"For the given a system of linear equations..." solved and explained

High School
Algebra
Algebra I
Forms of linear equations

Tammy Todd

Answered question

2021-02-19

For the given a system of linear equations
4x+y-5z=8
-2x+3y+z=12
3x-y+4z=5
Use matrix inversion to solve simultaneous equations.

Answer & Explanation

Fatema Sutton

Skilled2021-02-20Added 88 answers

This question is solved as follows:

Jeffrey Jordon

Expert2021-09-08Added 2605 answers

Given Linear equations

$4 x + y - 5 z = 8$

$- 2 x + 3 y + z = 12$

$3 x - y + 4 z = 5$

Now the matrix be 'A'

$A = [\begin{matrix} 4 & 1 & - 5 \\ - 2 & 3 & 1 \\ 3 & - 1 & 4 \end{matrix}]$

Let $A = | A | = 4 (12 + 1) - 1 (- 8 - 3) - 5 (2 - 9)$

$= 52 + 11 + 35$

$| A | = 98$

raw cofactors are found as

Cofactor of $4 = [\begin{matrix} 3 & 1 \\ - 1 & 4 \end{matrix}] = 12 + 1 = 13$

Cofactor of $1 = [\begin{matrix} - 2 & 1 \\ 3 & 4 \end{matrix}] = - 8 - 3 = - 11$

Cofactor of $- 5 = [\begin{matrix} - 2 & 3 \\ 3 & - 1 \end{matrix}] = 2 - 9 = - 7$

Cofactor of $- 2 = [\begin{matrix} 1 & - 5 \\ - 1 & 4 \end{matrix}] = 4 - 5 = - 1$

Cofactor of $1 = [\begin{matrix} 4 & 1 \\ 3 & - 1 \end{matrix}] = - 4 - 3 = - 7$

Cofactor of $3 = [\begin{matrix} 1 & - 5 \\ 3 & 1 \end{matrix}] = 1 + 15 = 16$

Cofactor of $1 = [\begin{matrix} 4 & - 5 \\ - 2 & 1 \end{matrix}] = 4 - 10 = - 6$

Cofactor of $4 = [\begin{matrix} 4 & 1 \\ - 2 & 3 \end{matrix}] = 12 + 2 = 14$

$∴ A^{'} = \frac{1}{| A |} a d j A = \frac{1}{98} [\begin{matrix} 13 & - 11 & - 7 \\ - 1 & 31 & - 7 \\ 16 & - 6 & 14 \end{matrix}]$

The system of equation can be written as AX=B

$[\begin{matrix} 4 & 1 & - 5 \\ - 2 & 3 & 1 \\ 3 & - 1 & 4 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 8 \\ 12 \\ 5 \end{matrix}]$

$A x = B \Rightarrow x = A^{- 1} B$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{98} [\begin{matrix} 13 & - 11 & - 7 \\ - 1 & 31 & 7 \\ 16 & - 6 & 14 \end{matrix}] [\begin{matrix} 8 \\ 12 \\ 5 \end{matrix}]$

$= \frac{1}{98} [\begin{matrix} 104 & - 132 & - 35 \\ - 8 & + 372 & - 35 \\ 128 & - 72 & 70 \end{matrix}]$

$= \frac{1}{98} [\begin{matrix} - 63 \\ 329 \\ 126 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} - 63 / 98 \\ 329 / 98 \\ 129 / 98 \end{matrix}]$

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