Math 211 Fall 2007: Solutions: HW #12
Instructor: S. Cautis
1. section 7.7, #6
Note that det(A) = ad − bc.
Suppose we add a multiple k of the first row to the second row. Then we
get
a + kc b + kd
det
= (a + kc)(d) − (b + kd)(c) = ad − bc = det(A)
c
d
Similarly if we add a multiple of the second row to the first. This proves
prop 7.7 part 1.
If we interchange the two rows then we get
c d
det
= bc − ad = − det(A)
a b
which proves part 2.
If we multiply the first row by k we get
ak bk
det
= akd − bkc = k(ad − bc) = k det(A)
c
d
and similarly with the second row. This proves part 3 of prop 7.7.
2. section 7.7, #8
We add the second row to the first then the third row to the second to get


1 −3 −4
0 3
3
4 6
5
Then we subtract four times the first row from the third to get


1 −3 −4
0 3
3
0 18 21
Now we subtract 6 times the second row from the third to get


1 −3 −4
0 3
3
0 0
3
Then the determinant is 1 · 3 · 3 = 9.
3. section 7.7, #10
We add twice the first row to the

1
0
0
third to get

2 −3
6 −2
7 −4
we then add −7/6 times the second row to the third to get


1 2
−3
0 6
−2 
7
0 0 3 −4
So the determinant is 1 · 6 · (7/3 − 4) = 14 − 24 = −10
4. section 7.7, #14
(a) If row i is a linear combination of the preceding rows then we can
subtract that linear combination from row i to get a row of all zeros.
Exchanging this row with the first row we get that the first row is full of
zeros. Hence the determinant is zero.
Similarly, if column i is a linear combination of preceding columns then
the determinant is zero.
(b) In the top left matrix the sum of the first two rows equals the third.
In the top right matrix the second row plus twice the third row equals the
first.
In the bottom left matrix the second row plus twice the first equals the
third.
In the bottom right matrix the second row plus twice the third equals the
first.
5. section 9.1, #4
p(λ) = (−4 − λ)(1 − λ) − 1 · (−2) = λ2 + 3λ − 2
so the eigenvalues are
√
−3± 17
.
2
6. section 9.1, #8
p(λ) = (6 − λ)(−9 − λ) − 10 · (−5) = λ2 + 3λ − 4
so the eigenvalues are
√
−3± 25
2
or 1 and −4.
7. section 9.1, #10
p(λ) = (1 − λ) det
3−λ
2
= (1 − λ)(λ2 − 1) = −(λ − 1)2 (λ + 1)
−4 −3 − λ
So the eigenvalues are −1 (with multiplicity one) and 1 (with multiplicity
two).
8. section 9.1, #14
We calculate
A2 =
−7 −8
16 17
Hence
2
A + 2A − 3I
−7 −8
10
8
−3 0
=
+
+
16 17
−16 −14
0 −3
0 0
=
0 0
so p(A) = 0. This result generalizes for any matrix of any size but is not
easy to prove (Cayley-Hamilton theorem).