We start with an augmented matrix table:
R1 |
6 |
2 |
-2 |
1 |
0 |
0 |
R2 |
2 |
5 |
0 |
0 |
1 |
0 |
R3 |
-2 |
0 |
7 |
0 |
0 |
1 |
This table is made up of the given matrix A in columns 2-4 and the identity matrix I in columns 5-7. The Gauss-Jordan method is a row-manipulation method resulting in I in columns 2-4 and the inverse matrix B=A-1 in columns 5-7, by applying the following steps:
- R1/6→R1, R2/5→R2, R3/7→R3 (this step initially produces in columns 2-4 the diagonal 1s of the identity matrix, but this result becomes temporarily distorted by the steps that follow)
- R1-R2/3→R1, R2-2R1/5→R2, R3+2R1/7→R3 (from now on the steps attempt to get zeroes in all the other cells in columns 2-4)
- 15R1/13→R1, 15R2/13→R2, 21R3/19→R3
- 5R2/2+R1→R1, R2-2R3/13→R2, R3-2R2/19→R3
- R1-1235R2/486→R1, 247R2/243→R2, 247R3/243→R3
These steps on the three rows of the matrix give the inverse B from columns 5-7:
⎛ 35/162 -7/81 5/81 ⎞
⎜ -7/81 19/81 -2/81 ⎟
⎝ 5/81 -2/81 13/81 ⎠
which can be written:
⎛ 35 -14 10 ⎞
⎜ -14 38 -4 ⎟ ➗ 162
⎝ 10 -4 26 ⎠
The matrix product AB gives us:
⎛ 6×35+2×(-14)-2×10=162 6×(-14)+2×38-2×(-4)=0 6×10+2×(-4)-2×26=0 ⎞
⎜ 2×35+5×(-14)+0×10=0 2×(-14)+5×38+0×(-4)=162 2×10+5×(-4)+0×26=0 ⎟ ➗ 162 = I
⎝ -2×35+0×(-14)+7×10=0 -2×(-14)+0×38+7×(-4)=0 -2×10+0×(-4)+7×26=162 ⎠
Therefore AB=I, which confirms that B is the inverse of A.