# Solve by matrix inversion method following system of linear equation$\;x-3y-8z+10=0\;,3x+y=4\;,2x+5y+6z=13$

Toolbox:
• Let $A = [ a_{ij} ]$ be a square matrix x. Let $A_{ij}$ be the cofactor of $a_{ij}$. Then $[ A_{ij}]$ is the matrix of cofactors and $adj\: A$ ( or adjoint of the matrix A) is given by $adj\: A=[A_{ij}]^T$
• 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 can be written in matrix form as $AX = B$
$A = \begin{bmatrix} 1 & -3 & -8 \\ 3 & 1 & 0 \\ 2 & 5 & 6 \end{bmatrix} , X = \begin{bmatrix}x \\ y \\ z \end{bmatrix} , B \begin{bmatrix}-10 \\ 4 \\ 13 \end{bmatrix}$
$|A| = \begin{vmatrix} 1 & -3 & -8 \\ 3 & 1 & 0 \\ 2 & 5 & 6 \end{vmatrix} = 1(6-0)+3(18-0)-8(15-2)$
$= 6+54-104=-44 \neq 0$
The system can be solved by matrix inversion $X = A^{-1}B$
Step 2
To find $A^{-1}$
$[A_{ij}] =$ the cofactor matrix = $\begin{bmatrix} (6-0) & -(18-0) & (15-2) \\ -(18+40) & (6+16) & -(5+6) \\ (0+8) & -(0+24) & (1+9) \end{bmatrix}$
$= \begin{bmatrix} 6 & -18 & 13 \\ -22 & 22 & -11 \\ 8 & -24 & 10 \end{bmatrix}$
$A^{-1} = \large\frac{1}{|A|} adj\: A = \large\frac{1}{|A|} [A_{ij}]^T = \large\frac{1}{-44} \begin{bmatrix} 6 & -22 & 8 \\ -18 & 22 & -24 \\ 13 & -11 & 10 \end{bmatrix}$
$X = A^{-1}B = \large\frac{-1}{44} \begin{bmatrix} 6 & -22 & 8 \\ -18 & 22 & -24 \\ 13 & -11 & 10 \end{bmatrix} \begin{bmatrix} -10 \\ 4 \\ 13 \end{bmatrix}$
$= \large\frac{-1}{44} \begin{bmatrix} -60 & -88 & +104 \\ 180 & +88 & -312 \\ -130 & -44 & +130 \end{bmatrix}$
$= \large\frac{-1}{44} \begin{bmatrix}-44 \\ -44 \\ -44 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$
$\therefore x = 1, y=1, z=1$

edited May 27, 2013