Chapter 3: Matrices

The information has three factories (I, II, III) and two categories of workers (men, women). We take the rows to represent the factories and the columns to represent the type of worker. Since there are 3 factories and 2 types of workers, the matrix will be of order $$3 \times 2$$.

Placing the men workers in the first column and the women workers in the second column, the required matrix is

$$A = \begin{bmatrix} 30 & 25 \\ 25 & 31 \\ 27 & 26 \end{bmatrix}$$

Here the rows correspond to factories I, II, III (top to bottom) and the columns correspond to men workers and women workers respectively.

The entry in the third row and second column is $$a_{32} = 26$$. Since the third row stands for factory III and the second column stands for women workers, this entry represents the number of women workers in factory III.

A matrix of order $$m \times n$$ has exactly $$mn$$ elements. So to find the possible orders of a matrix with 8 elements, we need all ordered pairs $$(m, n)$$ of positive integers for which

$$m \times n = 8$$

We therefore list all the ways of writing 8 as a product of two natural numbers (order matters, since $$m \times n$$ and $$n \times m$$ are different orders):

$$8 = 1 \times 8 = 2 \times 4 = 4 \times 2 = 8 \times 1$$

Hence the possible orders are

$$1 \times 8,\quad 8 \times 1,\quad 2 \times 4,\quad 4 \times 2$$

So there are 4 possible orders.

A $$3 \times 2$$ matrix has the form $$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}$$, where the row index $$i$$ takes values $$1, 2, 3$$ and the column index $$j$$ takes values $$1, 2$$.

We compute each element using $$a_{ij} = \frac{1}{2}|i - 3j|$$.

$$a_{11} = \frac{1}{2}|1 - 3(1)| = \frac{1}{2}|-2| = 1$$

$$a_{12} = \frac{1}{2}|1 - 3(2)| = \frac{1}{2}|-5| = \frac{5}{2}$$

$$a_{21} = \frac{1}{2}|2 - 3(1)| = \frac{1}{2}|-1| = \frac{1}{2}$$

$$a_{22} = \frac{1}{2}|2 - 3(2)| = \frac{1}{2}|-4| = 2$$

$$a_{31} = \frac{1}{2}|3 - 3(1)| = \frac{1}{2}|0| = 0$$

$$a_{32} = \frac{1}{2}|3 - 3(2)| = \frac{1}{2}|-3| = \frac{3}{2}$$

Therefore the required matrix is

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

Two matrices are equal if and only if they have the same order and their corresponding elements are equal. Both matrices here are of order $$3 \times 3$$, so we equate the entries in matching positions.

Comparing position $$(1,1)$$: $$x + 3 = 0 \Rightarrow x = -3$$

Comparing position $$(1,2)$$: $$z + 4 = 6 \Rightarrow z = 2$$

Comparing position $$(1,3)$$: $$2y - 7 = 3y - 2$$

$$\Rightarrow -7 + 2 = 3y - 2y \Rightarrow -5 = y \Rightarrow y = -5$$

Comparing position $$(2,2)$$: $$a - 1 = -3 \Rightarrow a = -2$$

Comparing position $$(2,3)$$: $$0 = 2c + 2 \Rightarrow 2c = -2 \Rightarrow c = -1$$

Comparing position $$(3,1)$$: $$b - 3 = 2b + 4$$

$$\Rightarrow -3 - 4 = 2b - b \Rightarrow b = -7$$

(The remaining positions $$(2,1), (3,2), (3,3)$$ already match, since $$-6 = -6$$, $$-21 = -21$$ and $$0 = 0$$.)

Hence $$a = -2,\ b = -7,\ c = -1,\ x = -3,\ y = -5,\ z = 2$$.

Since the two matrices are equal, their corresponding elements must be equal. Equating entries in matching positions gives four equations:

$$2a + b = 4 \quad \text{...(1)}$$

$$a - 2b = -3 \quad \text{...(2)}$$

$$5c - d = 11 \quad \text{...(3)}$$

$$4c + 3d = 24 \quad \text{...(4)}$$

Solving for $$a$$ and $$b$$: Multiply equation (1) by 2: $$4a + 2b = 8 \quad \text{...(5)}$$

Adding (5) and (2): $$(4a + 2b) + (a - 2b) = 8 + (-3)$$

$$5a = 5 \Rightarrow a = 1$$

Substituting $$a = 1$$ in (1): $$2(1) + b = 4 \Rightarrow b = 2$$

Solving for $$c$$ and $$d$$: Multiply equation (3) by 3: $$15c - 3d = 33 \quad \text{...(6)}$$

Adding (6) and (4): $$(15c - 3d) + (4c + 3d) = 33 + 24$$

$$19c = 57 \Rightarrow c = 3$$

Substituting $$c = 3$$ in (3): $$5(3) - d = 11 \Rightarrow 15 - d = 11 \Rightarrow d = 4$$

Hence $$a = 1,\ b = 2,\ c = 3,\ d = 4$$.

Both $$A$$ and $$B$$ are matrices of order $$2 \times 3$$, so their sum is defined. The sum $$A + B$$ is obtained by adding the corresponding (matching) elements.

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

$$= \begin{bmatrix} \sqrt{3}+2 & 1+\sqrt{5} & -1+1 \\ 2+(-2) & 3+3 & 0+\frac{1}{2} \end{bmatrix}$$

$$= \begin{bmatrix} \sqrt{3}+2 & 1+\sqrt{5} & 0 \\ 0 & 6 & \frac{1}{2} \end{bmatrix}$$

First multiply matrix $$A$$ by the scalar 2, which means multiplying every element of $$A$$ by 2:

$$2A = 2\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \end{bmatrix}$$

Now subtract $$B$$ from $$2A$$ by subtracting corresponding elements (both are of order $$2 \times 3$$):

$$2A - B = \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}$$

$$= \begin{bmatrix} 2-3 & 4-(-1) & 6-3 \\ 4-(-1) & 6-0 & 2-2 \end{bmatrix} = \begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}$$

We treat the matrix equation like an ordinary algebraic equation. Starting from

$$2A + 3X = 5B$$

Subtract $$2A$$ from both sides: $$3X = 5B - 2A$$

Multiply both sides by the scalar $$\frac{1}{3}$$: $$X = \frac{1}{3}(5B - 2A)$$

Now compute $$5B$$ and $$2A$$:

$$5B = 5\begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix} = \begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix}$$

$$2A = 2\begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix} = \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix}$$

So

$$5B - 2A = \begin{bmatrix} 10-16 & -10-0 \\ 20-8 & 10-(-4) \\ -25-6 & 5-12 \end{bmatrix} = \begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}$$

Therefore

$$X = \frac{1}{3}\begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix} = \begin{bmatrix} -2 & -\frac{10}{3} \\ 4 & \frac{14}{3} \\ -\frac{31}{3} & -\frac{7}{3} \end{bmatrix}$$

We are given the two matrix equations

$$X + Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} \quad \text{...(1)}$$

$$X - Y = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} \quad \text{...(2)}$$

Adding (1) and (2): the $$Y$$ terms cancel, giving $$2X$$:

$$2X = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}$$

$$\Rightarrow X = \frac{1}{2}\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}$$

Subtracting (2) from (1): the $$X$$ terms cancel, giving $$2Y$$:

$$2Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}$$

$$\Rightarrow Y = \frac{1}{2}\begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}$$

First carry out the scalar multiplication on the left side:

$$2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} = \begin{bmatrix} 2x & 10 \\ 14 & 2y-6 \end{bmatrix}$$

Now add the second matrix:

$$\begin{bmatrix} 2x & 10 \\ 14 & 2y-6 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2x+3 & 6 \\ 15 & 2y-4 \end{bmatrix}$$

So the equation becomes

$$\begin{bmatrix} 2x+3 & 6 \\ 15 & 2y-4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$$

Equating corresponding elements:

$$2x + 3 = 7 \Rightarrow 2x = 4 \Rightarrow x = 2$$

$$2y - 4 = 14 \Rightarrow 2y = 18 \Rightarrow y = 9$$

Hence $$x = 2$$ and $$y = 9$$.

Matrix $$A$$ has order $$2 \times 2$$ and matrix $$B$$ has order $$2 \times 3$$. Since the number of columns of $$A$$ (which is 2) equals the number of rows of $$B$$ (which is 2), the product $$AB$$ is defined and its order is $$2 \times 3$$.

Each entry of $$AB$$ is found by multiplying a row of $$A$$ with a column of $$B$$ (the sum of products of corresponding terms).

$$AB = \begin{bmatrix} 6 & 9 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 2 & 6 & 0 \\ 7 & 9 & 8 \end{bmatrix}$$

$$= \begin{bmatrix} 6(2)+9(7) & 6(6)+9(9) & 6(0)+9(8) \\ 2(2)+3(7) & 2(6)+3(9) & 2(0)+3(8) \end{bmatrix}$$

$$= \begin{bmatrix} 12+63 & 36+81 & 0+72 \\ 4+21 & 12+27 & 0+24 \end{bmatrix} = \begin{bmatrix} 75 & 117 & 72 \\ 25 & 39 & 24 \end{bmatrix}$$

Here $$A$$ is of order $$2 \times 3$$ and $$B$$ is of order $$3 \times 2$$.

Computing $$AB$$: columns of $$A$$ (3) = rows of $$B$$ (3), so $$AB$$ is defined and has order $$2 \times 2$$.

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

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

$$= \begin{bmatrix} 2-8+6 & 3-10+3 \\ -8+8+10 & -12+10+5 \end{bmatrix} = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$$

Computing $$BA$$: columns of $$B$$ (2) = rows of $$A$$ (2), so $$BA$$ is defined and has order $$3 \times 3$$.

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

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

$$= \begin{bmatrix} 2-12 & -4+6 & 6+15 \\ 4-20 & -8+10 & 12+25 \\ 2-4 & -4+2 & 6+5 \end{bmatrix} = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}$$

Since $$AB$$ is a $$2 \times 2$$ matrix while $$BA$$ is a $$3 \times 3$$ matrix, they are not even of the same order. Hence $$AB \neq BA$$.

Both $$A$$ and $$B$$ are $$2 \times 2$$ matrices, so both products $$AB$$ and $$BA$$ are defined and are $$2 \times 2$$.

Computing $$AB$$:

$$AB = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1(0)+0(1) & 1(1)+0(0) \\ 0(0)+(-1)(1) & 0(1)+(-1)(0) \end{bmatrix}$$

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

Computing $$BA$$:

$$BA = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 0(1)+1(0) & 0(0)+1(-1) \\ 1(1)+0(0) & 1(0)+0(-1) \end{bmatrix}$$

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

Comparing the two results, the $$(1,2)$$ entry is $$1$$ in $$AB$$ but $$-1$$ in $$BA$$ (and similarly for the $$(2,1)$$ entry). Since corresponding elements are not all equal,

$$AB \neq BA$$

Both matrices are of order $$2 \times 2$$, so $$AB$$ is defined and is a $$2 \times 2$$ matrix.

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

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

$$= \begin{bmatrix} 0+0 & 0+0 \\ 0+0 & 0+0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

So $$AB = O$$, the zero matrix. This is an important observation: the product of two matrices can be the zero matrix even though neither $$A$$ nor $$B$$ is the zero matrix.

The orders are: $$A$$ is $$3 \times 3$$, $$B$$ is $$3 \times 2$$, $$C$$ is $$2 \times 4$$. All the products below are therefore defined.

Step 1: Compute $$BC$$ — order $$(3 \times 2)(2 \times 4) = 3 \times 4$$.

$$BC = \begin{bmatrix} 1 & 3 \\ 0 & 2 \\ -1 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 1+6 & 2+0 & 3-6 & -4+3 \\ 0+4 & 0+0 & 0-4 & 0+2 \\ -1+8 & -2+0 & -3-8 & 4+4 \end{bmatrix}$$

$$= \begin{bmatrix} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{bmatrix}$$

Step 2: Compute $$A(BC)$$ — order $$(3 \times 3)(3 \times 4) = 3 \times 4$$.

$$A(BC) = \begin{bmatrix} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{bmatrix}\begin{bmatrix} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 7 & -2 & -11 & 8 \end{bmatrix}$$

Row 1: $$[\,7+4-7,\ 2+0+2,\ -3-4+11,\ -1+2-8\,] = [4, 4, 4, -7]$$

Row 2: $$[\,14+0+21,\ 4+0-6,\ -6+0-33,\ -2+0+24\,] = [35, -2, -39, 22]$$

Row 3: $$[\,21-4+14,\ 6-0-4,\ -9+4-22,\ -3-2+16\,] = [31, 2, -27, 11]$$

$$A(BC) = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}$$

Step 3: Compute $$AB$$ — order $$(3 \times 3)(3 \times 2) = 3 \times 2$$.

$$AB = \begin{bmatrix} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 0 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 1+0+1 & 3+2-4 \\ 2+0-3 & 6+0+12 \\ 3+0-2 & 9-2+8 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{bmatrix}$$

Step 4: Compute $$(AB)C$$ — order $$(3 \times 2)(2 \times 4) = 3 \times 4$$.

$$(AB)C = \begin{bmatrix} 2 & 1 \\ -1 & 18 \\ 1 & 15 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{bmatrix}$$

Row 1: $$[\,2+2,\ 4+0,\ 6-2,\ -8+1\,] = [4, 4, 4, -7]$$

Row 2: $$[\,-1+36,\ -2+0,\ -3-36,\ 4+18\,] = [35, -2, -39, 22]$$

Row 3: $$[\,1+30,\ 2+0,\ 3-30,\ -4+15\,] = [31, 2, -27, 11]$$

$$(AB)C = \begin{bmatrix} 4 & 4 & 4 & -7 \\ 35 & -2 & -39 & 22 \\ 31 & 2 & -27 & 11 \end{bmatrix}$$

Comparing Step 2 and Step 4, the two matrices are identical. Hence $$(AB)C = A(BC)$$, which illustrates the associative property of matrix multiplication.

$$A$$ and $$B$$ are $$3 \times 3$$ matrices and $$C$$ is a $$3 \times 1$$ matrix, so every product below is defined and gives a $$3 \times 1$$ matrix.

Computing $$AC$$:

$$AC = \begin{bmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{bmatrix}\begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} = \begin{bmatrix} 0(2)+6(-2)+7(3) \\ -6(2)+0(-2)+8(3) \\ 7(2)+(-8)(-2)+0(3) \end{bmatrix}$$

$$= \begin{bmatrix} 0-12+21 \\ -12+0+24 \\ 14+16+0 \end{bmatrix} = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix}$$

Computing $$BC$$:

$$BC = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}\begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} = \begin{bmatrix} 0(2)+1(-2)+1(3) \\ 1(2)+0(-2)+2(3) \\ 1(2)+2(-2)+0(3) \end{bmatrix}$$

$$= \begin{bmatrix} 0-2+3 \\ 2+0+6 \\ 2-4+0 \end{bmatrix} = \begin{bmatrix} 1 \\ 8 \\ -2 \end{bmatrix}$$

Computing $$(A + B)C$$: first add $$A$$ and $$B$$:

$$A + B = \begin{bmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0 \end{bmatrix}$$

$$(A+B)C = \begin{bmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0 \end{bmatrix}\begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} = \begin{bmatrix} 0-14+24 \\ -10+0+30 \\ 16+12+0 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$$

Verification:

$$AC + BC = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix} + \begin{bmatrix} 1 \\ 8 \\ -2 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix}$$

This equals $$(A+B)C$$. Hence $$(A + B)C = AC + BC$$, illustrating the distributive property.

Step 1: Compute $$A^2 = A \cdot A$$.

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

Row 1: $$[\,1+6+12,\ 2-4+6,\ 3+2+3\,] = [19, 4, 8]$$

Row 2: $$[\,3-6+4,\ 6+4+2,\ 9-2+1\,] = [1, 12, 8]$$

Row 3: $$[\,4+6+4,\ 8-4+2,\ 12+2+1\,] = [14, 6, 15]$$

$$A^2 = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix}$$

Step 2: Compute $$A^3 = A^2 \cdot A$$.

$$A^3 = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}$$

Row 1: $$[\,19+12+32,\ 38-8+16,\ 57+4+8\,] = [63, 46, 69]$$

Row 2: $$[\,1+36+32,\ 2-24+16,\ 3+12+8\,] = [69, -6, 23]$$

Row 3: $$[\,14+18+60,\ 28-12+30,\ 42+6+15\,] = [92, 46, 63]$$

$$A^3 = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix}$$

Step 3: Compute $$23A$$ and $$40I$$.

$$23A = \begin{bmatrix} 23 & 46 & 69 \\ 69 & -46 & 23 \\ 92 & 46 & 23 \end{bmatrix}, \qquad 40I = \begin{bmatrix} 40 & 0 & 0 \\ 0 & 40 & 0 \\ 0 & 0 & 40 \end{bmatrix}$$

Step 4: Compute $$A^3 - 23A - 40I$$.

$$A^3 - 23A - 40I = \begin{bmatrix} 63-23-40 & 46-46-0 & 69-69-0 \\ 69-69-0 & -6+46-40 & 23-23-0 \\ 92-92-0 & 46-46-0 & 63-23-40 \end{bmatrix}$$

$$= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$$

Hence $$A^3 - 23A - 40I = O$$.

The amount spent in a city is (number of contacts of each type) $$\times$$ (cost per contact), summed over the three types. This is exactly a row of $$B$$ multiplied by the column matrix $$A$$. So the total amounts are given by the product $$BA$$.

Here $$B$$ has order $$2 \times 3$$ and $$A$$ has order $$3 \times 1$$, so $$BA$$ is defined and has order $$2 \times 1$$.

$$BA = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10{,}000 \end{bmatrix}\begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix}$$

$$= \begin{bmatrix} 1000(40)+500(100)+5000(50) \\ 3000(40)+1000(100)+10{,}000(50) \end{bmatrix}$$

$$= \begin{bmatrix} 40{,}000+50{,}000+2{,}50{,}000 \\ 1{,}20{,}000+1{,}00{,}000+5{,}00{,}000 \end{bmatrix} = \begin{bmatrix} 3{,}40{,}000 \\ 7{,}20{,}000 \end{bmatrix}$$

So the group spent $$3{,}40{,}000$$ paise in city X and $$7{,}20{,}000$$ paise in city Y. Since $$100$$ paise $$= \text{₹}\, 1$$:

Amount in X $$= \text{₹}\, 3400$$ and amount in Y $$= \text{₹}\, 7200$$.

Total amount spent $$= \text{₹}\, 3400 + \text{₹}\, 7200 = \text{₹}\, 10{,}600$$.

$$A$$ is of order $$3 \times 1$$ and $$B$$ is of order $$1 \times 3$$, so $$AB$$ is defined and has order $$3 \times 3$$.

Step 1: Compute $$AB$$.

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

$$= \begin{bmatrix} -2 & -6 & 12 \\ 4 & 12 & -24 \\ 5 & 15 & -30 \end{bmatrix}$$

Step 2: Take the transpose, $$(AB)'$$.

$$(AB)' = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$$

Step 3: Compute $$B'A'$$. We have $$B' = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix}$$ (order $$3 \times 1$$) and $$A' = \begin{bmatrix} -2 & 4 & 5 \end{bmatrix}$$ (order $$1 \times 3$$). So $$B'A'$$ has order $$3 \times 3$$.

$$B'A' = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix}\begin{bmatrix} -2 & 4 & 5 \end{bmatrix} = \begin{bmatrix} 1(-2) & 1(4) & 1(5) \\ 3(-2) & 3(4) & 3(5) \\ -6(-2) & -6(4) & -6(5) \end{bmatrix}$$

$$= \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$$

Step 2 and Step 3 give the same matrix, so $$(AB)' = B'A'$$ is verified.

Any square matrix $$B$$ can be written as $$B = \dfrac{1}{2}(B + B') + \dfrac{1}{2}(B - B')$$, where $$\dfrac{1}{2}(B + B')$$ is symmetric and $$\dfrac{1}{2}(B - B')$$ is skew symmetric.

Step 1: Write down $$B'$$.

$$B' = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$$

Step 2: Form the symmetric part $$P = \dfrac{1}{2}(B + B')$$.

$$B + B' = \begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix}$$

$$P = \frac{1}{2}(B + B') = \begin{bmatrix} 2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix}$$

Since $$P' = P$$, the matrix $$P$$ is symmetric.

Step 3: Form the skew symmetric part $$Q = \dfrac{1}{2}(B - B')$$.

$$B - B' = \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix}$$

$$Q = \frac{1}{2}(B - B') = \begin{bmatrix} 0 & -\frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix}$$

Since $$Q' = -Q$$, the matrix $$Q$$ is skew symmetric.

Step 4: Verify $$P + Q = B$$.

$$P + Q = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = B$$

Hence $$B$$ is expressed as the sum of the symmetric matrix $$P$$ and the skew symmetric matrix $$Q$$.

A matrix of order $$m \times n$$ has $$mn$$ elements. To find all possible orders we list every way of writing the number of elements as an ordered product $$m \times n$$ of natural numbers.

Case 1: 24 elements. We need $$m \times n = 24$$. The factor pairs of 24 are:

$$1 \times 24,\ 2 \times 12,\ 3 \times 8,\ 4 \times 6,\ 6 \times 4,\ 8 \times 3,\ 12 \times 2,\ 24 \times 1$$

So there are 8 possible orders: $$1 \times 24,\ 24 \times 1,\ 2 \times 12,\ 12 \times 2,\ 3 \times 8,\ 8 \times 3,\ 4 \times 6,\ 6 \times 4$$.

Case 2: 13 elements. We need $$m \times n = 13$$. Since 13 is a prime number, its only factorisations are $$1 \times 13$$ and $$13 \times 1$$.

So there are 2 possible orders: $$1 \times 13$$ and $$13 \times 1$$.

A matrix of order $$m \times n$$ has $$mn$$ elements. The possible orders correspond to all ordered factorisations of the number of elements.

Case 1: 18 elements. We need $$m \times n = 18$$. The factor pairs of 18 are:

$$1 \times 18,\ 2 \times 9,\ 3 \times 6,\ 6 \times 3,\ 9 \times 2,\ 18 \times 1$$

So there are 6 possible orders: $$1 \times 18,\ 18 \times 1,\ 2 \times 9,\ 9 \times 2,\ 3 \times 6,\ 6 \times 3$$.

Case 2: 5 elements. We need $$m \times n = 5$$. Since 5 is a prime number, its only factorisations are $$1 \times 5$$ and $$5 \times 1$$.

So there are 2 possible orders: $$1 \times 5$$ and $$5 \times 1$$.

Since the two matrices are equal, their corresponding elements are equal. This gives four equations:

$$a - b = -1 \quad \text{...(1)}$$

$$2a + c = 5 \quad \text{...(2)}$$

$$2a - b = 0 \quad \text{...(3)}$$

$$3c + d = 13 \quad \text{...(4)}$$

Solving for $$a$$ and $$b$$: Subtract equation (1) from equation (3):

$$(2a - b) - (a - b) = 0 - (-1) \Rightarrow a = 1$$

Substitute $$a = 1$$ in (1): $$1 - b = -1 \Rightarrow b = 2$$

Solving for $$c$$: Substitute $$a = 1$$ in (2): $$2(1) + c = 5 \Rightarrow c = 3$$

Solving for $$d$$: Substitute $$c = 3$$ in (4): $$3(3) + d = 13 \Rightarrow 9 + d = 13 \Rightarrow d = 4$$

Hence $$a = 1,\ b = 2,\ c = 3,\ d = 4$$.

A matrix is called a square matrix when its number of rows is equal to its number of columns.

For the matrix $$A = [a_{ij}]_{m \times n}$$, the number of rows is $$m$$ and the number of columns is $$n$$.

So $$A$$ is a square matrix precisely when $$m = n$$.

Hence the correct option is (C).

If the two matrices are equal, then all corresponding elements must be equal. Equating the entries:

Position $$(1,1)$$: $$3x + 7 = 0 \Rightarrow x = -\dfrac{7}{3}$$

Position $$(2,2)$$: $$2 - 3x = 4 \Rightarrow -3x = 2 \Rightarrow x = -\dfrac{2}{3}$$

From the first condition $$x = -\dfrac{7}{3}$$, but from the second condition $$x = -\dfrac{2}{3}$$. A single value of $$x$$ cannot equal both $$-\dfrac{7}{3}$$ and $$-\dfrac{2}{3}$$ at the same time.

Since the conditions on $$x$$ contradict each other, no values of $$x$$ and $$y$$ can make the matrices equal.

Hence the correct option is (B).

A matrix of order $$3 \times 3$$ has $$3 \times 3 = 9$$ entries.

Each of these 9 entries can be filled independently in 2 ways — it can be either $$0$$ or $$1$$.

By the multiplication principle of counting, the total number of such matrices is

$$\underbrace{2 \times 2 \times \cdots \times 2}_{9 \text{ times}} = 2^9 = 512$$

Hence the correct option is (D).

All three matrices are of order $$3 \times 3$$, so every sum and difference below is defined.

Computing $$A + B$$:

$$A + B = \begin{bmatrix} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix} = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix}$$

Computing $$B - C$$:

$$B - C = \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0+2 & 3-3 \end{bmatrix} = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}$$

Verification:

$$A + (B - C) = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$$

$$(A + B) - C = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$$

Both sides are equal, so $$A + (B - C) = (A + B) - C$$ is verified.

First multiply $$A$$ by the scalar 3. Each fraction with denominator 3 gets multiplied by 3, which cancels the denominator:

$$3A = \begin{bmatrix} 3 \cdot \frac{2}{3} & 3 \cdot 1 & 3 \cdot \frac{5}{3} \\ 3 \cdot \frac{1}{3} & 3 \cdot \frac{2}{3} & 3 \cdot \frac{4}{3} \\ 3 \cdot \frac{7}{3} & 3 \cdot 2 & 3 \cdot \frac{2}{3} \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$$

Next multiply $$B$$ by the scalar 5. Each fraction with denominator 5 gets multiplied by 5:

$$5B = \begin{bmatrix} 5 \cdot \frac{2}{5} & 5 \cdot \frac{3}{5} & 5 \cdot 1 \\ 5 \cdot \frac{1}{5} & 5 \cdot \frac{2}{5} & 5 \cdot \frac{4}{5} \\ 5 \cdot \frac{7}{5} & 5 \cdot \frac{6}{5} & 5 \cdot \frac{2}{5} \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$$

Now subtract:

$$3A - 5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

So $$3A - 5B = O$$, the zero matrix.

Carry out the two scalar multiplications first.

$$\cos\theta \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} = \begin{bmatrix} \cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix}$$

$$\sin\theta \begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix} = \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix}$$

Now add the two matrices element by element:

$$= \begin{bmatrix} \cos^2\theta+\sin^2\theta & \sin\theta\cos\theta-\sin\theta\cos\theta \\ -\sin\theta\cos\theta+\sin\theta\cos\theta & \cos^2\theta+\sin^2\theta \end{bmatrix}$$

Using the identity $$\sin^2\theta + \cos^2\theta = 1$$:

$$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

So the expression simplifies to the identity matrix $$I$$.

From $$2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$$, subtract $$Y$$ from both sides:

$$2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$$

$$2X = \begin{bmatrix} 1-3 & 0-2 \\ -3-1 & 2-4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}$$

Multiply both sides by the scalar $$\frac{1}{2}$$:

$$X = \frac{1}{2}\begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}$$

Carry out the scalar multiplication, then add:

$$2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}$$

$$\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix}$$

So the equation becomes

$$\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$$

Equating corresponding elements:

$$2 + y = 5 \Rightarrow y = 3$$

$$2x + 2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3$$

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

Perform the scalar multiplications on each side.

$$2\begin{bmatrix} x & z \\ y & t \end{bmatrix} = \begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix}, \qquad 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix}, \qquad 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$$

The equation becomes

$$\begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$$

$$\begin{bmatrix} 2x+3 & 2z-3 \\ 2y & 2t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$$

Equating corresponding elements:

$$2x + 3 = 9 \Rightarrow x = 3$$

$$2z - 3 = 15 \Rightarrow 2z = 18 \Rightarrow z = 9$$

$$2y = 12 \Rightarrow y = 6$$

$$2t + 6 = 18 \Rightarrow 2t = 12 \Rightarrow t = 6$$

Hence $$x = 3,\ y = 6,\ z = 9,\ t = 6$$.

Carry out the scalar multiplications and then add the two column matrices:

$$x\begin{bmatrix} 2 \\ 3 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 2x-y \\ 3x+y \end{bmatrix}$$

So $$\begin{bmatrix} 2x-y \\ 3x+y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$$. Equating corresponding elements:

$$2x - y = 10 \quad \text{...(1)}$$

$$3x + y = 5 \quad \text{...(2)}$$

Adding (1) and (2) eliminates $$y$$:

$$5x = 15 \Rightarrow x = 3$$

Substitute $$x = 3$$ in (1): $$2(3) - y = 10 \Rightarrow 6 - y = 10 \Rightarrow y = -4$$

Hence $$x = 3$$ and $$y = -4$$.

The left side, after scalar multiplication, is $$\begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}$$.

The right side, after adding the two matrices, is $$\begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix}$$.

So $$\begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix} = \begin{bmatrix} x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix}$$.

Equating corresponding elements:

Position $$(1,1)$$: $$3x = x + 4 \Rightarrow 2x = 4 \Rightarrow x = 2$$

Position $$(2,2)$$: $$3w = 2w + 3 \Rightarrow w = 3$$

Position $$(1,2)$$: $$3y = x + y + 6 \Rightarrow 2y = x + 6 = 2 + 6 = 8 \Rightarrow y = 4$$

Position $$(2,1)$$: $$3z = z + w - 1 \Rightarrow 2z = w - 1 = 3 - 1 = 2 \Rightarrow z = 1$$

Hence $$x = 2,\ y = 4,\ z = 1,\ w = 3$$.

We have $$F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ and $$F(y) = \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$$.

Both are $$3 \times 3$$, so we multiply them:

$$F(x)F(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Entry $$(1,1)$$: $$\cos x\cos y + (-\sin x)\sin y + 0 = \cos x\cos y - \sin x\sin y = \cos(x+y)$$

Entry $$(1,2)$$: $$\cos x(-\sin y) + (-\sin x)\cos y + 0 = -(\sin x\cos y + \cos x\sin y) = -\sin(x+y)$$

Entry $$(1,3)$$: $$\cos x(0) + (-\sin x)(0) + 0(1) = 0$$

Entry $$(2,1)$$: $$\sin x\cos y + \cos x\sin y + 0 = \sin(x+y)$$

Entry $$(2,2)$$: $$\sin x(-\sin y) + \cos x\cos y + 0 = \cos x\cos y - \sin x\sin y = \cos(x+y)$$

Entry $$(2,3)$$: $$0$$. Entries $$(3,1), (3,2)$$: $$0$$. Entry $$(3,3)$$: $$0+0+1 = 1$$.

Therefore

$$F(x)F(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(x+y)$$

Hence $$F(x)F(y) = F(x+y)$$ is proved.

Step 1: Compute $$A^2 = A \cdot A$$.

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

Row 1 $$[2,0,1]$$: $$[\,4+0+1,\ 0+0-1,\ 2+0+0\,] = [5, -1, 2]$$

Row 2 $$[2,1,3]$$: $$[\,4+2+3,\ 0+1-3,\ 2+3+0\,] = [9, -2, 5]$$

Row 3 $$[1,-1,0]$$: $$[\,2-2+0,\ 0-1+0,\ 1-3+0\,] = [0, -1, -2]$$

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

Step 2: Compute $$5A$$ and $$6I$$.

$$5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}, \qquad 6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$$

Step 3: Compute $$A^2 - 5A + 6I$$.

$$A^2 - 5A + 6I = \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+6 \end{bmatrix}$$

$$= \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$$

Step 1: Compute $$A^2 = A \cdot A$$.

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

Row 1 $$[1,0,2]$$: $$[\,1+0+4,\ 0+0+0,\ 2+0+6\,] = [5, 0, 8]$$

Row 2 $$[0,2,1]$$: $$[\,0+0+2,\ 0+4+0,\ 0+2+3\,] = [2, 4, 5]$$

Row 3 $$[2,0,3]$$: $$[\,2+0+6,\ 0+0+0,\ 4+0+9\,] = [8, 0, 13]$$

$$A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}$$

Step 2: Compute $$A^3 = A^2 \cdot A$$.

Row 1 $$[5,0,8]$$: $$[\,5+0+16,\ 0+0+0,\ 10+0+24\,] = [21, 0, 34]$$

Row 2 $$[2,4,5]$$: $$[\,2+0+10,\ 0+8+0,\ 4+4+15\,] = [12, 8, 23]$$

Row 3 $$[8,0,13]$$: $$[\,8+0+26,\ 0+0+0,\ 16+0+39\,] = [34, 0, 55]$$

$$A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}$$

Step 3: Compute $$6A^2$$, $$7A$$ and $$2I$$.

$$6A^2 = \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix},\quad 7A = \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix},\quad 2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$

Step 4: Combine.

$$A^3 - 6A^2 + 7A + 2I = \begin{bmatrix} 21-30+7+2 & 0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0 & 55-78+21+2 \end{bmatrix}$$

$$= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$$

Hence $$A^3 - 6A^2 + 7A + 2I = O$$ is proved.

Step 1: Compute $$A^2$$.

$$A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 3(3)+(-2)(4) & 3(-2)+(-2)(-2) \\ 4(3)+(-2)(4) & 4(-2)+(-2)(-2) \end{bmatrix}$$

$$= \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$$

Step 2: Express the right side $$kA - 2I$$.

$$kA - 2I = k\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}$$

Step 3: Equate $$A^2 = kA - 2I$$.

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

Comparing the $$(1,1)$$ entries: $$3k - 2 = 1 \Rightarrow 3k = 3 \Rightarrow k = 1$$.

Check with the other entries: $$-2k = -2 \Rightarrow k = 1$$; $$4k = 4 \Rightarrow k = 1$$; $$-2k - 2 = -4 \Rightarrow k = 1$$. All are consistent.

Hence $$k = 1$$.

For convenience write $$t = \tan\dfrac{\alpha}{2}$$. Then $$A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}$$.

Left side:

$$I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$$

Right side: first $$I - A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$$.

We use the half-angle formulas $$\cos\alpha = \dfrac{1-t^2}{1+t^2}$$ and $$\sin\alpha = \dfrac{2t}{1+t^2}$$.

$$(I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$$

$$= \begin{bmatrix} \cos\alpha + t\sin\alpha & -\sin\alpha + t\cos\alpha \\ -t\cos\alpha + \sin\alpha & t\sin\alpha + \cos\alpha \end{bmatrix}$$

Now evaluate each entry. Entry $$(1,1)$$:

$$\cos\alpha + t\sin\alpha = \frac{1-t^2}{1+t^2} + t\cdot\frac{2t}{1+t^2} = \frac{1-t^2+2t^2}{1+t^2} = \frac{1+t^2}{1+t^2} = 1$$

Entry $$(1,2)$$:

$$-\sin\alpha + t\cos\alpha = -\frac{2t}{1+t^2} + t\cdot\frac{1-t^2}{1+t^2} = \frac{-2t + t - t^3}{1+t^2} = \frac{-t-t^3}{1+t^2} = \frac{-t(1+t^2)}{1+t^2} = -t$$

Entry $$(2,1)$$:

$$-t\cos\alpha + \sin\alpha = -t\cdot\frac{1-t^2}{1+t^2} + \frac{2t}{1+t^2} = \frac{-t+t^3+2t}{1+t^2} = \frac{t+t^3}{1+t^2} = \frac{t(1+t^2)}{1+t^2} = t$$

Entry $$(2,2)$$: $$t\sin\alpha + \cos\alpha = 1$$ (same computation as entry $$(1,1)$$).

Therefore

$$(I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = I + A$$

Hence $$I + A = (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$$ is proved.

First convert dozens to numbers of books (1 dozen = 12):

Chemistry: $$10 \times 12 = 120$$ books; Physics: $$8 \times 12 = 96$$ books; Economics: $$10 \times 12 = 120$$ books.

Write the number of books as a row matrix and the selling prices as a column matrix. The total amount received is their product:

$$\text{Total} = \begin{bmatrix} 120 & 96 & 120 \end{bmatrix}\begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}$$

$$= \begin{bmatrix} 120(80) + 96(60) + 120(40) \end{bmatrix}$$

$$= \begin{bmatrix} 9600 + 5760 + 4800 \end{bmatrix} = \begin{bmatrix} 20{,}160 \end{bmatrix}$$

Hence the bookshop will receive a total of $$\text{₹}\, 20{,}160$$.

The orders are: $$P$$ is $$p \times k$$, $$Y$$ is $$3 \times k$$, $$W$$ is $$n \times 3$$.

Condition for $$PY$$ to be defined: the number of columns of $$P$$ must equal the number of rows of $$Y$$. Columns of $$P$$ $$= k$$, rows of $$Y$$ $$= 3$$. So we need $$k = 3$$. Then $$PY$$ has order $$p \times k = p \times 3$$.

Condition for $$WY$$ to be defined: columns of $$W$$ $$= 3$$ must equal rows of $$Y$$ $$= 3$$. This is automatically satisfied. Then $$WY$$ has order $$n \times k = n \times 3$$ (using $$k = 3$$).

Condition for the sum $$PY + WY$$: two matrices can be added only if they have the same order. Here $$PY$$ is $$p \times 3$$ and $$WY$$ is $$n \times 3$$, so we need $$p = n$$.

Hence the required restrictions are $$k = 3$$ and $$p = n$$, which is option (A).

The order of $$X$$ is $$2 \times n$$ and the order of $$Z$$ is $$2 \times p$$.

For the difference $$7X - 5Z$$ to be defined, $$X$$ and $$Z$$ must have the same order. We are given $$n = p$$, so $$Z$$ has order $$2 \times p = 2 \times n$$, the same as $$X$$.

Multiplying a matrix by a scalar does not change its order, so $$7X$$ and $$5Z$$ are both of order $$2 \times n$$.

Therefore $$7X - 5Z$$ is of order $$2 \times n$$, which is option (B).

We have to evaluate the transpose $$(A + 2B)'$$.
Use the property $$ (P + Q)' = P' + Q' $$ and $$ (kP)' = kP' $$ for any scalar $$k$$.

1. We are already given the transpose of $$A$$:

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

2. Compute the transpose of $$B$$:

   $$B = \begin{bmatrix}-1 & 0\\ 1 & 2\end{bmatrix} \;\;\Longrightarrow\;\; B' = \begin{bmatrix}-1 & 1\\ 0 & 2\end{bmatrix}$$

3. Multiply this transpose by 2:

   $$2B' = 2 \times \begin{bmatrix}-1 & 1\\ 0 & 2\end{bmatrix} = \begin{bmatrix}-2 & 2\\ 0 & 4\end{bmatrix}$$

4. Add $$A'$$ and $$2B'$$:

   \[\begin{aligned} A' + 2B' &= \begin{bmatrix}-2 & 3\\ 1 & 2\end{bmatrix} + \begin{bmatrix}-2 & 2\\ 0 & 4\end{bmatrix} \\[4pt] &= \begin{bmatrix}-2 + (-2) & 3 + 2\\ 1 + 0 & 2 + 4\end{bmatrix} \\[4pt] &= \begin{bmatrix}-4 & 5\\ 1 & 6\end{bmatrix} \end{aligned}\]

5. Hence

   $$ (A + 2B)' = \begin{bmatrix}-4 & 5\\ 1 & 6\end{bmatrix}. $$

The given matrix is

$$A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$$

Step 1 : Compute the transpose $$A'$$

Interchanging rows and columns of $$A$$ gives

$$A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}$$

Step 2 : Evaluate $$A + A'$$

Therefore

$$\tfrac12 (A + A') = \tfrac12 \times 0_{3\times3} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

Step 3 : Evaluate $$A - A'$$

Hence

$$\tfrac12 (A - A') = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$$

The matrix $$A$$ thus splits into its symmetric part (which is the zero matrix) and its skew-symmetric part (which is $$A$$ itself):

$$A = \underbrace{\tfrac12 (A + A')}_{\text{symmetric}} + \underbrace{\tfrac12 (A - A')}_{\text{skew-symmetric}}.$$

Let $$A$$ and $$B$$ be symmetric matrices of the same order, so by definition

$$A^T = A \\[4pt] B^T = B.$$

We need to investigate the nature of the matrix $$AB - BA.$$

Step 1: Compute its transpose.

$$\begin{aligned}(AB - BA)^T &= (AB)^T - (BA)^T \\[4pt] &= B^T A^T - A^T B^T \\[4pt] &= BA - AB.\end{aligned}$$

Step 2: Relate the transpose to the original matrix.

Notice that

$$BA - AB = - (AB - BA).$$

Therefore,

$$(AB - BA)^T = - (AB - BA).$$

This is precisely the condition for a skew symmetric matrix: a matrix $$M$$ is skew symmetric if and only if $$M^T = -M.$$

Conclusion. Since $$AB - BA$$ satisfies this condition, it is a skew symmetric matrix.

Hence, the correct option is (A).

The given matrix is

$$A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$$

Here $$A'$$ denotes the transpose of $$A$$, obtained by interchanging its rows and columns:

$$A' = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$$

Adding the two matrices element by element:

$$A + A' = \begin{bmatrix} \cos\alpha+\cos\alpha & -\sin\alpha+\sin\alpha \\ \sin\alpha-\sin\alpha & \cos\alpha+\cos\alpha \end{bmatrix} = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix}$$

The condition given is $$A + A' = I$$, where $$I$$ is the $$2 \times 2$$ identity matrix. So

$$\begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Equating the corresponding elements gives

$$2\cos\alpha = 1 \Rightarrow \cos\alpha = \frac{1}{2}$$

The value of $$\alpha$$ that satisfies $$\cos\alpha = \frac{1}{2}$$ is

$$\alpha = \frac{\pi}{3}$$

Hence the correct option is (B).

Let $$A$$ and $$B$$ be two square matrices of the same order.

Definition of inverse: A matrix $$B$$ is called the inverse of $$A$$ if and only if both of the following products exist and are equal to the identity matrix $$I$$ of the same order:

• $$AB = I$$
• $$BA = I$$

The word “inverse” in matrix algebra therefore simultaneously requires left-multiplication and right-multiplication by the same matrix to produce the identity.

Now compare this definition with the alternatives given in the question:

Hence matrices $$A$$ and $$B$$ are inverses of each other only when option (D) holds.

Given: $$A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$.

We must prove that for every $$n\in\mathbb{N}$$,

$$A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}$$.

Method – Mathematical Induction

Base Case ($$n = 1$$):
$$A^1 = A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix},$$ which matches the required form with $$n=1$$.

Inductive Hypothesis:
Assume that for some arbitrary $$k\in\mathbb{N}$$,

$$A^{k}=\begin{bmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{bmatrix}.$$

Inductive Step:
Show the result for $$n=k+1$$.

Because $$A^{k+1}=A^{k}A$$, we have

$$A^{k+1}=\begin{bmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{bmatrix}\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}.$$

Multiply the matrices entry-wise:

Therefore

$$A^{k+1}=\begin{bmatrix} \cos (k+1)\theta & \sin (k+1)\theta \\ -\sin (k+1)\theta & \cos (k+1)\theta \end{bmatrix},$$

which is precisely the required expression for $$n=k+1$$.

Since the base case is true and the inductive step holds, by the Principle of Mathematical Induction the statement is true for all $$n\in\mathbb{N}$$.

Hence, $$A^{n}=\begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}\,.$$

Given : $$A$$ and $$B$$ are symmetric matrices of the same order.
To prove : The product $$AB$$ is symmetric  iff  $$AB = BA$$ (that is, the two matrices commute).

Recall that for any matrix $$C$$, the product rule for transposes gives
$$\left(C_1 C_2\right)^T = C_2^T C_1^T$$.
Also, a matrix $$C$$ is said to be symmetric when $$C^T = C$$.

Part I : If $$AB$$ is symmetric, then $$AB = BA$$.

1. Because $$AB$$ is assumed symmetric, we have $$\left(AB\right)^T = AB.$$
2. Using the product–transpose rule, $$\left(AB\right)^T = B^T A^T.$$
3. Since $$A$$ and $$B$$ are individually symmetric, $$A^T = A\,,\; B^T = B$$. Hence $$\left(AB\right)^T = B A.$$
4. Combine steps 1 – 3: $$BA = \left(AB\right)^T = AB.$$

Thus $$AB = BA$$; the two matrices commute.

Part II : If $$AB = BA$$, then $$AB$$ is symmetric.

1. Assume $$AB = BA$$.
2. Find the transpose of the product: $$\left(AB\right)^T = B^T A^T$$ (product–transpose rule).
3. Again using symmetry of the factors, $$B^T = B\,,\; A^T = A$$, so $$\left(AB\right)^T = BA.$$
4. But by hypothesis $$BA = AB$$, hence $$\left(AB\right)^T = AB.$$

Therefore $$AB$$ satisfies $$\left(AB\right)^T = AB$$ and is symmetric.

Conclusion : Both implications are verified, so $$AB$$ is symmetric if and only if $$AB = BA$$ when $$A$$ and $$B$$ are symmetric matrices of the same order.

Let

$$A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix},\; B = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix},\; C = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}.$$

We require a matrix $$D$$ satisfying $$CD - AB = O_{2 \times 2}.$$ Since $$O_{2 \times 2}$$ is the null matrix, this is equivalent to

$$CD = AB.$$ If $$C$$ is non–singular we may premultiply both sides by $$C^{-1}$$ to get $$D = C^{-1}AB.$$ Hence we first compute $$AB$$ and $$C^{-1}.$$

1. Product $$AB$$

$$AB = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} = \begin{bmatrix} 2 \times 5 + (-1) \times 7 & 2 \times 2 + (-1) \times 4 \\ 3 \times 5 + 4 \times 7 & 3 \times 2 + 4 \times 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix}. $$

2. Inverse of $$C$$

The determinant of $$C$$ is

$$|C| = 2 \times 8 - 5 \times 3 = 16 - 15 = 1 \neq 0,$$ so $$C$$ is invertible.

For a $$2 \times 2$$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ the inverse is $$\dfrac1{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$$ Hence

$$C^{-1} = \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix}.$$ (The factor $$1/|C| = 1$$ need not be written separately.)

3. Required matrix $$D$$

$$D = C^{-1}AB = \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix}.$$

Multiply:

$$\begin{aligned} D_{11} &= 8 \times 3 + (-5) \times 43 = 24 - 215 = -191,\\ D_{12} &= 8 \times 0 + (-5) \times 22 = -110,\\ D_{21} &= (-3) \times 3 + 2 \times 43 = -9 + 86 = 77,\\ D_{22} &= (-3) \times 0 + 2 \times 22 = 44. \end{aligned}$$

Thus

$$D = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}.$$

4. Verification (optional)

Compute $$CD$$:

$$CD = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}\begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix} = \begin{bmatrix} 2(-191)+5(77) & 2(-110)+5(44) \\ 3(-191)+8(77) & 3(-110)+8(44) \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} = AB.$$

Therefore $$CD - AB = O_{2 \times 2},$$ confirming our answer.

Let $$A$$ and $$B$$ be two square matrices of the same order such that each of them is symmetric; that is,

$$A' = A \;\;\text{and}\;\; B' = B.$$

We have to show that the matrix $$AB - BA$$ is skew-symmetric, i.e.

$$ (AB - BA)' = - (AB - BA). $$

Step 1 · Take the transpose of the required expression.

Compute the transpose of $$AB - BA$$:

$$ (AB - BA)' = (AB)' - (BA)'. $$

Step 2 · Use the rule for the transpose of a product.

For any conformable matrices $$X, Y$$ we have $$(XY)' = Y'X'$$. Applying this rule to both terms:

$$ (AB)' = B'A',\qquad (BA)' = A'B'. $$

Therefore

$$ (AB - BA)' = B'A' - A'B'. $$

Step 3 · Replace $$A'$$ and $$B'$$ by $$A$$ and $$B$$ (since they are symmetric).

Because $$A' = A$$ and $$B' = B$$, the above expression becomes

$$ (AB - BA)' = BA - AB. $$

Step 4 · Factor a minus sign.

$$ BA - AB = - (AB - BA). $$

Step 5 · Conclude.

We have shown that

$$ (AB - BA)' = - (AB - BA), $$

which is precisely the defining property of a skew-symmetric matrix. Hence $$AB - BA$$ is skew symmetric.

Hence proved.

Let $$A$$ and $$B$$ be square matrices of the same order n. Throughout the proof the symbol $$'$$ denotes the transpose.

Consider the matrix $$B'AB$$. To examine its symmetry we calculate its transpose.

Using the rule $$(XYZ)' = Z'Y'X'$$ for any conformable matrices $$X ,Y ,Z$$ we obtain

$$ (B'AB)' = B'A'B. $$

(The steps are: $$ (B'AB)' = (B'\,A)B)' = B'(B'A)' = B'\,A'\,(B')' = B'A'B$$.)

Case I : $A$ is symmetric

If $$A' = A$$, substitute this in the above result:

$$ (B'AB)' = B'AB. $$

Hence $$B'AB$$ equals its transpose and is therefore symmetric.

Case II : $A$ is skew–symmetric

If $$A' = -A$$, then

$$ (B'AB)' = B'(-A)B = -B'AB. $$

Thus the transpose of $$B'AB$$ is the negative of the matrix itself, so $$B'AB$$ is skew–symmetric.

Hence $$B'AB$$ is symmetric when $$A$$ is symmetric and skew–symmetric when $$A$$ is skew–symmetric, as required.

Given matrix

$$A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$$ and the condition $$A'A = I_3$$, where $$A'$$ is the transpose of $$A$$.

1. Transpose of $$A$$

$$A' = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}$$

2. Product $$A'A$$

Compute each entry by the rule “row of $$A'$$ × column of $$A$$”.

3. Solving the diagonal equations

• From $$2x^2 = 1\;\Rightarrow\;x^2 = \dfrac12 \;\Rightarrow\; x = \pm \dfrac{1}{\sqrt2}$$
• From $$6y^2 = 1\;\Rightarrow\;y^2 = \dfrac16 \;\Rightarrow\; y = \pm \dfrac{1}{\sqrt6}$$
• From $$3z^2 = 1\;\Rightarrow\;z^2 = \dfrac13 \;\Rightarrow\; z = \pm \dfrac{1}{\sqrt3}$$

4. Verification of off-diagonal entries

All off-diagonal elements have already been found to be $$0$$ for every choice of signs, so no further restriction is imposed on the signs of $$x,y,z$$.

Hence, any triple

$$\bigl(x, y, z\bigr)=\left(\pm\dfrac{1}{\sqrt2},\;\pm\dfrac{1}{\sqrt6},\;\pm\dfrac{1}{\sqrt3}\right)$$

satisfies $$A'A = I_3$$. There are $$2^3 = 8$$ such combinations.

First write the three factors separately:

Row vector  $$A=[1\;2\;1]$$,  matrix  $$B=\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}$$,  column vector  $$C=\begin{bmatrix}0\\2\\x\end{bmatrix}$$.

The required product is $$A\,B\,C$$. Because matrix multiplication is associative, compute $$B\,C$$ first (order: $3\times3$ with $3\times1$):

\[\begin{aligned} B\,C&=\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix} =\begin{bmatrix} 1\cdot0+2\cdot2+0\cdot x\\ 2\cdot0+0\cdot2+1\cdot x\\ 1\cdot0+0\cdot2+2\cdot x \end{bmatrix} =\begin{bmatrix}4\\x\\2x\end{bmatrix}.\end{aligned}\]

Now multiply the row vector $$A$$ by this $3\times1$ column vector (dot product):

\[\begin{aligned} A\,(B\,C)&=[1\;2\;1]\begin{bmatrix}4\\x\\2x\end{bmatrix} =1\cdot4+2\cdot x+1\cdot2x =4+2x+2x =4+4x. \end{aligned}\]

The question demands this final result to be zero: $$4+4x=0$$.

Solve for $$x$$:

$$4x=-4\;\Rightarrow\;x=-1.$$

Hence the given product equals the zero element (denoted by $$O$$) only when $$x=-1$$.

We are given the matrix

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

and we have to verify that $$A^2 - 5A + 7I = O$$, where $$I$$ is the identity matrix and $$O$$ is the null (zero) matrix of the same order (here, order 2).

1. Compute $$A^2$$

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

Multiply the rows of the first matrix with the columns of the second:

• First row, first column: $$3\times3 + 1\times(-1) = 9 - 1 = 8$$
• First row, second column: $$3\times1 + 1\times2 = 3 + 2 = 5$$
• Second row, first column: $$(-1)\times3 + 2\times(-1) = -3 - 2 = -5$$
• Second row, second column: $$(-1)\times1 + 2\times2 = -1 + 4 = 3$$

Thus

$$A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$

2. Compute $$5A$$

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

3. Compute $$7I$$

The identity matrix of order 2 is $$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$, so

$$7I = 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$

4. Evaluate $$A^2 - 5A + 7I$$

First, subtract $$5A$$ from $$A^2$$:

$$A^2 - 5A = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \, - \, \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} = \begin{bmatrix} 8-15 & 5-5 \\ -5-(-5) & 3-10 \end{bmatrix} = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}$$

Now, add $$7I$$ to the result:

$$\begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} -7+7 & 0+0 \\ 0+0 & -7+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$$

5. Conclusion

We have shown that $$A^2 - 5A + 7I = O$$. Hence the required relation is verified.

Let

$$A = \begin{bmatrix} x & -5 & -1 \end{bmatrix},\; B = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix},\; C = \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix}.$$

The given expression is $$A B C = O,$$ i.e. the scalar (a $1\times1$ matrix) obtained from this product must be zero.

Step 1: Multiply $B$ and $C$.

$$B C = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = \begin{bmatrix} 1\cdot x + 0\cdot4 + 2\cdot1 \\ 0\cdot x + 2\cdot4 + 1\cdot1 \\ 2\cdot x + 0\cdot4 + 3\cdot1 \end{bmatrix} = \begin{bmatrix} x + 2 \\ 9 \\ 2x + 3 \end{bmatrix}.$$

Step 2: Multiply $A$ with the result.

$$A\,(B C) = \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} x + 2 \\ 9 \\ 2x + 3 \end{bmatrix}$$

$$= x(x+2) + (-5)(9) + (-1)(2x+3).$$

Simplify term by term:

$$x(x+2) = x^2 + 2x,$$

$$(-5)(9) = -45,$$

$$(-1)(2x+3) = -2x - 3.$$

Adding these:

$$A\,(B C) = (x^2 + 2x) + (-45) + (-2x - 3) = x^2 - 48.$$

Step 3: Set the result equal to 0.

Because the product equals the zero scalar,

$$x^2 - 48 = 0.$$

Step 4: Solve for $x$.

$$x^2 = 48 \;\;\Rightarrow\;\; x = \pm\sqrt{48} = \pm 4\sqrt{3}.$$

Hence the required values of $x$ are $$x = 4\sqrt{3} \text{ or } x = -4\sqrt{3}.$$

Let the required matrix be denoted by $$X$$ and let

$$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}, \qquad B = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}.$$

We are asked to find $$X$$ such that $$X A = B.$$

1. Decide the order of $$X$$

The matrix $$A$$ has order $$2 \times 3$$ (2 rows, 3 columns).

• For the product $$X A$$ to be defined, the number of columns of $$X$$ must equal the number of rows of $$A$$, i.e. 2.
• The result $$B$$ has order $$2 \times 3$$, so the product must also have 2 rows. Hence $$X$$ must have 2 rows.

Therefore $$X$$ is a $$2 \times 2$$ matrix. Write

$$X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$

2. Multiply $$X$$ and $$A$$ symbolically

Compute $$X A$$:

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a\cdot1 + b\cdot4 & a\cdot2 + b\cdot5 & a\cdot3 + b\cdot6 \\ c\cdot1 + d\cdot4 & c\cdot2 + d\cdot5 & c\cdot3 + d\cdot6 \end{bmatrix}.$$

3. Equate with $$B$$ and set up equations

Since $$X A = B$$, match corresponding entries:

4. Solve for $$a$$ and $$b$$

From $$a + 4b = -7 \;\;(i)$$ and $$2a + 5b = -8 \;\;(ii)$$:

Multiply (i) by 2: $$2a + 8b = -14$$.
Subtract from (ii): $$(2a + 5b) - (2a + 8b) = -8 - (-14)$$ gives $$-3b = 6$$, so $$b = -2$$.

Substitute in (i): $$a + 4(-2) = -7 \Rightarrow a - 8 = -7 \Rightarrow a = 1.$$

Check with the third equation: $$3a + 6b = 3(1) + 6(-2) = 3 - 12 = -9,$$ which matches, so $$a = 1,\; b = -2$$ are correct.

5. Solve for $$c$$ and $$d$$

From $$c + 4d = 2 \;\;(iv)$$ and $$2c + 5d = 4 \;\;(v)$$:

Multiply (iv) by 2: $$2c + 8d = 4$$.
Subtract from (v): $$(2c + 5d) - (2c + 8d) = 4 - 4$$ gives $$-3d = 0$$, so $$d = 0$$.

Substitute in (iv): $$c + 4(0) = 2 \Rightarrow c = 2.$$

Check with the sixth equation: $$3c + 6d = 3(2) + 6(0) = 6,$$ matching the required value.

6. Write the matrix $$X$$

$$X = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}.$$

Thus the matrix that satisfies $$X A = B$$ is $$\begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$$.

Given matrix
$$A = \begin{bmatrix}\alpha & \beta\\ \gamma & -\alpha\end{bmatrix}$$
and it is specified that $$A^2 = I_2$$, the 2 × 2 identity matrix.

Step 1: Compute $$A^2$$.

Write the product explicitly:

$$A^2 = A\,A = \begin{bmatrix}\alpha & \beta\\ \gamma & -\alpha\end{bmatrix}\begin{bmatrix}\alpha & \beta\\ \gamma & -\alpha\end{bmatrix}.$$

Carry out entry-wise multiplication.

• First row, first column:
  $$a_{11} = \alpha\cdot\alpha + \beta\cdot\gamma = \alpha^2 + \beta\gamma.$$
• First row, second column:
  $$a_{12} = \alpha\cdot\beta + \beta\,(-\alpha) = \alpha\beta - \alpha\beta = 0.$$
• Second row, first column:
  $$a_{21} = \gamma\cdot\alpha + (-\alpha)\cdot\gamma = \alpha\gamma - \alpha\gamma = 0.$$
• Second row, second column:
  $$a_{22} = \gamma\cdot\beta + (-\alpha)(-\alpha) = \gamma\beta + \alpha^2 = \alpha^2 + \beta\gamma.$$

Thus

$$A^2 = \begin{bmatrix}\alpha^2 + \beta\gamma & 0\\ 0 & \alpha^2 + \beta\gamma\end{bmatrix}.$$

Step 2: Impose the condition $$A^2 = I_2.$$ That means

$$\begin{bmatrix}\alpha^2 + \beta\gamma & 0\\ 0 & \alpha^2 + \beta\gamma\end{bmatrix} = \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}.$$

Equating corresponding entries gives one equation:

$$\alpha^2 + \beta\gamma = 1.$$

Step 3: Rearrange the relation.

Bring all terms to one side:

$$1 - \alpha^2 - \beta\gamma = 0.$$

This matches option (C).

Hence, the correct statement is $$\boxed{1 - \alpha^2 - \beta\gamma = 0}.$$

Recall the definitions for a matrix $$A$$:

• $$A$$ is symmetric if $$A' = A$$.
• $$A$$ is skew symmetric if $$A' = -A$$.

Suppose $$A$$ is both symmetric and skew symmetric. Then both conditions hold at the same time:

$$A' = A \quad \text{and} \quad A' = -A$$

Since both expressions are equal to $$A'$$, they must be equal to each other:

$$A = -A$$

Adding $$A$$ to both sides:

$$A + A = O \Rightarrow 2A = O$$

where $$O$$ is the zero matrix. Dividing both sides by the scalar $$2$$:

$$A = O$$

So every entry of $$A$$ is $$0$$, that is, $$A$$ is the zero matrix. Hence the correct option is (B).

Let $$I$$ denote the identity matrix of the same order as $$A$$.

Because $$A^2 = A$$ (that is, $$A$$ is idempotent), expand $$(I + A)^3$$ using the binomial theorem; $$I$$ commutes with every matrix, so ordinary binomial coefficients apply:

$$ (I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3 $$

Simplify each term:

• $$I^3 = I$$
• $$I^2A = IA = A$$
• $$A^2 = A$$ (given)
• $$A^3 = A^2\!A = A\!A = A^2 = A$$

Substituting these results,

$$ (I + A)^3 = I + 3A + 3A + A = I + 7A. $$

Now evaluate the required expression:

$$ (I + A)^3 - 7A = (I + 7A) - 7A = I. $$

Thus the value is the identity matrix, corresponding to option (C).