Given A = begin{bmatrix} 1 & 3 2 & 2 end{bma | vedclass

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(B) matrix is singular if its determinant is equal to $0$. Given $A = \begin{bmatrix} 1 & 3 \ 2 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \

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$\lambda^2 - 3\lambda - 4 = 0$

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(B) matrix is singular if its determinant is equal to $0$. Given $A = \begin{bmatrix} 1 & 3 \ 2 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$. Then $A - \lambda I = \begin{bmatrix} 1 & 3 \ 2 & 2 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix} = \begin{bmatrix} 1 - \lambda & 3 \ 2 & 2 - \lambda \end{bmatrix}$. For $A - \lambda I$ to be a singular matrix,$\det(A - \lambda I) = 0$. $\det(A - \lambda I) = (1 - \lambda)(2 - \lambda) - (3)(2) = 0$. $2 - \lambda - 2\lambda + \lambda^2 - 6 = 0$. $\lambda^2 - 3\lambda - 4 = 0$. Thus,the correct option is $B$.

Given $A = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. If $A - \lambda I$ is a singular matrix,then:

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