If A = begin{bmatrix} 1 & 2 -3 & 0 end{bmatr | vedclass

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(C) First,we calculate $A^2$: $A^2 = \begin{bmatrix} 1 & 2 \ -3 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \ -3 & 0 \end{bmatrix} = \begin{bmatrix} (1)

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$AB 
eq BA$

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(C) First,we calculate $A^2$: $A^2 = \begin{bmatrix} 1 & 2 \ -3 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \ -3 & 0 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(-3) & (1)(2) + (2)(0) \ (-3)(1) + (0)(-3) & (-3)(2) + (0)(0) \end{bmatrix} = \begin{bmatrix} -5 & 2 \ -3 & -6 \end{bmatrix} 
eq A$. Next,we calculate $B^2$: $B^2 = \begin{bmatrix} -1 & 0 \ 2 & 3 \end{bmatrix} \begin{bmatrix} -1 & 0 \ 2 & 3 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (0)(2) & (-1)(0) + (0)(3) \ (2)(-1) + (3)(2) & (2)(0) + (3)(3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 4 & 9 \end{bmatrix} 
eq B$. Now,we calculate $AB$: $AB = \begin{bmatrix} 1 & 2 \ -3 & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \ 2 & 3 \end{bmatrix} = \begin{bmatrix} (1)(-1) + (2)(2) & (1)(0) + (2)(3) \ (-3)(-1) + (0)(2) & (-3)(0) + (0)(3) \end{bmatrix} = \begin{bmatrix} 3 & 6 \ 3 & 0 \end{bmatrix}$. Finally,we calculate $BA$: $BA = \begin{bmatrix} -1 & 0 \ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \ -3 & 0 \end{bmatrix} = \begin{bmatrix} (-1)(1) + (0)(-3) & (-1)(2) + (0)(0) \ (2)(1) + (3)(-3) & (2)(2) + (3)(0) \end{bmatrix} = \begin{bmatrix} -1 & -2 \ -7 & 4 \end{bmatrix}$. Since $AB = \begin{bmatrix} 3 & 6 \ 3 & 0 \end{bmatrix}$ and $BA = \begin{bmatrix} -1 & -2 \ -7 & 4 \end{bmatrix}$,it is clear that $AB 
eq BA$.

If $A = \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 2 & 3 \end{bmatrix}$,then:

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