Thread: Gauss-Jordan reduction to find the inverse of a matrix

Use Gauss-Jordan reduction of (A:I) to find A^-1.

The matrix is: 1 1 0/1 1 1/0 2 1 (ie 3x3 matrix)

I know you add the matrix on the right side with the diagonal 1's. I keep getting confused as to what I have to do. For example, I tried (Row3*-1/2)+row 1, then did (Row 1*2)+Row 2 and now I'm stuck. Help would be greatly appreciated!

The best thing to do is work one column at a time, working from left to right. You want to get a "1" at the pivot position (the main diagonal element for that particular column) so divide the entire row by that number. (Unless, of course, it is 0. If so, swap with a lower row. If the pivot and all lower numbers in that column are 0, the matrix is not invertible). Then, to get a 0 in that column in rows both above and below that pivot row, subtract that number times the pivot row from the row.

Here, you have


Since there is already a 1 in the pivot position (first row first column) you don't have to do anything to the first row (divide by 1). There is a 1 in second row, first column so subtract (1 times) the first row from the second column. There is already a 0 in the first column of the third row so you don't have to do anything to the third row (subtract 0 times the first row from the third row).



Now, we have a 0 in the pivot position (second row of the second column) so we have to swap the second and third rows


Since there is a 2 in the pivot position now, divide the second row by 2. Since there is a 1 in the first row of the second column, subtract 1 times the new second row from the first row. Since there is a 0 in the third row of the second column, subtract 0 times the new second row from the first row (in other words, don't change it).



The pivot position is now the third row of the third column. There is a 1 there so we don't have to do anything to the third row. There is a in the first row of the third column so we add the third row to the first row. There is a in the second row of the third column so we subtract the third row from the second row.



That is, the inverse to is

Originally Posted by HallsofIvy

The best thing to do is work one column at a time, working from left to right. You want to get a "1" at the pivot position (the main diagonal element for that particular column) so divide the entire row by that number. (Unless, of course, it is 0. If so, swap with a lower row. If the pivot and all lower numbers in that column are 0, the matrix is not invertible). Then, to get a 0 in that column in rows both above and below that pivot row, subtract that number times the pivot row from the row.

Here, you have


Since there is already a 1 in the pivot position (first row first column) you don't have to do anything to the first row (divide by 1). There is a 1 in second row, first column so subtract (1 times) the first row from the second column. There is already a 0 in the first column of the third row so you don't have to do anything to the third row (subtract 0 times the first row from the third row).



Now, we have a 0 in the pivot position (second row of the second column) so we have to swap the second and third rows


Since there is a 2 in the pivot position now, divide the second row by 2. Since there is a 1 in the first row of the second column, subtract 1 times the new second row from the first row. Since there is a 0 in the third row of the second column, subtract 0 times the new second row from the first row (in other words, don't change it).



The pivot position is now the third row of the third column. There is a 1 there so we don't have to do anything to the third row. There is a in the first row of the third column so we add the third row to the first row. There is a in the second row of the third column so we subtract the third row from the second row.



That is, the inverse to is

WOW! What a great tutorial! Great job. I'm gonna print this out and keep as a reference. Thank you so very much for your time!!