A First Course in Linear Algebra, 2017a

548 Selected Exercise Answers
3.1.23 ∣ ∣∣∣∣∣ 2 1 3
2 4 2
1 4 −5
∣ = −32
3.1.24 One can row reduce this using only row operation 3 to
⎡
⎤
1 2 1 2
0 −5 −5 −3
⎢
9
⎣
0 0 2 ⎥ 5 ⎦
0 0 0 − 63
10
and therefore, the determinant is −63.
∣
1 2 1 2
3 1 −2 3
−1 0 3 1
2 3 2 −2
= 63
∣
3.1.25 One can row reduce this using only row operation 3 to
⎡
⎤
1 4 1 2
0 −10 −5 −3
⎢
19
⎣ 0 0 2 ⎥ 5 ⎦
0 0 0 − 211
20
Thus the determinant is given by ∣ ∣∣∣∣∣∣∣ 1 4 1 2
3 2 −2 3
−1 0 3 3
2 1 2 −2
⎡
3.2.1 det⎣
1 2 3
0 2 1
3 1 0
⎤
= 211
∣
⎦ = −13 and so it has an inverse. This inverse is
⎡
2 1
1 0
1
−13
−
2 3
⎢ 1 0
⎣
∣ 2 3
2 1 ∣
−
0 1
3 0
1 3
3 0
−
∣ 1 3
0 1 ∣
0 2
⎤
3 1
−
1 2
3 1
∣ 1 2
⎥
⎦
0 2 ∣
T
=
=
⎡
1
⎣
−13
⎡
⎢
⎣
−1 3 −6
3 −9 5
−4 −1 2
1
13
−
13
3
− 3
13
9
13
4
13
1
13
6
13
− 5
13
− 2
13
⎤
⎥
⎦
⎤
⎦
T