### Inverse

Exercise Find invertible matrices and compute the inverse.

1)

 [ 3 -2 ] 5 -3

Answer By the conclusion about the inverse of 2 by 2 matrices, since 3×(-3) - 5×(-2) ≠ 0, the matrix is invertible. The inverse is

 (3×(-3) - 5×(-2))-1[ -3 2 ] = [ -3 2 ] -5 3 -5 3

2)

 [ 3 5 ] -2 -3

Answer Since 3×(-3) - (-2)×5 ≠ 0, the matrix is invertible. The inverse is

 (3×(-3) - 5×(-2))-1[ -3 -5 ] = [ -3 -5 ] 2 3 2 3

3)

 [ 6 4 ] 9 6

Answer Since 6×6 - 4×9 = 0, the matrix is not invertible.

4)

 [ 6 4 ] 9 7

Answer Since 6×7 - 4×9 ≠ 0, the matrix is invertible. The inverse is

 (6×7 - 4×9)-1[ 7 -4 ] = [ 7/6 -2/3 ] -9 6 -3/2 1

5 - 8)

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

Answer The first three matrices are obtained by selecting columns from the fourth matrix. In an earlier exercise, we have computed the row echelon form

 [ 1 1 -1 0 -1 ] 0 1 -2 1 -2 0 0 0 3 -1

for the fourth matrix. The first matrix, being [col 1], [col 2], [col 4], can be row changed to

 [ 1 1 0 ] 0 1 1 0 0 3

which shows that the first matrix is invertible. To actually compute the inverse, we use this method. So we append the first matrix by the identity matrix.

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

The same row operations as in the earlier exercise give us

 [ 1 1 0 0 1 0 ] 0 1 1 0 0 1 0 0 3 1 -2 2

Then the operations 1/3[row 3], - [row 3] + [row 2], - [row 2] + [row 1] gives us

 [ 1 0 0 1/3 1/3 -1/3 ] 0 1 0 -1/3 2/3 1/3 0 0 1 1/3 -2/3 2/3

From this we conclude

 [ 2 0 1 ]-1 = [ 1/3 1/3 -1/3 ] 1 1 0 -1/3 2/3 1/3 0 1 1 1/3 -2/3 2/3

Similarly, the second matrix, being [col 1], [col 3], [col 4], has the following row echelon form

 [ 1 -1 0 ] 0 -2 1 0 0 3

which shows that the first matrix is invertible. To compute the inverse, we again append the identity matrix.

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

The same row operations give us

 [ 1 -1 0 0 1 0 ] 0 -2 1 0 0 1 0 0 3 1 -2 2

Then the operations 1/3[row 3], - [row 3] + [row 2], (-1/2)[row 2], [row 2] + [row 1] gives us

 [ 1 0 0 1/6 2/3 -1/6 ] 0 1 0 1/6 -1/3 -1/6 0 0 1 1/3 -2/3 2/3

From this we conclude

 [ 2 0 1 ]-1 = [ 1/6 2/3 -1/6 ] 1 1 0 1/6 -1/3 -1/6 0 1 1 1/3 -2/3 2/3

The third matrix has the following row echelon form

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

which shows that the matrix is not invertible. Finally, the fourth matrix is not invertible because it is not a square matrix.

9)

 [ 1 2 1 1 ] 2 1 1 0 0 3 1 0 3 0 1 0

Answer In an earlier exercise, we computed the row echelon form.

 [ 1 2 1 1 ] 0 3 1 0 0 0 0 1 0 0 0 0

Since [col 3] (and [row 4]) are not pivot, by the criterion for the invertibility, the matrix is not invertible.

10)

 [ 1 2 3 4 ] 0 1 2 3 0 0 1 2 0 0 0 1

Answer Since this is a square matrix with all rows (and columns) pivot, the matrix is invertible. To compute the inverse, we form

 [ 1 2 3 4 1 0 0 0 ] 0 1 2 3 0 1 0 0 0 0 1 2 0 0 1 0 0 0 0 1 0 0 0 1

Then -2[row 4] + [row 3], -3[row 4] + [row 2], -4[row 4] + [row 1] give us

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

The operations -2[row 3] + [row 2], -3[row 3] + [row 1] further give us

 [ 1 2 0 0 1 0 -3 2 ] 0 1 0 0 0 1 -2 1 0 0 1 0 0 0 1 -2 0 0 0 1 0 0 0 1

The operations -2[row 2] + [row 1] finally give us

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

We conclude

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

11)

 [ 10.2 0.3 -5.4 ] -7.9 8.2 12.4 0.8 -0.9 1.6 33.5 -44.9 38.4 -5.9 -7.3 0.1

Answer Since the matrix is not square, it is not invertible.

12)

 [ 1 1 1 1 1 ] 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0

Answer We append the identity matrix

 [ 1 1 1 1 1 1 0 0 0 0 ] 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 0 1 0 0 1 1 1 0 1 0 0 0 1 0 1 1 1 1 0 0 0 0 0 1

By -[row 1] + [row 2], -[row 1] + [row 3], -[row 1] + [row 4], -[row 1] + [row 5], we get

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

By adding [row 2] through [row 5] to [row 1], we get

 [ 1 0 0 0 0 -3 1 1 1 1 ] 0 -1 0 0 0 -1 1 0 0 0 0 0 -1 0 0 -1 0 1 0 0 0 0 0 -1 0 -1 0 0 1 0 0 0 0 0 -1 -1 0 0 0 1

By multiplying -1 to [row 2] through [row 5], we get

 [ 1 0 0 0 0 -3 1 1 1 1 ] 0 1 0 0 0 1 -1 0 0 0 0 0 1 0 0 1 0 -1 0 0 0 0 0 1 0 1 0 0 -1 0 0 0 0 0 1 1 0 0 0 -1

Thus we conclude

 [ 1 1 1 1 1 ]-1 = [ -3 1 1 1 1 ] 1 0 1 1 1 1 -1 0 0 0 1 1 0 1 1 1 0 -1 0 0 1 1 1 0 1 1 0 0 -1 0 1 1 1 1 0 1 0 0 0 -1

Remark One can see the general pattern and conclude

 [ 1 1 1 . . 1 ]-1 = [ 2 - n 1 1 . . 1 ] 1 0 1 . . 1 1 -1 0 . . 0 1 1 0 . . 1 1 0 -1 . . 0 : : : : : : : : 1 1 1 . . 0 1 0 0 . . -1