Question

Solve the system of equations, using matrix method

$4x-3y=3,3x-5y=7$

Answer 1

Given system of equations

$4x-3y=3$
$3x-5y=7$

This can be written as

$AX=B$

where $A=\left[\begin{array}{cc}4& -3\\ 3& -5\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}3\\ 7\end{array}\right]$

Here, $|A|=-20+9=-11$

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 $A$.

$C_{11}=(-1{)}^{1+1}-5=-5$

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

$C_{21}=(-1{)}^{2+1}-3=3$

$C_{22}=(-1{)}^{2+2}4=4$

So, the co-factor matrix is $\left[\begin{array}{cc}-5& -3\\ 3& 4\end{array}\right]$

$\Rightarrow adjA=C^T=\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]$

$\Rightarrow A^{-1}=\frac{adjA}{|A|}=\frac{1}{-11}\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]$

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

$\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-11}\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]\left[\begin{array}{c}3\\ 7\end{array}\right]$

$=\frac{1}{-11}\left[\begin{array}{c}-15+21\\ -9+28\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-6/11\\ -19/5\end{array}\right]$

Hence, $x=-\frac{6}{11},y=-\frac{19}{11}$