If A = begin{bmatrix} -1 & 2 2 & -1 end{bmat | vedclass

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(B) Given $A = \begin{bmatrix} -1 & 2 \ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 \ 1 \end{bmatrix}$.From $AX = B$,we have $X = A^{-1}B$.Firs

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$\frac{1}{3} \begin{bmatrix} 5 \ 7 \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} -1 & 2 \ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 \ 1 \end{bmatrix}$.From $AX = B$,we have $X = A^{-1}B$.First,find the determinant $|A| = (-1)(-1) - (2)(2) = 1 - 4 = -3$.The adjoint of $A$ is $adj(A) = \begin{bmatrix} -1 & -2 \ -2 & -1 \end{bmatrix}$.Thus,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-3} \begin{bmatrix} -1 & -2 \ -2 & -1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix}$.Now,calculate $X = A^{-1}B = \frac{1}{3} \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \ 1 \end{bmatrix}$.$X = \frac{1}{3} \begin{bmatrix} (1)(3) + (2)(1) \ (2)(3) + (1)(1) \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 + 2 \ 6 + 1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 5 \ 7 \end{bmatrix}$.

List-$I$	List-$II$
$(A)$ $\lambda=8, \mu \neq 15$	$1$. Infinitely many solutions
$(B)$ $\lambda \neq 8, \mu \in R$	$2$. No solution
$(C)$ $\lambda=8, \mu=15$	$3$. Unique solution

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

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