### 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