Question

Express the following equations in the matrix form and solve them by method of reduction:
$2c-y+z=1$; $x+2y+3z=8$; $3x+y-4z=1$.

Answer 1

The given equations are:

$2x-y+z=1$

$x+2y+3z=8$

$3x+y-4z=1$

Writing it in matrix form,

$\left[\begin{array}{ccc}2& -1& 1\\ 1& 2& 3\\ 3& 1& -4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 8\\ 1\end{array}\right]$

Perform $R_2\to R_2-\frac{1}{2}{R}_1$ and $R_3\to R_3-\frac{3}{2}{R}_1$

$\left[\begin{array}{ccc}2& -1& 1\\ 0& \frac{5}{2}& \frac{5}{2}\\ 0& \frac{5}{2}& -\frac{11}{2}\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ \frac{15}{2}\\ -\frac{1}{2}\end{array}\right]$

Perform $R_3\to R_3-R_2$

$\left[\begin{array}{ccc}2& -1& 1\\ 0& \frac{5}{2}& \frac{5}{2}\\ 0& 0& -8\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ \frac{15}{2}\\ -8\end{array}\right]$

Hence, the new equations are,

$2x-y+z=1$ ....$\left(i\right)$

$\frac{5}{2}(y+z)=\frac{15}{2}\Rightarrow y+z=3$ ....$\left(ii\right)$

$-8z=-8\Rightarrow z=1$ ....$\left(iii\right)$

Substitute value of $z$ from $\left(iii\right)$ in equation $\left(ii\right)$

$y+1=3\Rightarrow y=2$

Substitute values of $y$ and $z$ in $\left(i\right)$

$2x-2+1=1\Rightarrow 2x=2\Rightarrow x=1$

Hence, $x=1,y=2,z=1$ is the solution of the given simultaneous equations.