If for AX = B, B = begin{bmatrix} 9 52 0 en | vedclass

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(A) Given the matrix equation $AX = B$. Multiplying both sides by $A^{-1}$ on the left,we get $A^{-1}(AX) = A^{-1}B$. Since $A^{-1}A = I$,we have $IX

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$\begin{bmatrix} 1 \ 3 \ 5 \end{bmatrix}$

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(A) Given the matrix equation $AX = B$. Multiplying both sides by $A^{-1}$ on the left,we get $A^{-1}(AX) = A^{-1}B$. Since $A^{-1}A = I$,we have $IX = A^{-1}B$,which simplifies to $X = A^{-1}B$. Now,substitute the given matrices: $X = \begin{bmatrix} 3 & -\frac{1}{2} & -\frac{1}{2} \ -4 & \frac{3}{4} & \frac{5}{4} \ 2 & -\frac{1}{4} & -\frac{3}{4} \end{bmatrix} \begin{bmatrix} 9 \ 52 \ 0 \end{bmatrix}$ Calculating the product: $X = \begin{bmatrix} (3 	imes 9) + (-\frac{1}{2} 	imes 52) + (-\frac{1}{2} 	imes 0) \ (-4 	imes 9) + (\frac{3}{4} 	imes 52) + (\frac{5}{4} 	imes 0) \ (2 	imes 9) + (-\frac{1}{4} 	imes 52) + (-\frac{3}{4} 	imes 0) \end{bmatrix}$ $X = \begin{bmatrix} 27 - 26 + 0 \ -36 + 39 + 0 \ 18 - 13 + 0 \end{bmatrix} = \begin{bmatrix} 1 \ 3 \ 5 \end{bmatrix}$.

If for $AX = B,$ $B = \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & -\frac{1}{2} & -\frac{1}{2} \\ -4 & \frac{3}{4} & \frac{5}{4} \\ 2 & -\frac{1}{4} & -\frac{3}{4} \end{bmatrix},$ then $X$ is equal to

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