The value of x for which the matrix $A=\begin{vmatrix}1 & -2 & 3\\1 & 2 & 1\\x & 2&-3 \end{vmatrix}$ is singular :

-1

Given matrix: $A = \begin{pmatrix} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{pmatrix}$

A matrix is singular if $|A| = 0$

Compute the determinant using cofactor expansion along the first row:

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

Compute 2×2 determinants:

$\begin{vmatrix} 2 & 1 \\ 2 & -3 \end{vmatrix} = 2(-3) - 1*2 = -6 - 2 = -8$

$\begin{vmatrix} 1 & 1 \\ x & -3 \end{vmatrix} = 1*(-3) - 1*x = -3 - x = -(x+3)$

$\begin{vmatrix} 1 & 2 \\ x & 2 \end{vmatrix} = 1*2 - 2*x = 2 - 2x$

Substitute back:

$|A| = 1*(-8) - (-2)*(-(x+3)) + 3*(2-2x)$

$|A| = -8 - 2*(x+3) + 6 - 6x$

$|A| = -8 - 2x - 6 + 6 - 6x = -8 - 8x + 0 = -8x - 8$

Set $|A| = 0$:

$-8x - 8 = 0 \Rightarrow -8x = 8 \Rightarrow x = -1$

Answer: $x = -1$