Answer to Question #207182 in Linear Algebra for James

Question

Use GAUSS-JORDAN INVERSE METHOD to solve these systems of Linear Equations.

y-10z=-8

2х - бу=8

x+2z=7

Answer 1

we are given

y-10z=-8

2х - бу=8

x+2z=7

this can be rewritten as

0x+y-10z=-8

2х - бу+0z=8

x+0y+2z=7

representing above equations in matrix form, we have

$\begin{pmatrix} 0 & 1&-10 \\ 2& -6&0\\ 1&0&2 \end{pmatrix}$ $\begin{pmatrix} x \\ y\\ z \end{pmatrix}$ =$\begin{pmatrix} -8 \\ 8\\ 7 \end{pmatrix}$

let A=$\begin{pmatrix} 0 & 1&-10 \\ 2& -6&0\\ 1&0&2 \end{pmatrix}$ and b= $\begin{pmatrix} -8 \\ 8\\ 7 \end{pmatrix}$ and X=$\begin{pmatrix} x\\ y\\ z \end{pmatrix}$

find inverse of matrix A using GAUSS-JORDAN INVERSE METHOD by reducing the following matrix to row echelon form

= $\begin{pmatrix} 0 & 1&-10 \mid&1&0&0\\ 2& -6& 0 \mid&0&1&0\\ 1&0&2 \mid&0&0&1 \end{pmatrix}$

$R_1\leftrightarrow R_3$

= $\begin{pmatrix} 1&0&2 \mid&0&0&1\\ 2& -6& 0 \mid&0&1&0\\ 0 & 1&-10 \mid&1&0&0 \end{pmatrix}$

$R_2\to R_2-2R_1$

=$\begin{pmatrix} 1&0&2 \mid&0&0&1\\ 0& -6& -4\mid&0&1&-2\\ 0 & 1&-10 \mid&1&0&0 \end{pmatrix}$

$R_2 \to- {1\over 6}R_2$

$R_3 \to 6R_3 +R_2$

=$\begin{pmatrix} 1&0&2 \mid&0&0&1\\ 0& 1& {4\over 6}\mid&0&-{1\over 6}&{2\over 6}\\ 0 &0&-64 \mid&6&1&-2 \end{pmatrix}$

$R_3 \to -{1\over 64}R_3$

=$\begin{pmatrix} 1&0&2 \mid&0&0&1\\ 0& 1& {4\over 6}\mid&0&-{1\over 6}&{2\over 6}\\ 0 &0&1 \mid&-{6\over 64}&-{1\over 64}&{2\over 64} \end{pmatrix}$

$R_2 \to R_2-{4\over 6}R_3$

$R_1 \to R_1-2R_3$

=$\begin{pmatrix} 1&0&0 \mid&{12\over 64}&{2\over 64}&{15\over 16}\\ 0& 1& 0\mid&{4\over 64}&-{5\over 32}&{5\over1 6}\\ 0 &0&1 \mid&-{6\over 64}&-{1\over 64}&{2\over 64} \end{pmatrix}$

$\implies$ $A^{-1}$ = $\begin{pmatrix} {12\over 64}&{2\over 64}&{15\over 16}\\ {4\over 64}&-{5\over 32}&{5\over 16}\\ -{6\over 64}&-{1\over 64}&{2\over 64} \end{pmatrix}$

A^-1AX=A^-1b

$\begin{pmatrix} {12\over 64}&{2\over 64}&{15\over 16}\\ {4\over 64}&-{5\over 32}&{5\over 16}\\ -{6\over 64}&-{1\over 64}&{2\over 64} \end{pmatrix}$ $\begin{pmatrix} 0 & 1&-10 \\ 2& -6&0\\ 1&0&2 \end{pmatrix}$ $\begin{pmatrix} x\\ y\\ z \end{pmatrix}$ =$\begin{pmatrix} {12\over 64}&{2\over 64}&{15\over 16}\\ {4\over 64}&-{5\over 32}&{5\over 16}\\ -{6\over 64}&-{1\over 64}&{2\over 64} \end{pmatrix}$ $\begin{pmatrix} -8 \\ 8\\ 7 \end{pmatrix}$

$\begin{pmatrix} 1& 0&0 \\ 0& 1&0\\ 0&0&1 \end{pmatrix}$ $\begin{pmatrix} x\\ y\\ z \end{pmatrix}$ =$\begin{pmatrix} 5.31250\\ 0.43750\\ 0.84375 \end{pmatrix}$

$\therefore$ $x=5.31250$

$y=0.43750$

$z=0.84375$