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# Solve system of linear equations, using matrix method: $4x-3y=3$ $3x-5y=7$

$\begin{array}{1 1} X=-6/11,Y=-19/11 \\X=12/11,Y=19/11 \\ X=-6/11,Y=19/11 \\ X=-12/11,Y=-19/11 \end{array}$

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.

$\begin{bmatrix}4 & -3\\3 & -5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}$

Where $A=\begin{bmatrix}4 & -3\\3 & -5\end{bmatrix}\;X=\begin{bmatrix}x\\y\end{bmatrix}\;B=\begin{bmatrix}3\\7\end{bmatrix}$

Let us see if matrix A is singular and non-singular.Let us find the value of the determinant |A|

$|A|=4\times -5-3\times -3=-20+9=-11\neq 0$

Hence it is non -singular.

So it is invertible.

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

Now let us find out adj 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 $adj A=\begin{bmatrix}-5 & 3\\-3 & 4\end{bmatrix}$ and |A|=-11.

$A^{-1}=\frac{1}{-11}\begin{bmatrix}-5 & 3\\-3 & 4\end{bmatrix}$

Matrix multiplication can be obtained by multiplying the rows of A with the column of B.

Now AX=B$\Rightarrow X=A^{-1}B.$

$X=\frac{1}{11}\begin{bmatrix}-5 & 3\\-3 & 4\end{bmatrix}\begin{bmatrix}3\\7\end{bmatrix}$

$\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{11}\begin{bmatrix}15-21\\9-28\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-6\\-19\end{bmatrix}$

$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}-6/11\\-19/11\end{bmatrix}$

Hence x=-6/11 and y=-19/11.

edited Feb 27, 2013