2\bigl(2\times (-2) - (-4)\times 1\bigr) -(-3)\bigl(3\times (-2) - (-4)\times 1\bigr) + 5\bigl(3\times 1 - 2\times 1\bigr)$$
$$|A| = 2\,( -4 + 4) -(-3)\,(-6+4) + 5\,(3-2)$$
$$|A| = 0 - (-3)\,(-2) + 5\,\times 1 = 0 - 6 + 5 = -1$$
Since $$|A|\neq 0$$, $$A$$ is invertible.

Step 3 : Cofactor matrix

Thus
$$\text{Cofactor}(A)=\begin{bmatrix}0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13\end{bmatrix}$$

Step 4 : Adjoint and inverse
Adjoint is its transpose:
$$\operatorname{adj}A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}$$
Finally
$$A^{-1}=\dfrac{1}{|A|}\,\operatorname{adj}A = (-1)\times \operatorname{adj}A = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$$

Step 5 : Solve the system $$AX=B$$
Take $$B = \begin{bmatrix}11\\-5\\-3\end{bmatrix}$$ and $$X = \begin{bmatrix}x\\y\\z\end{bmatrix}$$.
Compute $$X = A^{-1}B$$:

• $$x = 0\cdot11 + 1\cdot(-5) + (-2)\cdot(-3)= -5+6 = 1$$
• $$y = (-2)\cdot11 + 9\cdot(-5) + (-23)\cdot(-3)= -22-45+69 = 2$$
• $$z = (-1)\cdot11 + 5\cdot(-5) + (-13)\cdot(-3)= -11-25+39 = 3$$

Step 6 : Result
The inverse matrix is
$$A^{-1}=\begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$$
and the unique solution of the given system is
$$x = 1,\; y = 2,\; z = 3.$$

Step 1 : Formulate the simultaneous linear equations

Let  $$x$$, $$y$$ and $$z$$ (in rupees) be the per-kilogram costs of onion, wheat and rice respectively. The statements in the question translate to

• $$4x+3y+2z = 60$$
• $$2x+4y+6z = 90$$
• $$6x+2y+3z = 70$$

Step 2 : Write the system in matrix form

$$ \begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}60\\90\\70\end{bmatrix} $$

Denote  $$A=\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}, \; X=\begin{bmatrix}x\\y\\z\end{bmatrix}, \; B=\begin{bmatrix}60\\90\\70\end{bmatrix}$$. We have  $$AX=B$$.

Step 3 : Evaluate the determinant $$\Delta = |A|$$

$$\begin{aligned} \Delta &= 4\,(4\cdot3-6\cdot2) \; -\; 3\,(2\cdot3-6\cdot6) \; +\; 2\,(2\cdot2-4\cdot6)\\[2pt] &=4\,(12-12)-3\,(6-36)+2\,(4-24)\\[2pt] &=0+90-40\\[2pt] &=50\neq0. \end{aligned}$$

Because $$\Delta\neq0$$, a unique solution exists and we may apply Cramer’s Rule.

Step 4 : Compute $$\Delta_x,\;\Delta_y,\;\Delta_z$$

(a) Replace the first column of $$A$$ by $$B$$ to get $$\Delta_x$$ :

$$\Delta_x = \begin{vmatrix}60&3&2\\90&4&6\\70&2&3\end{vmatrix} =60(4\cdot3-6\cdot2)-3(90\cdot3-6\cdot70)+2(90\cdot2-4\cdot70) \\[2pt] =60(12-12)-3(270-420)+2(180-280) \\[2pt] =0+450-200=250.$$

(b) Replace the second column of $$A$$ by $$B$$ to get $$\Delta_y$$ :

$$\Delta_y = \begin{vmatrix}4&60&2\\2&90&6\\6&70&3\end{vmatrix} =4(90\cdot3-6\cdot70)-60(2\cdot3-6\cdot6)+2(2\cdot70-90\cdot6)\\[2pt] =4(270-420)-60(6-36)+2(140-540)\\[2pt] =-600+1800-800=400.$$

(c) Replace the third column of $$A$$ by $$B$$ to get $$\Delta_z$$ :

$$\Delta_z = \begin{vmatrix}4&3&60\\2&4&90\\6&2&70\end{vmatrix} =4(4\cdot70-90\cdot2)-3(2\cdot70-90\cdot6)+60(2\cdot2-4\cdot6)\\[2pt] =4(280-180)-3(140-540)+60(4-24)\\[2pt] =400+1200-1200=400.$$

Step 5 : Obtain $$x,y,z$$

Using Cramer’s Rule,

$$x = \dfrac{\Delta_x}{\Delta}=\dfrac{250}{50}=5,\qquad y = \dfrac{\Delta_y}{\Delta}=\dfrac{400}{50}=8,\qquad z = \dfrac{\Delta_z}{\Delta}=\dfrac{400}{50}=8.$$

Step 6 : Interpretation

The required per-kilogram costs are therefore :

• Onion : ₹ 5 per kg
• Wheat : ₹ 8 per kg
• Rice   : ₹ 8 per kg

Let the coefficient matrix, column–variable matrix and constant column be denoted by

$$A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \qquad B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$$

The linear system is $$AX = B.$$

The question supplies the product of $$A$$ with another square matrix

$$C = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}.$$

We first verify the product.

Step 1: Multiply $$A$$ and $$C.$*

For the first row of $$A$$:

• First column: $$1(-2) + (-1)9 + 2\,6 = -2-9+12 = 1$$
• Second column: $$1\cdot0 + (-1)2 + 2\cdot1 = -2+2 = 0$$
• Third column: $$1\cdot1 + (-1)(-3) + 2(-2) = 1+3-4 = 0$$

This gives the row $$(1,0,0).$$

Repeating for the second row of $$A$$:

• First column: $$0(-2) + 2\,9 + (-3)6 = 18-18 = 0$$
• Second column: $$0\cdot0 + 2\,2 + (-3)1 = 4-3 = 1$$
• Third column: $$0\cdot1 + 2(-3) + (-3)(-2) = -6+6 = 0$$

This gives $$(0,1,0).$$

Finally for the third row of $$A$$:

• First column: $$3(-2) + (-2)9 + 4\,6 = -6-18+24 = 0$$
• Second column: $$3\cdot0 + (-2)2 + 4\,1 = -4+4 = 0$$
• Third column: $$3\cdot1 + (-2)(-3) + 4(-2) = 3+6-8 = 1$$

This gives $$(0,0,1).$$

Putting the three rows together,

$$AC = I_{3},$$

the $$3\times3$$ identity matrix. Hence $$C = A^{-1}.$*

Step 2: Solve for $$X.$$

Because $$C = A^{-1},$$ multiply both sides of $$AX=B$$ on the left by $$C$$:

$$X = C B.$$

Compute $$CB$$:

$$\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} -2(1) + 0(1) + 1(2) \\ 9(1) + 2(1) - 3(2) \\ 6(1) + 1(1) - 2(2) \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix}.$$

Thus $$x = 0, \; y = 5, \; z = 3.$$

The given system has the unique solution $$ (x,y,z) = (0,5,3).$$

Let the given determinant be denoted by $$\Delta$$:

$$\Delta=\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$$

Expand along the first row (cofactor expansion):

$$\Delta = x\begin{vmatrix}-x & 1\\1 & x\end{vmatrix} - \sin\theta\begin{vmatrix}-\sin\theta & 1\\\cos\theta & x\end{vmatrix} + \cos\theta\begin{vmatrix}-\sin\theta & -x\\\cos\theta & 1\end{vmatrix}$$

Evaluate each 2 × 2 minor:

• $$\begin{vmatrix}-x & 1\\1 & x\end{vmatrix}=(-x)(x)-(1)(1)=-x^{2}-1$$
• $$\begin{vmatrix}-\sin\theta & 1\\\cos\theta & x\end{vmatrix}=(-\sin\theta)(x)-(1)(\cos\theta)=-x\sin\theta-\cos\theta$$
• $$\begin{vmatrix}-\sin\theta & -x\\\cos\theta & 1\end{vmatrix}=(-\sin\theta)(1)-(-x)(\cos\theta)=-\sin\theta+x\cos\theta$$

Substitute these back:

$$\Delta = x(-x^{2}-1) - \sin\theta(-x\sin\theta-\cos\theta) + \cos\theta(-\sin\theta+x\cos\theta)$$

Simplify each product:

• First term: $$x(-x^{2}-1)=-x^{3}-x$$
• Second term: $$-\sin\theta(-x\sin\theta-\cos\theta)=x\sin^{2}\theta+\sin\theta\cos\theta$$
• Third term: $$\cos\theta(-\sin\theta+x\cos\theta)=-\sin\theta\cos\theta+x\cos^{2}\theta$$

Add the three results:

$$\begin{aligned} \Delta &=(-x^{3}-x)+(x\sin^{2}\theta+\sin\theta\cos\theta)+(-\sin\theta\cos\theta+x\cos^{2}\theta)\\[4pt] &=-x^{3}-x + x(\sin^{2}\theta+\cos^{2}\theta)\\[4pt] &=-x^{3}-x + x\\[4pt] &=-x^{3}. \end{aligned}$$

Because $$\Delta=-x^{3}$$ contains no $$\theta$$, the determinant is independent of $$\theta$$.

Hence proved.

Let the given determinant be denoted by $$\triangle$$.

\[\triangle = \begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha\\[2pt] -\sin\beta & \cos\beta & 0\\[2pt] \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}\]

For a 3 × 3 determinant

$$\begin{vmatrix}a & b & c\\ d & e & f\\ g & h & i\end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg).$$

Here we identify

Compute each minor exactly.

1. First minor
   $$ei-fh=\cos\beta\,\cos\alpha-0\;(\sin\alpha\sin\beta)=\cos\alpha\cos\beta.$$

   Thus
   $$a(ei-fh)=(\cos\alpha\cos\beta)(\cos\alpha\cos\beta)=\cos^{2}\!\alpha\,\cos^{2}\!\beta.$$

2. Second minor
   $$di-fg=(-\sin\beta)\,\cos\alpha-0\,(\sin\alpha\cos\beta)=-\sin\beta\cos\alpha.$$

   Hence
   $$-b(di-fg)=-\bigl(\cos\alpha\sin\beta\bigr)(-\sin\beta\cos\alpha)=\cos^{2}\!\alpha\,\sin^{2}\!\beta.$$

3. Third minor
   $$dh-eg=(-\sin\beta)(\sin\alpha\sin\beta)-\cos\beta\,(\sin\alpha\cos\beta)$$ $$=-\sin\alpha\sin^{2}\!\beta-\sin\alpha\cos^{2}\!\beta$$ $$=-\sin\alpha(\sin^{2}\!\beta+\cos^{2}\!\beta)$$ $$=-\sin\alpha.$$

   Therefore
   $$c(dh-eg)=(-\sin\alpha)(-\sin\alpha)=\sin^{2}\!\alpha.$$

Now add the three contributions:

$$\triangle=\cos^{2}\!\alpha\,\cos^{2}\!\beta+\cos^{2}\!\alpha\,\sin^{2}\!\beta+\sin^{2}\!\alpha.$$

Factorising $$\cos^{2}\!\alpha$$ from the first two terms:

$$\triangle=\cos^{2}\!\alpha\bigl(\cos^{2}\!\beta+\sin^{2}\!\beta\bigr)+\sin^{2}\!\alpha.$$

Using the Pythagorean identity $$\cos^{2}\!\beta+\sin^{2}\!\beta=1$$, we get

$$\triangle=\cos^{2}\!\alpha+\sin^{2}\!\alpha=1.$$

Hence the value of the given determinant is $$1$$.

We have to determine $$(AB)^{-1}$$ when we know $$A^{-1}$$ and $$B$$.

Step 1 : Use the inverse–product rule
For any two square matrices $$A$$ and $$B$$ of the same order, $$ (AB)^{-1}=B^{-1}A^{-1} $$. Thus we only need $$B^{-1}$$, because $$A^{-1}$$ is already given.

Step 2 : Find the determinant of $$B$$
$$\det(B)=\begin{vmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{vmatrix}$$

Expand along the first row:

$$\det(B)=1\,(3\cdot1-0\cdot(-2)) -2\,((-1)\cdot1-0\cdot0) +(-2)\,((-1)(-2)-3\cdot0)=3+2-4=1.$$

Because $$\det(B)=1\neq0$$, $$B$$ is invertible. When the determinant equals 1, $$B^{-1}=\operatorname{adj}(B)$$.

Step 3 : Cofactor matrix of $$B$$

Hence the cofactor matrix is $$\begin{bmatrix}3&1&2\\2&1&2\\6&2&5\end{bmatrix}.$$

Step 4 : Adjoint and inverse of $$B$$
$$\operatorname{adj}(B)=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\quad\Longrightarrow\quad B^{-1}=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}.$$

Step 5 : Multiply $$B^{-1}$$ by $$A^{-1}$$

Given $$A^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix},\qquad B^{-1}=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}.$$

Compute $$B^{-1}A^{-1}$$ element-wise:

• $$(1,1):\;3\cdot3+2\cdot(-15)+6\cdot5=9-30+30=9$$
• $$(1,2):\;3\cdot(-1)+2\cdot6+6\cdot(-2)=-3+12-12=-3$$
• $$(1,3):\;3\cdot1+2\cdot(-5)+6\cdot2=3-10+12=5$$
• $$(2,1):\;1\cdot3+1\cdot(-15)+2\cdot5=3-15+10=-2$$
• $$(2,2):\;1\cdot(-1)+1\cdot6+2\cdot(-2)=-1+6-4=1$$
• $$(2,3):\;1\cdot1+1\cdot(-5)+2\cdot2=1-5+4=0$$
• $$(3,1):\;2\cdot3+2\cdot(-15)+5\cdot5=6-30+25=1$$
• $$(3,2):\;2\cdot(-1)+2\cdot6+5\cdot(-2)=-2+12-10=0$$
• $$(3,3):\;2\cdot1+2\cdot(-5)+5\cdot2=2-10+10=2$$

Therefore $$B^{-1}A^{-1}=\begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}.$$

Step 6 : Conclude
$$ (AB)^{-1}=B^{-1}A^{-1}=\begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}. $$

Let $$\Delta = \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$$.

Expand along the first row:

$$\Delta = x \begin{vmatrix} x+y & x \\ x & y \end{vmatrix} - y \begin{vmatrix} y & x \\ x+y & y \end{vmatrix} + (x+y) \begin{vmatrix} y & x+y \\ x+y & x \end{vmatrix}.$$

Evaluate the minors:

• $$\begin{vmatrix} x+y & x \\ x & y \end{vmatrix} = (x+y)y - x^{2} = xy + y^{2} - x^{2}.$$
• $$\begin{vmatrix} y & x \\ x+y & y \end{vmatrix} = y^{2} - x(x+y) = y^{2} - x^{2} - xy.$$
• $$\begin{vmatrix} y & x+y \\ x+y & x \end{vmatrix} = yx - (x+y)^{2} = -x^{2} - xy - y^{2}.$$

Substitute these values:

$$\begin{aligned}\Delta &= x(xy + y^{2} - x^{2}) - y(y^{2} - x^{2} - xy) + (x+y)(-x^{2} - xy - y^{2})\\&= x^{2}y + xy^{2} - x^{3} - y^{3} + x^{2}y + xy^{2} - (x+y)(x^{2}+xy+y^{2}).\end{aligned}$$

Now, $$ (x+y)(x^{2}+xy+y^{2}) = x^{3} + 2x^{2}y + 2xy^{2} + y^{3}. $$

Thus,

$$\begin{aligned}\Delta &= 2x^{2}y + 2xy^{2} - x^{3} - y^{3} - x^{3} - 2x^{2}y - 2xy^{2} - y^{3}\\&= -2x^{3} - 2y^{3}.\end{aligned}$$

Factorising,

$$\Delta = -2(x^{3}+y^{3}) = -2(x+y)(x^{2}-xy+y^{2}).$$

Hence the value of the determinant is $$-2(x+y)(x^{2}-xy+y^{2}).$$

Let $$\Delta = \begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}.$$

Apply the row operations $$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 - R_1$$ (these do not change the value of the determinant):

$$\Delta = \begin{vmatrix} 1 & x & y \\ 0 & (x+y)-x & y-y \\ 0 & x-x & (x+y)-y \end{vmatrix} = \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix}.$$

The resulting matrix is upper-triangular, so the determinant equals the product of the diagonal elements:

$$\Delta = 1 \times y \times x = xy.$$

Hence, $$\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix} = xy.$$

Step 1 : Transform the equations into a linear system

Introduce new variables $$u = \dfrac1x, \; v = \dfrac1y, \; w = \dfrac1z$$.
The given equations become

$$2u + 3v + 10w = 4$$
$$4u - 6v + 5w = 1$$
$$6u + 9v - 20w = 2$$

Now we have a linear system in $$u, v, w$$.

Step 2 : Write the coefficient matrix and its determinant

$$A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}$$

Compute $$\Delta = |A|$$:

$$\begin{aligned} \Delta &= 2\begin{vmatrix}-6 & 5 \\ 9 & -20\end{vmatrix} - 3\begin{vmatrix}4 & 5 \\ 6 & -20\end{vmatrix} + 10\begin{vmatrix}4 & -6 \\ 6 & 9\end{vmatrix}\\[4pt] &= 2(120 - 45) - 3(-80 - 30) + 10(36 + 36)\\[4pt] &= 2(75) + 3(110) + 10(72)\\[4pt] &= 150 + 330 + 720 = 1200. \end{aligned}$$

Since $$\Delta = 1200 \neq 0$$, a unique solution exists and we can use Cramer’s rule.

Step 3 : Find $$u$$

Replace the first column of $$A$$ by the constants to get $$\Delta_u$$:

$$\Delta_u = \begin{vmatrix} 4 & 3 & 10 \\ 1 & -6 & 5 \\ 2 & 9 & -20 \end{vmatrix}$$

$$\begin{aligned} \Delta_u &= 4\begin{vmatrix}-6 & 5 \\ 9 & -20\end{vmatrix} - 3\begin{vmatrix}1 & 5 \\ 2 & -20\end{vmatrix} + 10\begin{vmatrix}1 & -6 \\ 2 & 9\end{vmatrix}\\[4pt] &= 4(75) - 3(-30) + 10(21)\\[4pt] &= 300 + 90 + 210 = 600. \end{aligned}$$

Therefore $$u = \dfrac{\Delta_u}{\Delta} = \dfrac{600}{1200} = \dfrac12 \;\Rightarrow\; x = 2.$$

Step 4 : Find $$v$$

Replace the second column of $$A$$ by the constants:

$$\Delta_v = \begin{vmatrix} 2 & 4 & 10 \\ 4 & 1 & 5 \\ 6 & 2 & -20 \end{vmatrix}$$

$$\begin{aligned} \Delta_v &= 2\begin{vmatrix}1 & 5 \\ 2 & -20\end{vmatrix} - 4\begin{vmatrix}4 & 5 \\ 6 & -20\end{vmatrix} + 10\begin{vmatrix}4 & 1 \\ 6 & 2\end{vmatrix}\\[4pt] &= 2(-30) - 4(-110) + 10(2)\\[4pt] &= -60 + 440 + 20 = 400. \end{aligned}$$

Thus $$v = \dfrac{\Delta_v}{\Delta} = \dfrac{400}{1200} = \dfrac13 \;\Rightarrow\; y = 3.$$

Step 5 : Find $$w$$

Replace the third column of $$A$$ by the constants:

$$\Delta_w = \begin{vmatrix} 2 & 3 & 4 \\ 4 & -6 & 1 \\ 6 & 9 & 2 \end{vmatrix}$$

$$\begin{aligned} \Delta_w &= 2\begin{vmatrix}-6 & 1 \\ 9 & 2\end{vmatrix} - 3\begin{vmatrix}4 & 1 \\ 6 & 2\end{vmatrix} + 4\begin{vmatrix}4 & -6 \\ 6 & 9\end{vmatrix}\\[4pt] &= 2(-12 - 9) - 3(8 - 6) + 4(36 + 36)\\[4pt] &= 2(-21) - 3(2) + 4(72)\\[4pt] &= -42 - 6 + 288 = 240. \end{aligned}$$

Therefore $$w = \dfrac{\Delta_w}{\Delta} = \dfrac{240}{1200} = \dfrac15 \;\Rightarrow\; z = 5.$$

Step 6 : Conclusion

The solution of the given system is $$x = 2,\; y = 3,\; z = 5.$$

Step 1 : Write the given matrix

$$A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}, \;\; x , y , z \ne 0$$

Step 2 : Check that an inverse can exist

The determinant of a diagonal matrix equals the product of its diagonal entries:

$$\det (A) = x\,y\,z.$$

Because $$x , y , z \ne 0,$$ we have $$\det (A) \ne 0,$$ so $$A^{-1}$$ exists.

Step 3 : Recall the inverse of a diagonal matrix

For any invertible diagonal matrix $$\operatorname{diag}(d_1,d_2,\dots ,d_n),$$ the inverse is obtained by replacing every diagonal entry by its reciprocal:

$$\bigl[\operatorname{diag}(d_1,d_2,\dots ,d_n)\bigr]^{-1}=\operatorname{diag}\Bigl(\tfrac1{d_1},\tfrac1{d_2},\dots ,\tfrac1{d_n}\Bigr).$$

Applying this fact to $$A$$ gives

$$A^{-1}=\begin{bmatrix} \dfrac1x & 0 & 0 \\ 0 & \dfrac1y & 0 \\ 0 & 0 & \dfrac1z \end{bmatrix}.$$

Step 4 : Verify by multiplication (optional check)

$$A\,A^{-1}=\begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}\begin{bmatrix} \tfrac1x & 0 & 0 \\ 0 & \tfrac1y & 0 \\ 0 & 0 & \tfrac1z \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=I_3.$$

Likewise $$A^{-1}A=I_3,$$ confirming the result.

Step 5 : Choose the correct option

The inverse obtained matches option (A).

Let $$s=\sin \theta$$. Then $$A=\begin{bmatrix} 1 & s & 1 \\ -s & 1 & s \\ -1 & -s & 1 \end{bmatrix}$$.

Using expansion along the first row:

$$\det(A)=1\cdot\Bigl|\begin{matrix} 1 & s \\ -s & 1 \end{matrix}\Bigr|-s\cdot\Bigl|\begin{matrix}-s & s \\ -1 & 1\end{matrix}\Bigr|+1\cdot\Bigl|\begin{matrix}-s & 1 \\ -1 & -s\end{matrix}\Bigr|.$$

Compute each 2 × 2 determinant:

• $$\Bigl|\begin{matrix} 1 & s \\ -s & 1 \end{matrix}\Bigr|=1\cdot1-s(-s)=1+s^{2}.$$
• $$\Bigl|\begin{matrix}-s & s \\ -1 & 1\end{matrix}\Bigr|=(-s)(1)-s(-1)=-s+s=0.$$
• $$\Bigl|\begin{matrix}-s & 1 \\ -1 & -s\end{matrix}\Bigr|=(-s)(-s)-1(-1)=s^{2}+1.$$

Substituting back,

$$\det(A)=1\cdot(1+s^{2})-s\cdot0+1\cdot(1+s^{2})=2(1+s^{2}).$$

Now $$s^{2}=\sin^{2}\theta\in[0,1]$$ for $$0\le\theta\le2\pi$$. Therefore

$$\det(A)=2(1+\sin^{2}\theta)\in[2,4].$$

Hence among the given options, only (D) $$\det(A)\in[2,4]$$ is correct.