# Solve system of linear equations, using matrix method: $5x+2y=3$ $3x+2y=5$

$\begin{array}{1 1} x = 1, y = 4\\ x = -1, y = 4\\ x = 4, y = 1 \\ x = -4, y = -1 \end{array}$

Toolbox:
• A matrix is said to be singular if |A|= 0.
• A matrix is said to be invertible if |A|$\neq 0$.
• If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$
• Using this we can solve the system of equation which has unique solution.
This can be written in the form AX=B.

$\begin{bmatrix}5 & 2\\3 & 2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\5\end{bmatrix}$

Where $A=\begin{bmatrix}5 & 2\\3 & 2\end{bmatrix}\;X=\begin{bmatrix}x\\y\end{bmatrix}\;B=\begin{bmatrix}3\\5\end{bmatrix}$

Let us now check whether A is a singular or non-singular matrix,by finding the value of the determinant

$|A|=(5\times 2-2\times 3)$

$\;\;\;=10-6=4\neq 0$

Hence it is non -singular and therefore inverse exists.

To find the inverse $A^{-1}$,let us first find the adjoint A.

This can be obtained by interchanging the element of $a_{11}$ and $a_{22}$ and interchanging the symbols of elements of $a_{21}$ and $a_{12}$

Therefore $adj A=\begin{bmatrix}2 & -2\\-3 & 5\end{bmatrix}$ and |A|=4.

We know $A^{-1}=\frac{1}{|A|}(adj \;A)$

$\qquad\qquad=\frac{1}{4}\begin{bmatrix}2 & -2\\-3 & 5\end{bmatrix}$

Since AX=B

X=$A^{-1}B$

Let us find $A^{-1}B=\frac{1}{4}\begin{bmatrix}2 &-2\\-3 & 5\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix}$

We can do matrix multiplication by multiplying the rows of matrix A by the column of matrix B

$A^{-1}B=\frac{1}{4}\begin{bmatrix}6-10\\-9+25\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$

$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \frac{1}{4}\begin{bmatrix}-4\\16\end{bmatrix}$

$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}-1\\4\end{bmatrix}$

Hence x=-1 and y=4.

edited Feb 27, 2013