# Solve by matrix inversion method following system of linear equation $\;7x\;+\;3y\;=\;-1\;,\;2x\;+\;y\;=\;0$

Toolbox:
• Let A be a square matrix of order 2. $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Then $adj\: A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ ( shortcut for finding the adjoint )
• The inverse of a non-singular square matrix A is given by $A^{-1} = \frac{1}{|A|} adj\: A.$ A non-singular matrix is one whose determinant value is nonzero.
• A system of x linear non-homogeneous equations in x unknowns given by
• $a_{11}x_1 +a_{12}x_2 \dotsc + a_{1n}x_n = b_1$
• $a_{22}x_1 +a_{22}x_2 \dotsc + a_{2n}x_n = b_2$
• $\: \: \: \: \vdots \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \vdots$
• $a_{n1}x_1 +a_{n2}x_2 \dotsc + a_{nn}x_n = b_n$ can be written as $AX = B$ in matrix form where $A = [a_{ij}], x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$ and $B = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix}$ If A is non-singular $( |A| \neq 0 )$ the inverse $A^{-1}$ exists and the system can be solved by matrix inversion $X = A^{-1}B$
Step 1
The system is of the matrix form $AX = B$
$A = \begin{bmatrix} 7 & 3 \\ 2 & 1 \end{bmatrix} , X = \begin{bmatrix} x \\ y \end{bmatrix} , B = \begin{bmatrix} -1 \\ 0 \end{bmatrix}$
$|A| = \begin{bmatrix} 7 & 3 \\ 2 & 1 \end{bmatrix} = -7-6=1 \neq 0 \: \: A^{-1}$ exists.
By matrix inversion, $X = A^{-1}B$
Step 2
To find $A^{-1}$
$adj\: A = \begin{bmatrix} 1 & -3 \\ -2 & 7 \end{bmatrix}$
$A^{-1} = \large\frac{1}{|A|} adj\: A = \begin{bmatrix} 1 & -3 \\ -2 & 7 \end{bmatrix}$
Step 3
$X = A^{-1}B = \begin{bmatrix} 1 & -3 \\ -2 & 7 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -1 & +0 \\ 2 & +0 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$
$\therefore x = -1, y=2$

edited May 27, 2013