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Exercise For what choices of the parameters, the following matrices are invertible?

1)

[ a + 1 a - 1 ]
a a + 2

Answer The condition for the matrix to be invertible is (a + 1)(a + 2) - a(a - 1) ≠ 0. Since (a + 1)(a + 2) - a(a - 1) = 4a + 2, we see that the condition is a ≠ -1/2.

2)

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

Answer The matrix appears as [col 1], [col 2], [col 4], [col 5] of the matrix in an earlier exercise. According to the computation there, row operations change the matrix into

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

By the criterion, we need all columns (and rows) to be pivot in order for the matrix to be invertible. Thus the condition is a ≠ 3/2.

3)

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

Answer The matrix appears as the coefficient matrix of the system in an earlier exercise. According to the computation there, row operations change the matrix into

[ 1 1 -1 ]
0 -1 1 - a
0 0 2a - a2

The condition for the matrix to be invertible is that all columns (and rows) are pivot, which means 2a - a2 ≠ 0, i.e., a ≠ 0 or 2.

4)

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

Answer According to the computation in an earlier exercise, row operations change the matrix into

[ 1 0 a 1 ]
0 1 1 a
0 0 -a - 1 -a - 1
0 0 0 -a2 + a + 2

The condition for the matrix to be invertible is - a - 1 ≠ 0 and - a2 + a + 2 ≠ 0, which means a ≠ -1 or 2.

5)

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

Answer By (-a)[row 1] + [row 2], (-a)[row 2] + [row 3], (-a)[row 3] + [row 4], the matrix becomes the identity. Therefore the matrix is invertible for any a.

6)

[ 1 1 1 1 ]
a 1 1 1
a a 1 1
a a a 1

Answer By (-a)[row 1] + [row 2], (-a)[row 1] + [row 3], (-a)[row 1] + [row 4], the matrix becomes

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

Therefore the matrix is invertible for a ≠ 1.

7)

[ a b b b ]
a a b b
a a a b
a a a a

Answer By - [row 3] + [row 4], - [row 2] + [row 3], - [row 1] + [row 2], the matrix becomes

[ a b b b ]
0 a - b 0 0
0 0 a - b 0
0 0 0 a - b

Therefore the matrix is invertible for a ≠ 0, b.