Determinant of 3x3 Matrix

For the system to have no solution, we need the determinant of the coefficient matrix to be zero and the system to be inconsistent.

The determinant of the coefficient matrix is given by$$D = \begin{vmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix}$$

Expanding, we get $$D = 3(\alpha + 2) - 1(2 + 1) + 4(4 - \alpha) = 3\alpha + 6 - 3 + 16 - 4\alpha = -\alpha + 19$$.

For no solution: $$D = 0 \Rightarrow \alpha = 19$$.

With $$\alpha = 19$$, equation 2 becomes: $$2x + 19y - z = -3$$.

Subtracting 3 times equation 3 from equation 1 gives $$0 - 5y + z = -9$$, so $$-5y + z = -9$$.

Subtracting 2 times equation 3 from equation 2 gives $$0 + 15y - 3z = -11$$, so $$15y - 3z = -11$$.

From these, $$3(-5y + z) = -27 \Rightarrow -15y + 3z = -27$$.

Adding this to $$15y - 3z = -11$$ gives $$0 = -38 \neq 0$$, so the system is inconsistent. ✓

The answer is Option 1: 19.

We seek the values of n—obtained by rolling a fair die, so that n ∈ {1,2,3,4,5,6}—for which the system of equations

$$x - ny + z = 6,$$ $$x + (n-2)\,y + (n+1)\,z = 8,$$ $$(n-1)\,y + z = 1$$

has a unique solution. A linear system is uniquely solvable exactly when the determinant of its coefficient matrix is nonzero. The coefficient matrix here is

$$D = \begin{vmatrix} 1 & -n & 1 \\ 1 & n-2 & n+1 \\ 0 & n-1 & 1 \end{vmatrix}.$$

Expanding this determinant along the first row gives

$$D = 1\cdot\begin{vmatrix}n-2 & n+1 \\ n-1 & 1\end{vmatrix} \;-\;(-n)\cdot\begin{vmatrix}1 & n+1 \\ 0 & 1\end{vmatrix} \;+\;1\cdot\begin{vmatrix}1 & n-2 \\ 0 & n-1\end{vmatrix}.$$

Each of the 2×2 determinants evaluates as follows:
First minor: $$(n-2)\cdot1 - (n+1)(n-1) = n - 2 - (n^2 - 1) = -n^2 + n - 1;$$ Second minor: $$1\cdot1 - (n+1)\cdot0 = 1;$$ Third minor: $$1\cdot(n-1) - (n-2)\cdot0 = n - 1.$$

Substituting these into the expansion yields

$$D = 1\cdot(-n^2 + n - 1) + n\cdot1 + 1\cdot(n-1) = -n^2 + n - 1 + n + n - 1 = -n^2 + 3n - 2.$$

Factoring shows

$$D = -\bigl(n^2 - 3n + 2\bigr) = -(n - 1)(n - 2).$$

Hence the determinant is nonzero precisely when n ≠ 1 and n ≠ 2. Since a fair die roll gives n ∈ {1,2,3,4,5,6}, the system has a unique solution exactly for n ∈ {3,4,5,6}, which comprises 4 out of the 6 equally likely outcomes. The probability is therefore 4/6, which we write as k/6, giving k = 4.

Finally, summing k and all possible values of n that yield a unique solution gives

$$k + 3 + 4 + 5 + 6 = 4 + 18 = 22.$$

Hence the correct answer is 22.

We are given the system of linear equations:

$$x + y + z = 6 \quad \cdots (1)$$

$$2x + 5y + az = 36 \quad \cdots (2)$$

$$x + 2y + 3z = b \quad \cdots (3)$$

The coefficient matrix is:

$$A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix}$$

Its determinant is

$$\det(A) = 1(15 - 2a) - 1(6 - a) + 1(4 - 5) = 15 - 2a - 6 + a - 1 = 8 - a.$$

Since $$\det(A) = 8 - a\,, $$ for $$a = 8$$ we have $$\det(A) = 0$$, so the system does not have a unique solution and Options 1 and 3 are eliminated.

Substituting $$a = 8$$ into the augmented matrix gives

$$\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 2 & 5 & 8 & | & 36 \\ 1 & 2 & 3 & | & b \end{pmatrix}.$$

Applying $$R_2 \to R_2 - 2R_1$$ yields $$(0, 3, 6 \mid 24)$$ and then $$R_3 \to R_3 - R_1$$ yields $$(0, 1, 2 \mid b - 6).$$ Finally, $$R_3 \to R_3 - \tfrac{1}{3}R_2$$ gives $$(0, 0, 0 \mid b - 14).$$

For consistency (infinitely many solutions) we require $$b - 14 = 0\,, $$ so $$b = 14.$$ If instead $$b = 16\,, $$ then $$b - 14 = 2 \neq 0$$ and the system is inconsistent (no solution). Therefore the correct answer is Option 4: infinitely many solutions for $$a = 8$$ and $$b = 14$$.