If A = begin{bmatrix} 1 & 0 0 & -1 end{bmatr | vedclass

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(B) The statement that $AB 
eq BA$ does not hold for every pair of matrices $A$ and $B$ for which $AB$ and $BA$ are defined. For instance,if $A = \be

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No,matrix multiplication can be commutative for some pairs of matrices.

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(B) The statement that $AB 
eq BA$ does not hold for every pair of matrices $A$ and $B$ for which $AB$ and $BA$ are defined. For instance,if $A = \begin{bmatrix} 1 & 0 \ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 0 \ 0 & 4 \end{bmatrix}$,then: $AB = \begin{bmatrix} 1 	imes 3 + 0 	imes 0 & 1 	imes 0 + 0 	imes 4 \ 0 	imes 3 + 2 	imes 0 & 0 	imes 0 + 2 	imes 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 \ 0 & 8 \end{bmatrix}$ $BA = \begin{bmatrix} 3 	imes 1 + 0 	imes 0 & 3 	imes 0 + 0 	imes 2 \ 0 	imes 1 + 4 	imes 0 & 0 	imes 0 + 4 	imes 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 \ 0 & 8 \end{bmatrix}$ Since $AB = BA$,matrix multiplication is commutative in this case. Thus,the multiplication of diagonal matrices of the same order is commutative.

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