(A) matrix $A$ is not invertible if and only if its determinant is zero,i.e.,$|A| = 0$. Given the matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1

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(A) matrix $A$ is not invertible if and only if its determinant is zero,i.e.,$|A| = 0$. Given the matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$. Calculating the determinant $|A|$: $|A| = a(0 \times 2 - 1 \times 1) - (-1)(-3 \times 2 - 1 \times (-1)) + 4(-3 \times 1 - 0 \times (-1)) = 0$ $|A| = a(0 - 1) + 1(-6 + 1) + 4(-3 - 0) = 0$ $|A| = a(-1) + 1(-5) + 4(-3) = 0$ $-a - 5 - 12 = 0$ $-a - 17 = 0$ $a = -17$. Thus,the matrix is not invertible when $a = -17$.

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

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The matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is not invertible only if $a =$

MHT CET 2020 All MHT CET

A
-$17$
B
-$16$
C
$16$
D
$17$

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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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The matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is not invertible only if $a =$

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