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### Answers (1)

Answer:

$x=-1\; \; ,\; \; y=2$

Given:

$\left[\begin{array}{ll} 3& 4 \\ 1 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 5 \\ -3 \end{array}\right]$

Hint:

X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .

Solution:

$\left[\begin{array}{ll} 3& 4 \\ 1 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 5 \\ -3 \end{array}\right]$

\begin{aligned} &A X=B \\ &|A|=\left[\begin{array}{cc} 3 & 4 \\ 1 & -1 \end{array}\right]=-3-4=-7 \neq 0 \end{aligned}

$This\; has\; a\; unique\; solution\; given\; by\; X=A^{-1} B.\\ C_{i j}\; be\; the\; co\! -\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$

\begin{aligned} &C_{11}=(-1)^{1+1}(-1)=-1 \quad, \quad C_{12}=(-1)^{1+2}(1)=-1 \\ &C_{21}=(-1)^{2+1}(4)=-4 \quad, \quad C_{22}=(-1)^{2+2}(3)=3 \\ &\operatorname{adj} A=\left[\begin{array}{cc} -1 & -1 \\ -4 & 3 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} -1 & -4 \\ -1 & 3 \end{array}\right] \end{aligned}

\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-7}\left[\begin{array}{cc} -1 & -4 \\ -1 & 3 \end{array}\right] \\ X &=A^{-1} B \\ &=\frac{1}{-7}\left[\begin{array}{cc} -1 & -4 \\ -1 & 3 \end{array}\right]\left[\begin{array}{c} 5 \\ -3 \end{array}\right] \\ &=\frac{1}{-7}\left[\begin{array}{c} -5+12 \\ -5-9 \end{array}\right] \end{aligned}

\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{7}{-7} \\ \frac{-14}{-7} \end{array}\right]} \\ &\therefore x=-1 \quad, \quad \mathrm{y}=2 \end{aligned}

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