In April 2024
Answer to Question #188898 in Linear Algebra for Ausayd Ur Rahman U
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
solution:
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 2& 3 &2 &0 &0&1&0&0\\\\\n3& 1& 0& 3 &0&0&1&0\\\\\n2 &0& 2& 1& 0&0&0&1\n\\end{array}\\right]"
2 line ->2 line -2* 1 line
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 2-2*1& 3-2*2 &2-2 *(-1)&0-2*4 &0-2*1&1-2*0&0-2*0&0-2*0\\\\\n3& 1& 0& 3 &0&0&1&0\\\\\n2 &0& 2& 1& 0&0&0&1\n\\end{array}\\right]=\\\\\n=\n\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& -1 &4&-8 &-2&1&0&0\\\\\n3& 1& 0& 3 &0&0&1&0\\\\\n2 &0& 2& 1& 0&0&0&1\n\\end{array}\\right]"
3 line-> 3 line -3* 1 line
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& -1 &4&-8 &-2&1&0&0\\\\\n3-3*1& 1-3*2& 0-3*(-1)& 3-3*4 &0-3*1&0-3*0&1-3*0&0-3*0\\\\\n2 &0& 2& 1& 0&0&0&1\n\\end{array}\\right]=\\\\=\n\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& -1 &4&-8 &-2&1&0&0\\\\\n 0& -5& 3& -9 &-3&0&1&0\\\\\n2 &0& 2& 1& 0&0&0&1\n\\end{array}\\right]"
4 line-> 4 line -2* 1 line
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& -1 &4&-8 &-2&1&0&0\\\\\n 0& -5& 3& -9 &-3&0&1&0\\\\\n2 -2*1&0-2*2& 2-2*(-1)& 1-2*4& 0-2*1&0-2*0&0-2*0&1-2*0\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& -1 &4&-8 &-2&1&0&0\\\\\n 0& -5& 3& -9 &-3&0&1&0\\\\\n 0&-4& 4& -7& -2&0&0&1\n\\end{array}\\right]"
2 line -> (-1)*2 line
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0*(-1)& (-1)*(-1) &4*(-1)&(-8)*(-1) &(-2)*(-1)&1*(-1)&0*(-1)&0*(-1)\\\\\n 0& -5& 3& -9 &-3&0&1&0\\\\\n 0&-4& 4& -7& -2&0&0&1\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0& -5& 3& -9 &-3&0&1&0\\\\\n 0&-4& 4& -7& -2&0&0&1\n\\end{array}\\right]"
3 line -> 3 line+ 5* 2 line
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0+5*0& -5+5*1& 3+5*(-4)& -9+5*8 &-3+5*2&0+5*(-1)&1+5*0&0+5*0\\\\\n 0&-4& 4& -7& -2&0&0&1\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0& 0& -17& 31 &7&-5&1&0\\\\\n 0&-4& 4& -7& -2&0&0&1\n\\end{array}\\right]"
4 line -> 4 line +4* 2 line
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0& 0& -17& 31 &7&-5&1&0\\\\\n 0+4*0&-4+4*1& 4+4*(-4)& -7+4*8& -2+4*2&0+4*(-1)&0+4*(0)&1+4*0\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0& 0& -17& 31 &7&-5&1&0\\\\\n 0&0& -12& 25& 6&-4&0&1\n\\end{array}\\right]"
3 line->"\\frac{-1}{17}" * 3 line
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 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}\\\\\n 0&0& -12& 25& 6&-4&0&1\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0 & 0& 1& \\frac{-31}{17} &\\frac{-7}{17}&\\frac{5}{17}&\\frac{-1}{17}&0\\\\\n 0&0& -12& 25& 6&-4&0&1\n\\end{array}\\right]"
4 line-> 4 line +12 * 3 line
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0 & 0& 1& \\frac{-31}{17} &\\frac{-7}{17}&\\frac{5}{17}&\\frac{-1}{17}&0\\\\\n 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\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0 & 0& 1& \\frac{-31}{17} &\\frac{-7}{17}&\\frac{5}{17}&\\frac{-1}{17}&0\\\\\n 0&0& 0& \\frac{53}{17}& \\frac{18}{17}&\\frac{-8}{17}&\\frac{-12}{17}&\\frac{17}{17}\n\\end{array}\\right]"
4 line -> "\\frac{17}{53}" * 4 line
"\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0 & 0& 1& \\frac{-31}{17} &\\frac{-7}{17}&\\frac{5}{17}&\\frac{-1}{17}&0\\\\\n 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}\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0 & 0& 1& \\frac{-31}{17} &\\frac{-7}{17}&\\frac{5}{17}&\\frac{-1}{17}&0\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]"
3 line -> 3 line +"\\frac{31}{17}" * 4 line
"\\left[\n \\begin{array}{cccc|cccc}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 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}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4&8 &2&-1&0&0\\\\\n 0 & 0& 1& 0 &\\frac{11}{53}&\\frac{1}{53}&\\frac{-25}{53}&\\frac{31}{53}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]"
2 line ->2 line -8* 4 line
"\\left[\n \\begin{array}{cccc|cccc}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 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}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 &4 & 1 &0&0&0\\\\ \n 0& 1 &-4& 0 &\\frac{-38}{53}&\\frac{11}{53}&\\frac{96}{53}&\\frac{-136}{53}\\\\\n 0 & 0& 1& 0 &\\frac{11}{53}&\\frac{1}{53}&\\frac{-25}{53}&\\frac{31}{53}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]"
1 line ->1 line -4* 4 line
"\\left[\n \\begin{array}{cccc|cccc}\n 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}\\\\\n 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}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 & 0 & \\frac{-19}{53} &\\frac{32}{53}&\\frac{48}{53}&\\frac{-68}{53}\\\\ \n 0& 1 &-4& 0 &\\frac{-38}{53}&\\frac{11}{53}&\\frac{96}{53}&\\frac{-136}{53}\\\\\n 0 & 0& 1& 0 &\\frac{11}{53}&\\frac{1}{53}&\\frac{-25}{53}&\\frac{31}{53}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]"
2 line ->2 line +4* 3 line
"\\left[\n \\begin{array}{cccc|cccc}\n 1 &2& -1 & 0 & \\frac{-19}{53} &\\frac{32}{53}&\\frac{48}{53}&\\frac{-68}{53}\\\\ \n 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}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& -1 & 0 & \\frac{-19}{53} &\\frac{32}{53}&\\frac{48}{53}&\\frac{-68}{53}\\\\ \n 0& 1 &0& 0 &\\frac{6}{53}&\\frac{15}{53}&\\frac{-4}{53}&\\frac{-12}{53}\\\\\n 0 & 0& 1& 0 &\\frac{11}{53}&\\frac{1}{53}&\\frac{-25}{53}&\\frac{31}{53}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]"
1 line ->1 line +1* 3 line
"\\left[\n \\begin{array}{cccc|cccc}\n 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}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &2& 0 & 0 & \\frac{-8}{53} &\\frac{33}{53}&\\frac{23}{53}&\\frac{-37}{53}\\\\ \n 0& 1 &0& 0 &\\frac{6}{53}&\\frac{15}{53}&\\frac{-4}{53}&\\frac{-12}{53}\\\\\n 0 & 0& 1& 0 &\\frac{11}{53}&\\frac{1}{53}&\\frac{-25}{53}&\\frac{31}{53}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]"
1 line -> 1 line -2* 2 line
"\\left[\n \\begin{array}{cccc|cccc}\n 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}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]=\\\\=\\left[\n \\begin{array}{c c c c | c c c c}\n 1 &0& 0 & 0 & \\frac{-20}{53} &\\frac{3}{53}&\\frac{31}{53}&\\frac{-13}{53}\\\\ \n 0& 1 &0& 0 &\\frac{6}{53}&\\frac{15}{53}&\\frac{-4}{53}&\\frac{-12}{53}\\\\\n 0 & 0& 1& 0 &\\frac{11}{53}&\\frac{1}{53}&\\frac{-25}{53}&\\frac{31}{53}\\\\\n 0&0& 0& 1& \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]"
Answer:"\\left[\n \\begin{array}{c c c c }\n \\frac{-20}{53} &\\frac{3}{53}&\\frac{31}{53}&\\frac{-13}{53}\\\\ \n \\frac{6}{53}&\\frac{15}{53}&\\frac{-4}{53}&\\frac{-12}{53}\\\\\n \\frac{11}{53}&\\frac{1}{53}&\\frac{-25}{53}&\\frac{31}{53}\\\\\n \\frac{18}{53}&\\frac{-8}{53}&\\frac{-12}{53}&\\frac{17}{53}\n\\end{array}\\right]"
#339914 on Dec 2023
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