(B) matrix $A$ is singular if its determinant $|A| = 0$. Given $A = \begin{bmatrix} 2-k & 2 \\ 1 & 3-k \end{bmatrix}$. $|A| = (2-k)(3-k) - (2)(1) = 0$

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(B) matrix $A$ is singular if its determinant $|A| = 0$. Given $A = \begin{bmatrix} 2-k & 2 \\ 1 & 3-k \end{bmatrix}$. $|A| = (2-k)(3-k) - (2)(1) = 0$. Expanding the expression: $6 - 2k - 3k + k^2 - 2 = 0$. $k^2 - 5k + 4 = 0$. $k^2 - 5k = -4$. Multiplying by $-1$ on both sides: $5k - k^2 = 4$.

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 $A = \begin{bmatrix} 2-k & 2 \\ 1 & 3-k \end{bmatrix}$ is a singular matrix,then the value of $5k - k^2$ is equal to

KCET 2023 All KCET

A
$6$
B
$4$
C
$-6$
D
$-4$

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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 $A = \begin{bmatrix} 2-k & 2 \\ 1 & 3-k \end{bmatrix}$ is a singular matrix,then the value of $5k - k^2$ is equal to

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If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^{3}| = 125$,then $\alpha$ is equal to

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