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# Solve system of linear equations, using matrix method: $5x+2y=4$ $7x+3y = 5$

$\begin{array}{1 1} X=-2,Y=-3 \\ X=-2,Y=-3 \\ X=-2,Y=3 \\ X=2,Y=3\end{array}$

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.

$\begin{bmatrix}5 & 2\\7 & 3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4\\5\end{bmatrix}$

Where $A=\begin{bmatrix}5 & 2\\7 & 3\end{bmatrix}\;X=\begin{bmatrix}x\\y\end{bmatrix}\;B=\begin{bmatrix}4\\5\end{bmatrix}$

Let us now see if A is singular or non-singular,to obtain the value of determinant |A|,

|A|=$[5\times 3-2\times 7]$

$\;\;\;=15-14=1\neq 0$

Hence it is non-singular .Hence inverse exists,Now let us obtain $A^{-1}$

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

To calculate (adj A) let us interchange the elements of $a_{11}$ and $a_{22}$ and interchange the symbols of elements $a_{12}$ and $a_{21}$.

(adj A)=$\begin{bmatrix}3 & -2\\-7 & 5\end{bmatrix}$

We know |A|=1.

Therefore $A^{-1}=\frac{1}{1}\begin{bmatrix}3 & -2\\-7 & 5\end{bmatrix}$

$\Rightarrow A^{-1}=\begin{bmatrix}3 & -2\\-7 & 5\end{bmatrix}$

x=$A^{-1}B=\begin{bmatrix}3 & -2\\-7 & 5\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix}$

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

$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}12-10\\-28+25\end{bmatrix}$

$\Rightarrow x=2,y=-3$

edited Feb 27, 2013