Answer to Question #188898 in Linear Algebra for Ausayd Ur Rahman U

Question

Using GAUSS –JORDON Method, find the Inverse of the following matrix :

1 2 -1 4

2 3 2 0

3 1 0 3

2 0 2 1

Answer 1

solution:

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 2& 3 &2 &0 &0&1&0&0\\ 3& 1& 0& 3 &0&0&1&0\\ 2 &0& 2& 1& 0&0&0&1 \end{array}\right]$

2 line ->2 line -2* 1 line

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 2-2*1& 3-2*2 &2-2 *(-1)&0-2*4 &0-2*1&1-2*0&0-2*0&0-2*0\\ 3& 1& 0& 3 &0&0&1&0\\ 2 &0& 2& 1& 0&0&0&1 \end{array}\right]=\\ = \left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& -1 &4&-8 &-2&1&0&0\\ 3& 1& 0& 3 &0&0&1&0\\ 2 &0& 2& 1& 0&0&0&1 \end{array}\right]$

3 line-> 3 line -3* 1 line

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& -1 &4&-8 &-2&1&0&0\\ 3-3*1& 1-3*2& 0-3*(-1)& 3-3*4 &0-3*1&0-3*0&1-3*0&0-3*0\\ 2 &0& 2& 1& 0&0&0&1 \end{array}\right]=\\= \left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& -1 &4&-8 &-2&1&0&0\\ 0& -5& 3& -9 &-3&0&1&0\\ 2 &0& 2& 1& 0&0&0&1 \end{array}\right]$

4 line-> 4 line -2* 1 line

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& -1 &4&-8 &-2&1&0&0\\ 0& -5& 3& -9 &-3&0&1&0\\ 2 -2*1&0-2*2& 2-2*(-1)& 1-2*4& 0-2*1&0-2*0&0-2*0&1-2*0 \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& -1 &4&-8 &-2&1&0&0\\ 0& -5& 3& -9 &-3&0&1&0\\ 0&-4& 4& -7& -2&0&0&1 \end{array}\right]$

2 line -> (-1)*2 line

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0*(-1)& (-1)*(-1) &4*(-1)&(-8)*(-1) &(-2)*(-1)&1*(-1)&0*(-1)&0*(-1)\\ 0& -5& 3& -9 &-3&0&1&0\\ 0&-4& 4& -7& -2&0&0&1 \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0& -5& 3& -9 &-3&0&1&0\\ 0&-4& 4& -7& -2&0&0&1 \end{array}\right]$

3 line -> 3 line+ 5* 2 line

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0+5*0& -5+5*1& 3+5*(-4)& -9+5*8 &-3+5*2&0+5*(-1)&1+5*0&0+5*0\\ 0&-4& 4& -7& -2&0&0&1 \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0& 0& -17& 31 &7&-5&1&0\\ 0&-4& 4& -7& -2&0&0&1 \end{array}\right]$

4 line -> 4 line +4* 2 line

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0& 0& -17& 31 &7&-5&1&0\\ 0+4*0&-4+4*1& 4+4*(-4)& -7+4*8& -2+4*2&0+4*(-1)&0+4*(0)&1+4*0 \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0& 0& -17& 31 &7&-5&1&0\\ 0&0& -12& 25& 6&-4&0&1 \end{array}\right]$

3 line->$\frac{-1}{17}$ * 3 line

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0*\frac{-1}{17}& 0*\frac{-1}{17}& -17*\frac{-1}{17}& 31*\frac{-1}{17} &7*\frac{-1}{17}&-5*\frac{-1}{17}&1*\frac{-1}{17}&0*\frac{-1}{17}\\ 0&0& -12& 25& 6&-4&0&1 \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0 & 0& 1& \frac{-31}{17} &\frac{-7}{17}&\frac{5}{17}&\frac{-1}{17}&0\\ 0&0& -12& 25& 6&-4&0&1 \end{array}\right]$

4 line-> 4 line +12 * 3 line

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0 & 0& 1& \frac{-31}{17} &\frac{-7}{17}&\frac{5}{17}&\frac{-1}{17}&0\\ 0+12*0&0+12*0& -12+12*1& 25+12*\frac{-31}{17}& 6+12*\frac{-7}{17}&-4+12*\frac{5}{17}&0+12*\frac{-1}{17}&1+12*0 \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0 & 0& 1& \frac{-31}{17} &\frac{-7}{17}&\frac{5}{17}&\frac{-1}{17}&0\\ 0&0& 0& \frac{53}{17}& \frac{18}{17}&\frac{-8}{17}&\frac{-12}{17}&\frac{17}{17} \end{array}\right]$

4 line -> $\frac{17}{53}$ * 4 line

$\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0 & 0& 1& \frac{-31}{17} &\frac{-7}{17}&\frac{5}{17}&\frac{-1}{17}&0\\ 0*\frac{17}{53}&0*\frac{17}{53}& 0*\frac{17}{53}& \frac{53}{17}*\frac{17}{53}& \frac{18}{17}*\frac{17}{53}&\frac{-8}{17}*\frac{17}{53}&\frac{-12}{17}*\frac{17}{53}&\frac{17}{17}*\frac{17}{53} \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0 & 0& 1& \frac{-31}{17} &\frac{-7}{17}&\frac{5}{17}&\frac{-1}{17}&0\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]$

3 line -> 3 line +$\frac{31}{17}$ * 4 line

$\left[ \begin{array}{cccc|cccc} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0+\frac{31}{17} *0& 0+\frac{31}{17} *0& 1+\frac{31}{17} *0& \frac{-31}{17}+\frac{31}{17} *1 & \frac{-7}{17}+\frac{31}{17}*\frac{18}{53}&\frac{5}{17}+\frac{31}{17}*\frac{-8}{53}&\frac{-1}{17}+\frac{31}{17}*\frac{-12}{53}&0+\frac{31}{17}*\frac{17}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4&8 &2&-1&0&0\\ 0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]$

2 line ->2 line -8* 4 line

$\left[ \begin{array}{cccc|cccc} 1 &2& -1 &4 & 1 &0&0&0\\ 0-8*0& 1-8*0& -4-8*0& 8-8*1 & 2-8*\frac{18}{53} &-1-8*\frac{-8}{53}&0-8*\frac{-12}{53}&0-8*\frac{17}{53}\\0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 &4 & 1 &0&0&0\\ 0& 1 &-4& 0 &\frac{-38}{53}&\frac{11}{53}&\frac{96}{53}&\frac{-136}{53}\\ 0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]$

1 line ->1 line -4* 4 line

$\left[ \begin{array}{cccc|cccc} 1-4*0& 2-4*0& -1-4*0& 4-4*1 &1-4*\frac{18}{53} &0-4*\frac{-8}{53}&0-4*\frac{-12}{53}&0-4*\frac{17}{53}\\ 0& 1 &-4& 0 &\frac{-38}{53}&\frac{11}{53}&\frac{96}{53}&\frac{-136}{53}\\0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 & 0 & \frac{-19}{53} &\frac{32}{53}&\frac{48}{53}&\frac{-68}{53}\\ 0& 1 &-4& 0 &\frac{-38}{53}&\frac{11}{53}&\frac{96}{53}&\frac{-136}{53}\\ 0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]$

2 line ->2 line +4* 3 line

$\left[ \begin{array}{cccc|cccc} 1 &2& -1 & 0 & \frac{-19}{53} &\frac{32}{53}&\frac{48}{53}&\frac{-68}{53}\\ 0+4*0& 1+4*0& -4+4*1& 0+4*0 & \frac{-38}{53}+4*\frac{11}{53} &\frac{11}{53}+4*\frac{1}{53}&\frac{96}{53}+4*\frac{-25}{53}&\frac{-136}{53}+4*\frac{31}{53}\\0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& -1 & 0 & \frac{-19}{53} &\frac{32}{53}&\frac{48}{53}&\frac{-68}{53}\\ 0& 1 &0& 0 &\frac{6}{53}&\frac{15}{53}&\frac{-4}{53}&\frac{-12}{53}\\ 0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]$

1 line ->1 line +1* 3 line

$\left[ \begin{array}{cccc|cccc} 1+1*0& 2+1*0& -1+1*1& 0+1*0 & \frac{-19}{53}+1*\frac{11}{53} &\frac{32}{53}+1*\frac{1}{53}&\frac{48}{53}+1*\frac{-25}{53}&\frac{-68}{53}+1*\frac{31}{53}\\0& 1 &0& 0 &\frac{6}{53}&\frac{15}{53}&\frac{-4}{53}&\frac{-12}{53}\\0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &2& 0 & 0 & \frac{-8}{53} &\frac{33}{53}&\frac{23}{53}&\frac{-37}{53}\\ 0& 1 &0& 0 &\frac{6}{53}&\frac{15}{53}&\frac{-4}{53}&\frac{-12}{53}\\ 0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]$

1 line -> 1 line -2* 2 line

$\left[ \begin{array}{cccc|cccc} 1-2*0& 2-2*1& 0-2*0& 0-2*0 & \frac{-8}{53}-2*\frac{6}{53} &\frac{33}{53}-2*\frac{15}{53}&\frac{23}{53}+2*\frac{-4}{53}&\frac{-37}{53}-2*\frac{12}{53}\\0& 1 &0& 0 &\frac{6}{53}&\frac{15}{53}&\frac{-4}{53}&\frac{-12}{53}\\0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]=\\=\left[ \begin{array}{c c c c | c c c c} 1 &0& 0 & 0 & \frac{-20}{53} &\frac{3}{53}&\frac{31}{53}&\frac{-13}{53}\\ 0& 1 &0& 0 &\frac{6}{53}&\frac{15}{53}&\frac{-4}{53}&\frac{-12}{53}\\ 0 & 0& 1& 0 &\frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ 0&0& 0& 1& \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]$

Answer:$\left[ \begin{array}{c c c c } \frac{-20}{53} &\frac{3}{53}&\frac{31}{53}&\frac{-13}{53}\\ \frac{6}{53}&\frac{15}{53}&\frac{-4}{53}&\frac{-12}{53}\\ \frac{11}{53}&\frac{1}{53}&\frac{-25}{53}&\frac{31}{53}\\ \frac{18}{53}&\frac{-8}{53}&\frac{-12}{53}&\frac{17}{53} \end{array}\right]$