System of Linear Equations in Matrices

In maths, a system of the linear system is a set of two or more linear equation involving the same set of variables. For example : 2x – y = 1, 3x + 2y = 12 . It is a system of two equation in the two variables that is x and y which is called a two linear equation in two unknown x and y and solution to a linear equation is the value to the variables such that all the equations are fulfilled.

In the matrix, every equation in the system becomes a row and each variable in the system becomes a column and the variables are dropped and the coefficients are placed into a matrix.

A system of two linear equations in two unknown x and y are as follows: $\left.\begin{matrix} a_{11}x + a_{12}y = b_{1} & \\ a_{21}x + a_{22}y = b_{2} & \end{matrix}\right\}$

Let $A=\begin {bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}$ , $X=\begin {bmatrix} x\\ y\end{bmatrix}$ , $B=\begin {bmatrix} {b1}\\ {b2}\end{bmatrix}$ .

Then system of equation can be written in matrix form as:

$\begin {bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}$ $\begin {bmatrix} x\\ y\end{bmatrix}$ = $\begin {bmatrix} {b1}\\ {b2}\end{bmatrix}$ i.e. AX = B and X = $A^{-1}B$ .

If the R.H.S., namely B is 0 then the system is homogeneous, otherwise non-homogeneous.

$\left.\begin{matrix} x+6y=0 & \\ 7x-5y=0 & \end{matrix}\right\}$ is a homogeneous system of two eqations in two unknowns x and y.

$\left.\begin{matrix} 4x+6y=8 & \\ 9x-5y=0 & \end{matrix}\right\}$ is a non-homogenoeus system of equations.

A system of three linear equations in three unknown x, y, z are as follows:

$\left.\begin{matrix} a_{11}x + a_{12}y + a_{13}z = b_{1} & \\ a_{21}x + a_{22}y + a_{23}z = b_{2} & \\ a_{31}x + a_{32}y + a_{33}z = b_{3} \end{matrix}\right\}$ .

Let $A=\begin {bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{bmatrix}$ , $X=\begin {bmatrix} x\\ y \\ z\end{bmatrix}$ , $B=\begin {bmatrix} {b1}\\ {b2} \\ b3\end{bmatrix}$ .

Then system of equation can be written in matrix form as:

$\begin {bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{bmatrix}$ $\begin {bmatrix} x\\ y \\ z\end{bmatrix}$ = $\begin {bmatrix} {b1}\\ {b2} \\ b3\end{bmatrix}$ i.e. AX = B and X = $A^{-1}B$ .

Algorithm to solve the Linear Equation via Matrix

Write the given system in the form of matrix equation as AX = B.
Find the determinant of the matrix. If determinant |A| = 0, then $A^{-1}$ does not exist so that solution does not exist. Write “System is not consistent”.
If the determinant exist then find the inverse of the matrix i.e. $A^{-1}$ .
Find $X = A^{-1}B$ where $A^{-1}$ is the inverse of the matrix.
Solve the equation by the matrix method of linear equation with the formula $X = A^{-1}B$ and find the values of x,y,z.

Example 1: Solve the equation: 4x+7y-9 = 0 , 5x-8y+15 = 0

Solution: Given equation can be written in matrix form as : $A=\begin {bmatrix} 4 & 7\\ 5 &-8 \end {bmatrix}$ , $X=\begin {bmatrix} x\\ y\end{bmatrix}$ , $B=\begin {bmatrix} 9\\ -15\end{bmatrix}$

Given system can be written as : AX = B , where $X = A^{-1}B$ .

Let us find determinant : |A| = 4*(-8) – 5*7 = -32-35 = -67 So, solution exist.

Minor and Cofactor of matrix A are : $M_{11}$ = -8 $C_{11}$ = -8, $M_{12}$ = 5 $C_{12}$ = -5, $M_{21}$ = 7 $C_{21}$ = -7, $M_{22}$ = 4 $C_{22}$ = 4.

Cofactor matrix = $\begin {bmatrix} -8 & -5\\ -7 &4 \end {bmatrix}$ and Adj A = $\begin {bmatrix} -8 & -7\\ -5 &4 \end {bmatrix}$

$A^{-1} = \frac{1}{|A|} adj A$ $= \frac{-1}{67}\begin {bmatrix} -8 & -7\\ -5 &4 \end {bmatrix}$ .

$X = A^{-1}B$ = $\frac{-1}{67}\begin {bmatrix} -8 & -7\\ -5 &4 \end {bmatrix} \begin {bmatrix} 9\\ -15\end{bmatrix}$ = $\frac{-1}{67}\begin {bmatrix} -72+105 &\\-45-60 \end {bmatrix}$ = $\frac{-1}{67}\begin {bmatrix} 33 &\\-105 \end {bmatrix}$

x = $-\frac{33}{67}$ and y = $\frac{105}{67}$

Example 2: Solve the equation: 2x+y+3z = 1, x+z = 2, 2x+y+z = 3

Solution: Given equation can be written in matrix form as : $A=\begin {bmatrix} 2 &1 &3 \\ 1&0&1 \\ 2&1&1 \end {bmatrix}$ , $X=\begin {bmatrix} x\\ y \\ z\end{bmatrix}$ , $B=\begin {bmatrix} 1\\ 2 \\ 3\end{bmatrix}$ .

Given system can be written as : AX = B , where $X = A^{-1}B$ .

Let us find determinant : |A| = 2(0-1) – 1(1-2) + 3(1-0) = -2+1+3 = 2. So, solution exist.

Minor and Cofactor of matrix A are : $M_{11}$ = -1 $C_{11}$ = -1, $M_{12}$ = -1 $C_{12}$ = 1, $M_{13}$ = 1 $C_{13}$ = 1, $M_{21}$ = -2 $C_{21}$ = 2, $M_{22}$ = -4 $C_{22}$ = -4, $M_{23}$ = 0 $C_{23}$ = 0 $M_{31}$ = 1 $C_{31}$ = -1, $M_{32}$ = -1 $C_{32}$ = -1, $M_{33}$ = -1 $C_{33}$ = 1.

$Cofactor=\begin {bmatrix} -1 &1 &1 \\ 2&-4&0 \\ 1&1&-1 \end {bmatrix}$ and $Adj(A)=\begin {bmatrix} -1 &2&1 \\ 1&-4&1 \\ 1&0&-1 \end {bmatrix}$

$A^{-1} = \frac{1}{|A|} adj A$ $= \frac{1}{2}\begin {bmatrix} -1 &2&1 \\ 1&-4&1 \\ 1&0&-1 \end {bmatrix}$ .

$X = A^{-1}B$ = $\frac{1}{2}\begin {bmatrix} -1 &2&1 \\ 1&-4&1 \\ 1&0&-1 \end {bmatrix}$ $\begin {bmatrix} 1\\ 2 \\ 3\end{bmatrix}$ = $\frac{1}{2}\begin {bmatrix}-1+4+3\\1-8+3\\1+0-3 \end {bmatrix}$ = $\frac{1}{2}\begin {bmatrix}6\\-4\\-2 \end {bmatrix}$ = $\begin {bmatrix}3\\-2\\-1 \end {bmatrix}$ .