Question

Use matrices to solve the following system of equations
$x+y+z=6$

$x-y+z=2$
$2x+y-z=1$
Solving we get $x=k,y=m,z=n$. Find $n-(k\times m)$

Answer 1

Given system of equations can be written as
$AX=B$
where $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ 2& 1& -1\end{array}\right]$
$X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ ;$B=\left[\begin{array}{c}6\\ 2\\ 1\end{array}\right]$
Here, $|A|=6\ne 0$
Hence, $A^{-1}$ exists.
Now, $adjA=C^T={\left[\begin{array}{ccc}0& 3& 3\\ 2& -3& 1\\ 2& 0& -2\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{ccc}0& 2& 2\\ 3& -3& 0\\ 3& 1& -2\end{array}\right]$
$A^{-1}=\frac{adjA}{|A|}$
$\Rightarrow A^{-1}=\frac{1}{6}\left[\begin{array}{ccc}0& 2& 2\\ 3& -3& 0\\ 3& 1& -2\end{array}\right]$
Solution to the system is given by
$X=A^{-1}B$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{ccc}0& 2& 2\\ 3& -3& 0\\ 3& 1& -2\end{array}\right]\left[\begin{array}{c}6\\ 2\\ 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}6\\ 12\\ 18\end{array}\right]$
$\Rightarrow x=1,y=2,z=3$
Comparing with given values , we get
$k=1;m=2;n=3$
$n-(k\times m)=3-2=1$