(B) matrix $A$ is not invertible if its determinant is zero,i.e.,$|A| = 0$.Given the matrix $A = \begin{bmatrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \

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(B) matrix $A$ is not invertible if its determinant is zero,i.e.,$|A| = 0$.Given the matrix $A = \begin{bmatrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{bmatrix}$,we calculate the determinant:$|A| = 1(2 \times 1 - 5 \times 1) - a(1 \times 1 - 5 \times 2) + 2(1 \times 1 - 2 \times 2) = 0$$|A| = 1(2 - 5) - a(1 - 10) + 2(1 - 4) = 0$$|A| = 1(-3) - a(-9) + 2(-3) = 0$$-3 + 9a - 6 = 0$$9a - 9 = 0$$9a = 9$$a = 1$Thus,the matrix is not invertible when $a = 1$.

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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The matrix $\begin{bmatrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{bmatrix}$ is not invertible if $a$ has the value:

A
$2$
B
$1$
C
$0$
D
$-1$

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3-4. Matrices and Determinants

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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 $\begin{bmatrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{bmatrix}$ is not invertible if $a$ has the value:

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The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
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has non-trivial solutions is