(A) matrix $A$ is singular if its determinant $|A| = 0$.Given $A = \begin{bmatrix} 1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1 \end{bmatrix}$.Setting the d

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(A) matrix $A$ is singular if its determinant $|A| = 0$.Given $A = \begin{bmatrix} 1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1 \end{bmatrix}$.Setting the determinant to zero:$\begin{vmatrix} 1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1 \end{vmatrix} = 0$Expanding along the first row $(R_1)$:$1((x-2)(1) - (1)(1)) - 2((1)(1) - (1)(x)) + (-1)((1)(1) - (x)(x-2)) = 0$$1(x-2-1) - 2(1-x) - 1(1 - (x^2 - 2x)) = 0$$(x-3) - 2 + 2x - 1 + x^2 - 2x = 0$$x^2 + x - 6 = 0$Factoring the quadratic equation:$(x+3)(x-2) = 0$Therefore,$x = -3$ or $x = 2$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $\begin{bmatrix} 1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1 \end{bmatrix}$ is a singular matrix,then the value of $x$ is:

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$2$
B
$3$
C
$1$
D
$0$

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $\begin{bmatrix} 1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1 \end{bmatrix}$ is a singular matrix,then the value of $x$ is:

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