Question

Solve the system of equations, using matrix method

$5x+2y=3,3x+2y=5$

Answer 1

Given system of equations

$5x+2y=3$
$3x+2y=5$

This can be written as

$AX=B$

where $A=\left[\begin{array}{cc}5& 2\\ 3& 2\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}3\\ 5\end{array}\right]$

Here, $|A|=10-6=4$

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 $A$.

$C_{11}=(-1{)}^{1+1}2=2$

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

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

$C_{22}=(-1{)}^{2+2}5=5$

So, the co-factor matrix is $\left[\begin{array}{cc}2& -3\\ -2& 5\end{array}\right]$

$\Rightarrow adjA=C^T=\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]$

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

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

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

$\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}6-10\\ -9+25\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-1\\ 4\end{array}\right]$

Hence, $x=-1,y=4$