math111_logo Inverse


Exercise Compute the inverse of the following matrices. Then use your result to solve the following systems.

1)

[ 2 3 ]
7 9
2x1 + 3x2 = 2 , 2x1 + 3x2 = 1
7x1 + 9x2 = -1 7x1 + 9x2 = 0

Answer The inverse of the matrix is

(2×9 - 7×3)-1[ 9 -3 ] = [ -3 1 ]
-7 2 7/3 -2/3

Since

[ 2 3 ]-1[ 2 ] = [ -3 1 ] [ 2 ] = [ -7 ]
7 9 -1 7/3 -2/3 -1 16/3

the solution of the first system is x1 = -7, x2 = 4. Similarly, the solution of the second system is

[ x1 ] = [ 2 3 ]-1[ 1 ] = [ -3 ]
x2 7 9 0 7/3

2)

[ 1 0 2 ]
2 -1 3
4 -1 8
x1   + 2x3 = 1 , x1   + 2x3 = 0
2x1 - x2 + 3x3 = 0 2x1 - x2 + 3x3 = 2
4x1 - x2 + 8x3 = -1 4x1 - x2 + 8x3 = 1

Answer Row operations change

[ 1 0 2 1 0 0 ]
2 -1 3 0 1 0
4 -1 8 0 0 1

into

[ 1 0 0 5 2 -2 ]
0 1 0 4 0 -1
0 0 1 -2 -1 1

Thus the inverse of the matrix is

[ 5 2 -2 ]
4 0 -1
-2 -1 1

and the solution to the two systems are respectively

[ x1 ] = [ 5 2 -2 ] [ 1 ] = [ 7 ]
x2 4 0 -1 0 5
x3 -2 -1 1 -1 -3
[ x1 ] = [ 5 2 -2 ] [ 0 ] = [ 2 ]
x2 4 0 -1 2 -1
x3 -2 -1 1 1 -1

3)

[ 2 0 2 1 ]
1 1 -1 0
0 1 -2 -1
1 -1 2 1
2x1   + 2x3 + x4 = 1
x1 + x2 - x3   = 0
  x2 - 2x3 - x4 = 1
x1 - x2 + 2x3 + x4 = 0

Answer The inverse of the matrix is

[ 0 0 1 1 ]
1 0 -1 -2
1 -1 0 -1
-1 2 -2 0

The solution of the system is

[ x1 ] = [ 0 0 1 1 ] [ 1 ] = [ 1 ]
x2 1 0 -1 -2 0 0
x3 1 -1 0 -1 1 1
x4 -1 2 -2 0 0 -3