Name: Solving System of Linear Equations using Matrices
Uploaded: 2019-07-19
Duration: 11 min 30 s
Description: A system of linear equations can be represented in matrix form using its coefficient, variable and constant matrices. Learn about the inverse matrix method!

Question

Solve and find the unique solution of the following system of linear equation (whenever possible) by matrix method:

$x + z = 2$
$3 x + 4 y + 5 z = 0$
$2 x + 3 y + 4 z = - 5$

Accepted Answer

Given system of linear equations is

$x + z = 2$

$3 x + 4 y + 5 z = 0$

$2 x + 3 y + 4 z = - 5$

Here the system of linear equations may be written as $A X = B$ , where

$A = (\begin{matrix} 1 & 0 & 1 \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{matrix})$ , $X = (\begin{array}{l} x \\ y \\ z \end{array})$ , and $B = (\begin{array}{l} 2 \\ 0 \\ - 5 \end{array})$

Since $|A| = |\begin{matrix} 1 & 0 & 1 \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{matrix}|$

$= 1 (16 - 15) - 0 + 1 (9 - 8)$

$= 1 - 0 + 1$

$= 2 \neq 0$

So, $A^{- 1}$ exists. And, the given system of linear equations is consistent whose unique solution is given by

$X = A^{- 1} B$ , where $A^{- 1} = \frac{1}{| A |} (adj (A))$

To find $adj (A)$ , let us first find the cofactor matrix of $A$ .

Let the cofactor matrix be given by $C = (\begin{matrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{matrix})$ , where $A_{i j}$ is the cofactor of elements $a_{i j}$ in $A = [\begin{matrix} a_{i j} \end{matrix}]$ .

Here, $A = (\begin{matrix} 1 & 0 & 1 \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{matrix})$ , then

$A_{11} = (- 1)^{1 + 1} |\begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix}|$ $= [16 - 15] = 1$

$A_{12} = (- 1)^{1 + 2} |\begin{array}{l} 3 & 5 \\ 2 & 4 \end{array}|$ $= - [12 - 10] = - 2$

$A_{13} = (- 1)^{1 + 3} |\begin{array}{l} 3 & 4 \\ 2 & 3 \end{array}|$ $= (- 1)^{4} [9 - 8] = 1$

$A_{21} = (- 1)^{2 + 1} |\begin{matrix} 0 & 1 \\ 3 & 4 \end{matrix}|$ $= - [0 - 3] = 3$

$A_{22} = (- 1)^{2 + 2} |\begin{array}{l} 1 & 1 \\ 2 & 4 \end{array}| = 4 - 2 = 2$

$A_{23} = (- 1)^{2 + 3} |\begin{matrix} 1 & 0 \\ 2 & 3 \end{matrix}| = - (3 - 0) = - 3$

$A_{31} = (- 1)^{3 + 1} |\begin{array}{r} 0 & 1 \\ 4 & 5 \end{array}| = 0 - 4 = - 4$

$A_{32} = (- 1)^{3 + 2} |\begin{matrix} 1 & 1 \\ 3 & 5 \end{matrix}|$ $= - [5 - 3] = - 2$

$A_{33} = (- 1)^{3 + 3} |\begin{matrix} 1 & 0 \\ 3 & 4 \end{matrix}|$ $= (- 1)^{6} [4 - 0] = 4$

Thus, $C = (\begin{matrix} 1 & - 2 & 1 \\ 3 & 2 & - 3 \\ - 4 & - 2 & 4 \end{matrix})$

We know that $adj (A) = C^{T}$

$\Rightarrow adj (A) = (\begin{matrix} 1 & 3 & - 4 \\ - 2 & 2 & - 2 \\ 1 & - 3 & 4 \end{matrix})$

$∴ A^{- 1} = \frac{1}{2} (\begin{matrix} 1 & 3 & - 4 \\ - 2 & 2 & - 2 \\ 1 & - 3 & 4 \end{matrix})$ $[∵ A^{- 1} = \frac{1}{| A |} (adj (A))]$

Since $X = A^{- 1} B$

$∴ A^{- 1} B = \frac{1}{2} (\begin{matrix} 1 & 3 & - 4 \\ - 2 & 2 & - 2 \\ 1 & - 3 & 4 \end{matrix}) (\begin{array}{l} 2 \\ 0 \\ - 5 \end{array})$

$= \frac{1}{2} (\begin{matrix} 1 \times 2 + 0 + (- 4 \times - 5) \\ - 2 \times 2 + 0 + (- 2 \times - 5) \\ 2 \times 1 + 0 + (4 \times - 5) \end{matrix})$

$= \frac{1}{2} (\begin{matrix} 22 \\ 6 \\ - 18 \end{matrix})$

$\Rightarrow X = (\begin{matrix} 11 \\ 3 \\ - 9 \end{matrix})$

$\Rightarrow (\begin{array}{l} x \\ y \\ z \end{array}) = (\begin{array}{l} 11 \\ 3 \\ - 9 \end{array})$

Hence, $x = 11, y = 3,$ and $z = - 9$ is the unique solution of the given system of linear equations.

Solve and find the unique solution of the following system of linear equation (whenever possible) by matrix method:
$x + z = 2$
$3 x + 4 y + 5 z = 0$
$2 x + 3 y + 4 z = - 5$

Important Questions on Adjoint and Inverse of a Matrix

Learn and Prepare for any exam you want !

Solve and find the unique solution of the following system of linear equation (whenever possible) by matrix method: x+z=2 3x+4y+5z=0 2x+3y+4z=-5

Important Questions on Adjoint and Inverse of a Matrix

Learn and Prepare for any exam you want !

Solve and find the unique solution of the following system of linear equation (whenever possible) by matrix method:
$x + z = 2$
$3 x + 4 y + 5 z = 0$
$2 x + 3 y + 4 z = - 5$