Question

Solve the following equations using matrix inversion method: $2x-3y+6=0$ and $6x+y+8=0$.

Answer 1

Solve system of linear equations using matrix inversion method:

Step 1: Find the corresponding matrix equation.

The given equations are:

$2x-3y+6=0\Rightarrow 2x-3y=-6$

$6x+y+8=0\Rightarrow 6x+y=-8$

The corresponding matrix equation is

$\left[\begin{array}{cc}2& -3\\ 6& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-6\\ -8\end{array}\right]$

$\Rightarrow \mathrm{AX}=\mathrm{B}$ where $\mathrm{A}=\left[\begin{array}{cc}2& -3\\ 6& 1\end{array}\right]$, $\mathrm{X}=\left[\begin{array}{c}x\\ y\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{c}-6\\ -8\end{array}\right]$

Pre multiply the above matrix equation by ${\mathrm{A}}^{-1}$ on both sides:

$\begin{array}{rcl}{\mathrm{A}}^{-1}\left(\mathrm{AX}\right)& =& {\mathrm{A}}^{-1}\left(\mathrm{B}\right)\\ & \Rightarrow & \left({\mathrm{A}}^{-1}\mathrm{A}\right)\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\\ & \Rightarrow & \mathrm{IX}={\mathrm{A}}^{-1}\mathrm{B}\\ & \Rightarrow & \mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\end{array}$

Step 2: Find the inverse of $\mathrm{A}$.

$\mathrm{A}=\left[\begin{array}{cc}2& -3\\ 6& 1\end{array}\right]$

Determinant of $\mathrm{A}$ is

$\begin{array}{rcl}\left|\mathrm{A}\right|& =& 2\left(1\right)-6\left(-3\right)\\ & \Rightarrow & \left|\mathrm{A}\right|=2+18\\ & \Rightarrow & \left|\mathrm{A}\right|=20\end{array}$

Adjoint of $\mathrm{A}=\mathrm{adj}\mathrm{A}=\left[\begin{array}{cc}1& 3\\ -6& 2\end{array}\right]$

Therefore,

$\begin{array}{rcl}{\mathrm{A}}^{-1}& =& \frac{1}{\left|\mathrm{A}\right|}\left(\mathrm{adj}\mathrm{A}\right)\\ & \Rightarrow & {\mathrm{A}}^{-1}=\frac{1}{20}\left[\begin{array}{cc}1& 3\\ -6& 2\end{array}\right]\end{array}$

Step 3: Substitute the value of ${\mathrm{A}}^{-1}$ in the equation $\begin{array}{rcl}\mathrm{X}& =& {\mathrm{A}}^{-1}\mathrm{B}\end{array}$ and solve.

$\begin{array}{rcl}\mathrm{X}& =& \frac{1}{20}\left[\begin{array}{cc}1& 3\\ -6& 2\end{array}\right]\left[\begin{array}{c}-6\\ -8\end{array}\right]\\ & =& \frac{1}{20}\left[\begin{array}{c}1\left(-6\right)+3\left(-8\right)\\ \left(-6\right)\left(-6\right)+2\left(-8\right)\end{array}\right]\\ & & \\ & =& \frac{1}{20}\left[\begin{array}{c}-30\\ 20\end{array}\right]\\ & =& \left[\begin{array}{c}\frac{-30}{20}\\ \frac{20}{20}\end{array}\right]\\ & =& \left[\begin{array}{c}\frac{-3}{2}\\ 1\end{array}\right]\end{array}$

Hence,

$\begin{array}{rcl}\mathrm{X}& =& \left[\begin{array}{c}\frac{-3}{2}\\ 1\end{array}\right]\\ & \Rightarrow & \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{-3}{2}\\ 1\end{array}\right]\\ & \Rightarrow & x=\frac{-3}{2},\mathrm{y}=1\end{array}$

Hence, the solution of the equations $2x-3y+6=0$ and $6x+y+8=0$ is $x=\frac{-3}{2},y=1$.