Solutions To Linear Algebra Practice Problems Page 7
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1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18(d) We use cofactor expansion to compute this determinant:
0
0
0
3
0
2
0
0
0
2
0
0
2
0
8
1
0
2
8
1
0
7
2
=
3
1
2
2
2
1
2
2
3
2
2
2
3
4
2
2
3
6
4
1
0
2
=
3( 2)
2
2
2
2
3
4
2
2
2
2
=
3( 2)( 1)
3( 2)(2)
3
4
2
3
=
6(2) + 12(2) = 12
7. Let A be the following matrix:
1 1
1
1 2
t
A =
2
1 4 t
(a) Compute the determinant of A. (Your answer will be in terms of t.)
(b) For what values of t is A invertible?
Answer:
(a) We compute the determinant:
1 1
1
2
t
1
t
1
2
1 2
t
=
+
2
2
4
t
1
t
1
4
2
1 4 t
2
2
= (2t
4t)
(t
t) + 2
2
=
t
3t + 2
2
2
(b) A is invertible when t
3t+2 = 0. Since t
3t+2 = (t 2)(t 1),
A is invertible for t = 2, t = 1
7
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