Question

Solve the system of equations, using Matrix method

$2x-y=-2,3x-4y=3$

Answer 1

Given system of equations

$2x-y=-2$
$3x-4y=3$

This can be written as

$AX=B$

where $A=\left[\begin{array}{cc}2& -1\\ 3& -4\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}-2\\ 3\end{array}\right]$

Here, $|A|=-8+3=-5$

Since, $|A|\ne 0$

Hence, $A^{-1}$ exists and the system has a unique solution given by $X=A^{-1}B$

$A^{-1}=\frac{adjA}{|A|}$ and $adjA=C^T$

So, we will find the co-factors of each element of A.

$C_{11}=(-1{)}^{1+1}-4=-4$

$C_{12}=(-1{)}^{1+2}3=-3$

$C_{21}=(-1{)}^{2+1}-1=1$

$C_{22}=(-1{)}^{2+2}2=2$

So, the co-factor matrix is $\left[\begin{array}{cc}4& -3\\ 1& 2\end{array}\right]$

$\Rightarrow adjA=C^T=\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]$

$\Rightarrow A^{-1}=\frac{adjA}{|A|}=\frac{1}{-5}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]$

The solution is $X=A^{-1}B$

$\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-5}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\left[\begin{array}{c}-2\\ 3\end{array}\right]$

$=\frac{1}{-5}\left[\begin{array}{c}-8+3\\ 6+6\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}1\\ -12/5\end{array}\right]$

Hence, $x=1,y=-\frac{12}{5}$