Question
Solve the following system of equations by matrix method:
$3x + 7y = 4$
$x + 2y = -1$

Answer

The above system can be written in matrix form as:$\begin{bmatrix}3&7\\ 1&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}4\\ -1\end{bmatrix}$
Or AX = BWhere $\text{A}=\begin{bmatrix}3&7\\ 1&2\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}4\\ -1\end{bmatrix}$
Now,$\text{|A|}=-1\neq0$
So, the above system has a unique solution, given by $X=A^{-1} B$ Now, let $C_{i j}$ be the co-factors of $a_{i j}$ in $A$ $\text{C}_{11} = 2,\text{C}_{12} = -1$$\text{C}_{21} = -7,\text{C}_{22} = 3$
$\text{Adj A}=\begin{bmatrix}2&-1\\ -7&3\end{bmatrix}^\text{T}=\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$
$\therefore\text{A}^{-1}=\frac{1}{\text{|A|}}.\text{adj A}=\frac{1}{(-1)}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$
Now, $X = A^{-1}$​​​​​​​ B$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}\begin{bmatrix}4\\ -1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}15\\ -7\end{bmatrix}=\begin{bmatrix}-15\\ 7\end{bmatrix}$
Hence, x = -15 y = 7

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