A First Course in Linear Algebra, 2017a

550 Selected Exercise Answers
3.2.9
⎡
det⎣
e t cosht sinht
e t sinht cosht
e t cosht sinht
⎤
⎦ = 0
and so this matrix fails to have a nonzero determinant at any value of t.
3.2.10
⎡
det⎣
and so this matrix is always invertible.
e t e −t cost e −t sint
e t −e −t cost − e −t sint −e −t sint + e −t cost
e t 2e −t sint −2e −t cost
⎤
⎦ = 5e −t ≠ 0
3.2.11 If det(A) ≠ 0, then A −1 exists and so you could multiply on both sides on the left by A −1 and obtain
that X = 0.
3.2.12 You have 1 = det(A)det(B). Hence both A and B have inverses. Letting X be given,
A(BA − I)X =(AB)AX − AX = AX − AX = 0
and so it follows from the above problem that (BA − I)X = 0. Since X is arbitrary, it follows that BA = I.
3.2.13
Hence the inverse is
=
⎡
det⎣
e −3t ⎡
⎣
⎡
⎣
e t 0 0
0 e t cost e t sint
0 e t cost − e t sint e t cost + e t sint
⎤
⎦ = e 3t .
e 2t 0 0
0 e 2t cost + e 2t sint − ( e 2t cost − e 2t sin ) t
0 −e 2t sint e 2t cos(t)
e −t 0 0
⎤
0 e −t (cost + sint) −(sint)e −t ⎦
⎤
⎦
T
3.2.14
=
⎡
⎣
⎡
⎣
e t cost sint
e t −sint cost
e t −cost −sint
⎤
⎦
−1
1
2 e−t 1
0 2
e −t ⎤
1
⎦
2 sint − 1 2 cost cost − 1 2 cost − 1 2 sint
2 cost + 1 2 sint −sint 1
2
sint − 1 2 cost
1
3.2.15 The given condition is what it takes for the determinant to be non zero. Recall that the determinant
of an upper triangular matrix is just the product of the entries on the main diagonal.