find inverse of 3*3 matrx whose row elements are (6,2,-2)(2,5,0)(-2,0,7) using gausd jordan method

Question

Find the inverse and check it as A*A(inverse)=identity matrix

Answer 1

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:

1. 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)
2. 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)
3. 15R1/13→R1, 15R2/13→R2, 21R3/19→R3
4. 5R2/2+R1→R1, R2-2R3/13→R2, R3-2R2/19→R3
5. 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.