Question

Examine the consistency of the system of equations
$x+2y=2$, $2x+3y=3$

Answer 1

Given system of equations

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

This can be written as

$AX=B$

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

Here, $|A|=3-4=-1$

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 $A$.

$C_{11}=(-1{)}^{1+1}3=3$

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

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

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

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

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

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

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

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

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

$=\left[\begin{array}{c}-6+6\\ 4-3\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 1\end{array}\right]$

Hence, $x=0,y=1$