Linear Algebra Exercises-n-Answers.pdf

164 Linear Algebra, by Hefferon
Four.I.2.23 This is how the answer was given in the cited source. The value (1−a 4 ) 3 of the determinant
is independent of the values B, C, D. Hence operation (e) does not change the value of the determinant
but merely changes its appearance. Thus the element of likeness in (a), (b), (c), (d), and (e) is only
that the appearance of the principle entity is changed. The same element appears in (f) changing the
name-label of a rose, (g) writing a decimal integer in the scale of 12, (h) gilding the lily, (i) whitewashing
a politician, and (j) granting an honorary degree.
Subsection Four.I.3: The Permutation Expansion
Four.I.3.14
(a) This matrix is singular.
1 2 3
4 5 6
∣7 8 9∣ = (1)(5)(9) |P φ 1
| + (1)(6)(8) |P φ2 | + (2)(4)(9) |P φ3 |
+ (2)(6)(7) |P φ4 | + (3)(4)(8) |P φ5 | + (7)(5)(3) |P φ6 |
= 0
(b) This matrix is nonsingular.
2 2 1
3 −1 0
∣−2 0 5∣ = (2)(−1)(5) |P φ 1
| + (2)(0)(0) |P φ2 | + (2)(3)(5) |P φ3 |
+ (2)(0)(−2) |P φ4 | + (1)(3)(0) |P φ5 | + (−2)(−1)(1) |P φ6 |
= −42
Four.I.3.15 (a) Gauss’ method gives this
∣ 2 1
∣ ∣∣∣ 3 1∣ = 2 1
0 −1/2∣ = −1
and permutation expansion gives this.
∣ 2 1
∣ ∣ ∣ ∣ ∣∣∣ 3 1∣ = 2 0
∣∣∣ 0 1∣ + 0 1
∣∣∣ 3 0∣ = (2)(1) 1 0
∣∣∣ 0 1∣ + (1)(3) 0 1
1 0∣ = −1
(b) Gauss’ method gives this
∣ ∣ 0 1 4
∣∣∣∣∣ 1 5 1
∣∣∣∣∣ 0 2 3
∣1 5 1∣ = − 1 5 1
0 2 3
0 1 4∣ = − 0 2 3
0 0 5/2∣ = −5
and the permutation expansion gives this.
0 1 4
0 2 3
∣1 5 1∣ = (0)(2)(1) |P φ 1
| + (0)(3)(5) |P φ2 | + (1)(0)(1) |P φ3 |
+ (1)(3)(1) |P φ4 | + (4)(0)(5) |P φ5 | + (1)(2)(0) |P φ6 |
Four.I.3.16
= −5
Following Example 3.6 gives this.
∣ t 1,1 t 1,2 t 1,3∣∣∣∣∣
t 2,1 t 2,2 t 2,3 = t 1,1 t 2,2 t 3,3 |P φ1 | + t 1,1 t 2,3 t 3,2 |P φ2 |
∣t 3,1 t 3,2 t 3,3 + t 1,2 t 2,1 t 3,3 |P φ3 | + t 1,2 t 2,3 t 3,1 |P φ4 |
+ t 1,3 t 2,1 t 3,2 |P φ5 | + t 1,3 t 2,2 t 3,1 |P φ6 |
= t 1,1 t 2,2 t 3,3 (+1) + t 1,1 t 2,3 t 3,2 (−1)
+ t 1,2 t 2,1 t 3,3 (−1) + t 1,2 t 2,3 t 3,1 (+1)
+ t 1,3 t 2,1 t 3,2 (+1) + t 1,3 t 2,2 t 3,1 (−1)
Four.I.3.17 This is all of the permutations where φ(1) = 1
φ 1 = 〈1, 2, 3, 4〉 φ 2 = 〈1, 2, 4, 3〉 φ 3 = 〈1, 3, 2, 4〉
φ 4 = 〈1, 3, 4, 2〉 φ 5 = 〈1, 4, 2, 3〉 φ 6 = 〈1, 4, 3, 2〉
the ones where φ(1) = 1
φ 7 = 〈2, 1, 3, 4〉 φ 8 = 〈2, 1, 4, 3〉 φ 9 = 〈2, 3, 1, 4〉
φ 10 = 〈2, 3, 4, 1〉 φ 11 = 〈2, 4, 1, 3〉 φ 12 = 〈2, 4, 3, 1〉