Let A, B be two 3 times 3 matrices and C be a | vedclass

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(C) Let $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix

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Both Statement $I$ and Statement $II$ are true

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(C) Let $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$, and $C = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Then $AB = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$, so $AB-C = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$.Since $\det(AB-C) = -1 
eq 0$, $AB-C$ is non-singular and $D = (AB-C)^{-1} = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix} = -C$.For Statement $I$: $\det(BA) = \det(0) = 0$. $\det(BA-C) = \det(-C) = -1$. $\det(BDA) = \det(-CBA) = \det(0) = 0$. Thus $0 = (-1)(0)$ is true.For Statement $II$: $ABD = AB(-C) = -AB = 0$. $DAB = (-C)AB = -AB = 0$. Thus $ABD = DAB$ is true.Both statements are true.

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