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=199334It 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
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