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(A) Let $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix}$.From $A \begin{bmatrix} 0

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$-1$

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(A) Let $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix}$.From $A \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix}$,we get the second column of $A$ as $\begin{bmatrix} a_{12} \ a_{22} \ a_{32} \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix}$. Thus,$a_{12} = 0, a_{22} = 0, a_{32} = 1$.From $A \begin{bmatrix} 4 \ 1 \ 3 \end{bmatrix} = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}$,the second row equation is $4a_{21} + a_{22} + 3a_{23} = 1$. Since $a_{22} = 0$,we have $4a_{21} + 3a_{23} = 1$ (Equation $1$).From $A \begin{bmatrix} 2 \ 1 \ 2 \end{bmatrix} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$,the second row equation is $2a_{21} + a_{22} + 2a_{23} = 0$. Since $a_{22} = 0$,we have $2a_{21} + 2a_{23} = 0$,which implies $a_{21} = -a_{23}$ (Equation $2$).Substituting Equation $2$ into Equation $1$: $4(-a_{23}) + 3a_{23} = 1 \Rightarrow -4a_{23} + 3a_{23} = 1 \Rightarrow -a_{23} = 1 \Rightarrow a_{23} = -1$.

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