Question

The solution of the systems of linear equations $x-y+2z=7;3x+4y-5z=-5;2x-y+3z=12$ is:

• A. $(1,2,3)$
• B. $(2,1,3)$
• C. $(2,1,2)$
• D. $(1,2,1)$

Answer 1

The correct option is B $(2,1,3)$

Let $A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right]B=\left[\begin{array}{c}7\\ -5\\ 12\end{array}\right]X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

$\left|A\right|=\left|\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right|=1(12-5)+1(9+10)+2(-3-8)$
$\Rightarrow |A|=7+19-22=26-22=4\ne 0$

Now Cofactor matrix of A $=\left[\begin{array}{ccc}7& -19& -11\\ 1& -1& -1\\ -3& 11& 7\end{array}\right]$

$\therefore adjA={\left[\begin{array}{ccc}7& -19& -11\\ 1& -1& -1\\ -3& 11& 7\end{array}\right]}^T=\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$

$\therefore A^{-1}=\frac{adjA}{\left|A\right|}=\frac{1}{4}\left[\begin{array}{ccc}7& 1& 3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]$

$\therefore X=A^{-1}B=\frac{1}{4}\left[\begin{array}{ccc}7& 1& 3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\left[\begin{array}{c}7\\ -5\\ 12\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}49-5-36\\ -133+5+132\\ -77+5+84\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}8\\ 4\\ 12\end{array}\right]$

$\Rightarrow x=2;y=1;z=3$