Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

I had a question about the solution:/solution-player.aspx?solution_problem_id=19

ID: 2939072 • Letter: I

Question

I had a question about the solution:/solution-player.aspx?solution_problem_id=199334

It is basically using RREF of a certain matrix and the Identitymatrix (I3 to find the inverse of it.


I understand perfectly what it's doing, but when I try to find theRREF(reduced row echelon form) of the original matrix to find theinverse on the right side for the last two columns. I startgetting very large fractions. I think I'm doing it wrongcompletely. Is it at all possible for someone to maybe do the firstfew steps(or even all ), going towards RREF, so I can figure outwhere to start, and avoid the large fractions I've beengetting? Any help on this would be awesome :). Thanksin advance.

Explanation / Answer

Question Details: I had a question about thesolution: /solution-player.aspx?solution_problem_id=199334 It is basically using RREF of a certain matrix and the Identitymatrix (I3 to find the inverse of it. I understand perfectly what it's doing, but when I try to find theRREF(reduced row echelon form) of the original matrix to find theinverse on the right side for the last two columns. I startgetting very large fractions. I think I'm doing it wrongcompletely. Is it at all possible for someone to maybe do the firstfew steps(or even all ), going towards RREF, so I can figure outwhere to start, and avoid the large fractions I've beengetting? Any help on this would be awesome :). Thanksin advance. HOPE THE FOLLOWING EXAMPLE WILL SERVE YOUR PURPOSE MATRIX INVERSION                              A=                          2    2    3   3    1    0   0    0 2    3    3   2    0    1   0    0 5    3    7   9    0    0   1    0 3    2    4   7    0    0   0    1 NR1=R1/2…NR2=R2-R1….NR3=R3-2.5R1….NR4=R4-1.5R1                          1    1    1.5   1.5    0.5    0   0    0 0    1     0    -1       -1      1    0    0 0    -2    -0.5   1.5    -2.5    0   1    0 0    -1    -0.5   2.5    -1.5    0   0    1 NR1=R1-R2…NR3=R3+2E2…NR4=R4+R2                          1    0    1.5   2.5 1.5    -1   0    0 0    1   0      -1 -1       1    0    0 0    0    -0.5   -0.5 -4.5      2   1    0 0    0    -0.5   1.5 -2.5     1   0    1 NR1=R1+3R3…NR3=-2R3….NR4=R4-R3                          1    0    0    1    -12   5   3    0 0    1    0   -1    -1    1   0    0 0    0    1   1    9    -4   -2    0 0    0    0     2 2    -1   -1    1 NR1=R1-0.5R4…NR2=R2+0.5R4…NR3=R3-0.5R4…NR4=R4/2                          1    0    0   0    -13    5.5   3.5    -0.5 0    1    0   0      0   0.5    -0.5    0.5 0    0    1   0      8   -3.5    -1.5    -0.5 0    0    0   1      1   -0.5    -0.5    0.5 HENCE INVERSE OF A =                           -13    5.5    3.5   -0.5               0    0.5     -0.5    0.5               8    -3.5    -1.5   -0.5               1    -0.5    -0.5   0.5