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(D) Given $A = \begin{bmatrix} 0 & 1 & 2 \ 2 & 3 & 0 \ 4 & 0 & 3 \end{bmatrix}$. Let $B = \begin{bmatrix} x & y & z \ a & b & c \ u & v & w \end{b

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$\begin{bmatrix} 9 & -3 & -6 \ -6 & -8 & 4 \ -12 & 4 & -2 \end{bmatrix}$

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(D) Given $A = \begin{bmatrix} 0 & 1 & 2 \ 2 & 3 & 0 \ 4 & 0 & 3 \end{bmatrix}$. Let $B = \begin{bmatrix} x & y & z \ a & b & c \ u & v & w \end{bmatrix}$.Since $AB = BA$,we perform matrix multiplication on both sides.Calculating $AB = \begin{bmatrix} 0 & 1 & 2 \ 2 & 3 & 0 \ 4 & 0 & 3 \end{bmatrix} \begin{bmatrix} x & y & z \ a & b & c \ u & v & w \end{bmatrix} = \begin{bmatrix} a+2u & b+2v & c+2w \ 2x+3a & 2y+3b & 2z+3c \ 4x+3u & 4y+3v & 4z+3w \end{bmatrix}$.Calculating $BA = \begin{bmatrix} x & y & z \ a & b & c \ u & v & w \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \ 2 & 3 & 0 \ 4 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 2y+4z & x+3y & 2x+3z \ 2b+4c & a+3b & 2a+3c \ 2v+4w & u+3v & 2u+3w \end{bmatrix}$.By comparing the elements of $AB$ and $BA$,we check the given options.Testing option $D$: $B = \begin{bmatrix} 9 & -3 & -6 \ -6 & -8 & 4 \ -12 & 4 & -2 \end{bmatrix}$.Substituting these values into the matrix multiplication confirms that $AB = BA$ holds true for this matrix.

If $A = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix}$ and $B$ is a matrix such that $AB = BA$. If $AB$ is not an identity matrix,then the matrix that can be taken as $B$ is:

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