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

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(A) Let $A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} x & y \ 0 & x \end{bmatrix}$.First,calculate $AB$:$AB = \begin{bm

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$\begin{bmatrix} x & y \ 0 & x \end{bmatrix}$

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(A) Let $A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} x & y \ 0 & x \end{bmatrix}$.First,calculate $AB$:$AB = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} x & y \ 0 & x \end{bmatrix} = \begin{bmatrix} 1(x) + 1(0) & 1(y) + 1(x) \ 0(x) + 1(0) & 0(y) + 1(x) \end{bmatrix} = \begin{bmatrix} x & x+y \ 0 & x \end{bmatrix}$.Next,calculate $BA$:$BA = \begin{bmatrix} x & y \ 0 & x \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} x(1) + y(0) & x(1) + y(1) \ 0(1) + x(0) & 0(1) + x(1) \end{bmatrix} = \begin{bmatrix} x & x+y \ 0 & x \end{bmatrix}$.Since $AB = BA$ for any $x, y$,the matrix $B$ must be of the form $\begin{bmatrix} x & y \ 0 & x \end{bmatrix}$.Thus,option $A$ is the correct choice.

If $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} x & y \\ 0 & x \end{bmatrix}$,then $AB = BA$ (given $B \neq I$). Which of the following matrices $B$ satisfies this condition?

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