In solving a system of linear equations AX=B | vedclass

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(A) By Cramer's rule,the solution for $X = \left[\begin{array}{c}x \ y \ z\end{array}ight]$ is given by $x = \frac{\Delta_1}{\Delta}$,$y = \frac{\

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

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(A) By Cramer's rule,the solution for $X = \left[\begin{array}{c}x \ y \ z\end{array}ight]$ is given by $x = \frac{\Delta_1}{\Delta}$,$y = \frac{\Delta_2}{\Delta}$,and $z = \frac{\Delta_3}{\Delta}$.First,calculate $\Delta_1$: $\Delta_1 = -11(1 - (-2)) - 1(-4 - (-10)) - 7(-4 - 5) = -11(3) - 1(6) - 7(-9) = -33 - 6 + 63 = 24$.Next,calculate $\Delta_3$: $\Delta_3 = 4(5 - (-4)) - 1(5 - (-16)) - 11(1 - 4) = 4(9) - 1(21) - 11(-3) = 36 - 21 + 33 = 48$.Since $x = \frac{\Delta_1}{\Delta}$ and $z = \frac{\Delta_3}{\Delta}$,we have $x = \frac{24}{\Delta}$ and $z = \frac{48}{\Delta}$.This implies $z = 2x$. Looking at the options:Option $A$: $x = -1, z = 2$ (Satisfies $z = 2x$)Option $B$: $x = 2, z = -1$ (Does not satisfy)Option $C$: $x = 1, z = 2$ (Does not satisfy)Option $D$: $x = 1, z = -1$ (Does not satisfy)Thus,the correct option is $A$.

In solving a system of linear equations $AX=B$ by Cramer's rule,in the usual notation,if $\Delta_1=\left|\begin{array}{ccc}-11 & 1 & -7 \\ -4 & 1 & -2 \\ 5 & 1 & 1\end{array}\right|$ and $\Delta_3=\left|\begin{array}{ccc}4 & 1 & -11 \\ 1 & 1 & -4 \\ 4 & 1 & 5\end{array}\right|$,then $X=$

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