For a system of simultaneous linear equations | vedclass

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(A) Given $A X=\left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$ and $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \ 1 & 1 & -1 \ 1 &

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$\left[\begin{array}{c}-1 \ 0 \ 2\end{array}ight]$

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(A) Given $A X=\left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$ and $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \ 1 & 1 & -1 \ 1 & 1 & 1\end{array}ight]$.We know that $A^{-1} = \frac{1}{|A|} \operatorname{Adj} A$.Multiplying $A X = B$ by $A^{-1}$ on both sides, we get $X = A^{-1} B = \frac{1}{|A|} \operatorname{Adj} A \cdot B$.$X = \frac{1}{|A|} \left[\begin{array}{ccc}1 & -1 & -1 \ 1 & 1 & -1 \ 1 & 1 & 1\end{array}ight] \left[\begin{array}{l}1 \ 1 \ 2\end{array}ight]$.Performing matrix multiplication:$X = \frac{1}{|A|} \left[\begin{array}{c}1(1) + (-1)(1) + (-1)(2) \ 1(1) + 1(1) + (-1)(2) \ 1(1) + 1(1) + 1(2)\end{array}ight] = \frac{1}{|A|} \left[\begin{array}{c}1 - 1 - 2 \ 1 + 1 - 2 \ 1 + 1 + 2\end{array}ight] = \frac{1}{|A|} \left[\begin{array}{c}-2 \ 0 \ 4\end{array}ight]$.Since $|A| > 0$, we check the determinant of the given $\operatorname{Adj} A$. We know $|\operatorname{Adj} A| = |A|^{n-1}$, where $n=3$.$|\operatorname{Adj} A| = 1(1+1) - (-1)(1+1) + (-1)(1-1) = 2 + 2 + 0 = 4$.So, $|A|^2 = 4 \implies |A| = 2$ (since $|A| > 0$).Substituting $|A| = 2$ into the expression for $X$:$X = \frac{1}{2} \left[\begin{array}{c}-2 \ 0 \ 4\end{array}ight] = \left[\begin{array}{c}-1 \ 0 \ 2\end{array}ight]$.Thus, option $A$ is correct.

For a system of simultaneous linear equations, if $A X=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$, $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1\end{array}\right]$ and $\operatorname{det} A>0$, then $X=$

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