Find the matrix X such that X begin{bmatrix} | vedclass

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(A) Given the equation: $X \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \ 2 & 4 & 6 \end{bmatrix}$.The matrix

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$\begin{bmatrix} 1 & -2 \ 2 & 0 \end{bmatrix}$

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(A) Given the equation: $X \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \ 2 & 4 & 6 \end{bmatrix}$.The matrix on the right side is a $2 	imes 3$ matrix. Since the matrix on the left is a $2 	imes 3$ matrix,$X$ must be a $2 	imes 2$ matrix.Let $X = \begin{bmatrix} a & c \ b & d \end{bmatrix}$.Multiplying the matrices:$\begin{bmatrix} a & c \ b & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4c & 2a+5c & 3a+6c \ b+4d & 2b+5d & 3b+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \ 2 & 4 & 6 \end{bmatrix}$.Equating corresponding elements:$1$) $a+4c = -7$ and $2a+5c = -8$.From the first,$a = -7-4c$. Substituting into the second: $2(-7-4c) + 5c = -8 \Rightarrow -14 - 8c + 5c = -8 \Rightarrow -3c = 6 \Rightarrow c = -2$.Then $a = -7 - 4(-2) = 1$.$2$) $b+4d = 2$ and $2b+5d = 4$.From the first,$b = 2-4d$. Substituting into the second: $2(2-4d) + 5d = 4 \Rightarrow 4 - 8d + 5d = 4 \Rightarrow -3d = 0 \Rightarrow d = 0$.Then $b = 2 - 4(0) = 2$.Thus,$X = \begin{bmatrix} 1 & -2 \ 2 & 0 \end{bmatrix}$.

Find the matrix $X$ such that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.

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