MCQ
Solve system of linear equations, using matrix method. $x-y+z=4$ ; $2 x+y-3 z=0$ ; $x+y+z=2$
- A$x=-2, y=-1,z=1$
- B$x=-2, y=-1,z=-1$
- ✓$x=2, y=-1,z=1$
- D$x=-2, y=1,z=1$
$A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
Now,
$|A|=1(1+3)+1(2+3)+1(2-1)=4+5+1=10 \neq 0$
Thus $A$ is non-singular. Therefore, its inverse exists.
Now,
$A_{11}=4, A_{12}=-5, A_{13}=1$
$A_{21}=2, A_{22}=0, A_{33}=-2$
$A_{31}=2, A_{32}=5, A_{33}=3$
$\therefore A^{-1}=\frac{1}{|A|}(a d j A)=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
$\therefore X=A^{-1} B=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}16+0+4 \\ -20+0+10 \\ 4+0+6\end{array}\right]$
$=\frac{1}{10}\left[\begin{array}{c}20 \\ -10 \\ 10\end{array}\right]$
$=\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]$
Hence, $x=2, y=-1$ and $z=1$
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