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(D) Let $B = \begin{bmatrix} x & y \ z & w \end{bmatrix}$.Given $AB = O$,we have $\begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x & y

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$\begin{bmatrix} -1 & 0 \ 0 & 0 \end{bmatrix}$

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(D) Let $B = \begin{bmatrix} x & y \ z & w \end{bmatrix}$.Given $AB = O$,we have $\begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x & y \ z & w \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Performing matrix multiplication: $\begin{bmatrix} 0(x) + 1(z) & 0(y) + 1(w) \ 0(x) + 0(z) & 0(y) + 0(w) \end{bmatrix} = \begin{bmatrix} z & w \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Comparing the elements,we get $z = 0$ and $w = 0$,while $x$ and $y$ can be any real numbers.Checking the options,option $D$ gives $B = \begin{bmatrix} -1 & 0 \ 0 & 0 \end{bmatrix}$,which satisfies the condition $AB = O$ because $\begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} = O$.

If $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $AB = O$,then $B =$

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