Question

Solve the following system of linear equations, using matrix method

$5x+2y=4,7x+3y=5$

Answer 1

The given system can be written as AX=B, where
$A=\left[\begin{array}{cc}5& 2\\ 7& 3\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]andB=\left[\begin{array}{c}4\\ 5\end{array}\right]$
Here, $|A|=\left[\begin{array}{cc}5& 2\\ 7& 3\end{array}\right]=15-14=1\ne 0$
Thus, A is non singular. Therefore, its inverse exists.
Therefore, the given system is consistent and has a unique solution given by
$A^{-1}\left(AX\right)=A^{-1}B\Rightarrow X=A^{-1}B$
Cofactors of A are $A_{11}=3,A_{12}=-7,A_{21}=-2,A_{22}=5$
$\therefore adj\left(A\right)={\left[\begin{array}{cc}3& 7\\ -2& 5\end{array}\right]}^T=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]$
Now, $A^{-1}=\frac{1}{|A|}\left(adjA\right)=\frac{1}{1}\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]=,X=A^{-1}B=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\left[\begin{array}{c}4\\ 5\end{array}\right]=\left[\begin{array}{cc}3\times 4+(-2)\times 5& \\ -7\times 4+5\times 5& \end{array}\right]=\left[\begin{array}{c}2\\ -3\end{array}\right]\Rightarrow \left[\begin{array}{c}2\\ -3\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$
Hence, x=2 and y=-3.