Let A=left[begin{array}{rr}2 & -1 3 & 4end{a | vedclass

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(A) Since $A, B, C$ are all square matrices of order $2$,and $CD-AB$ is well-defined,$D$ must be a square matrix of order $2$.Let $D=\left[\begin{arra

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$\left[\begin{array}{cc}-191 & -110 \ 77 & 44\end{array}ight]$

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(A) Since $A, B, C$ are all square matrices of order $2$,and $CD-AB$ is well-defined,$D$ must be a square matrix of order $2$.Let $D=\left[\begin{array}{ll}a & b \ c & d\end{array}ight]$. Then $CD-AB=O$ gives:$\left[\begin{array}{ll}2 & 5 \ 3 & 8\end{array}ight]\left[\begin{array}{ll}a & b \ c & d\end{array}ight] = \left[\begin{array}{rr}2 & -1 \ 3 & 4\end{array}ight]\left[\begin{array}{ll}5 & 2 \ 7 & 4\end{array}ight]$Calculating the product $AB$:$AB = \left[\begin{array}{cc}2(5)+(-1)(7) & 2(2)+(-1)(4) \ 3(5)+4(7) & 3(2)+4(4)\end{array}ight] = \left[\begin{array}{cc}3 & 0 \ 43 & 22\end{array}ight]$Now,$CD = \left[\begin{array}{cc}2a+5c & 2b+5d \ 3a+8c & 3b+8d\end{array}ight] = \left[\begin{array}{cc}3 & 0 \ 43 & 22\end{array}ight]$By equality of matrices:$2a+5c=3$ and $3a+8c=43$ (Solving gives $a=-191, c=77$)$2b+5d=0$ and $3b+8d=22$ (Solving gives $b=-110, d=44$)Thus,$D = \left[\begin{array}{cc}-191 & -110 \ 77 & 44\end{array}ight]$.

Let $A=\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}5 & 2 \\ 7 & 4\end{array}\right], C=\left[\begin{array}{ll}2 & 5 \\ 3 & 8\end{array}\right]$. Find a matrix $D$ such that $CD-AB=O$.

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