(D) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$. Given $A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix}

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(D) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$. Given $A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix}$. Calculating the determinant $|A|$: $|A| = 0((-4)(-5) - (-6)(-3)) - k(k(-5) - (-6)(k)) + k(k(-3) - (-4)(k))$ $|A| = 0 - k(-5k + 6k) + k(-3k + 4k)$ $|A| = -k(k) + k(k)$ $|A| = -k^2 + k^2 = 0$. Since the determinant is $0$ for all real values of $k$,the matrix $A$ is singular for all real values of $k$.

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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If $A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix}$ is a singular matrix,then the value of $k$ is:

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A
$k = 2$ only
B
$k = \pm 2$ only
C
no real value of $k$
D
all real values of $k$

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3 and 4 .Determinants and Matrices

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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 $A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix}$ is a singular matrix,then the value of $k$ is:

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