TA: Geoff Williams
Tutorials: Thursday 1:30-2:20 in TSH-120
E-mail:williagg@math.mcmaster.ca
Office hours: Wednesday 1:30-3:30 at the Math Help Centre
Tutorial #5
Returned Test #1
Section 3.1
Exercise 6
b). Evaluate by inspection:


a
b
c
det  a + b 2b c + b  = det A
2
2
2
Solution:


a b c
0
0
A =  b b b  −R−2−−+−−R→1 A = A
2 2 2


a b c
00
00
0
bR2
A = 1 1 1 
2R3 A = A
1 1 1


a b c
000
00
000
A =  1 1 1  −R−3−−+−−R→2 A = A
0 0 0
000
0
det A = det A
00
2b det A = det A
000
0
0 = det A = det A
00
00
0
0 = 2b det A = 2b det A = det A = det A
Notice:
• determinant doesn’t change if one row is added/subtracted to a different row
• if a row (or column) of the matrix, A is multiplied by a constant u, the
determinant of the resulting matrix is u det A
• if A has row (or column) of zeros, det A = 0
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Section 3.2
Exercise 10
Explain what can be said about det A if:
c). A3 = A
Solution:
det A3 = det A
⇒ det A(det2 A − 1) = 0
det A = 0
or
det2 A = 1 ⇒
det A = 1 or
det A = −1
d). P A = P , P is invertible
Solution:
det(P A) = det P
⇒ det P det A − det P = 0
⇒ det P (det A − 1) = 0
det P 6= 0 since P is invertible, so det A = 1
f ). A = −AT , A is n × n
Solution:
det A = det(−AT )
⇒ det A = (−1)n det A = 0
if n is odd ⇒ det A = 0
if n is even ⇒ no information about det A
2
Exercise 12
If A and B are n × n matrices, AB = −BA and n is odd, show that either
A or B has no inverse
Solution:
AB = −BA
⇒ det(AB) = det(−BA)
⇒ det A det B = (−1)n det B det A
⇒ det A det B = − det B det A
⇒ 2 det A det B = 0
⇒ det A det B = 0
⇒ det A = 0 or det B = 0
A is not invertible or B is not invertible
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Section 3.3
1 c.)
For A find cA (x), all eigenvalues, all eigenvectors, and (if possible) an
invertible matrix P s.t. P −1 AP is diagonal.


7 0 −4
A= 0 5 0 
5 0 −2


λ−7
0
4
λ−5
0 
λI − A =  0
−5
0
λ+2
cA (x) = det(λI − A)
= (λ − 7)(λ − 5)(λ + 2) + 4(5(λ − 5))
= (λ − 5)((λ − 7)(λ + 2) + 20)
= (λ − 5)(λ2 − 5λ + 6)
= (λ − 5)(λ − 2)(λ − 3)
This gives us that the 3 eigenvalues are:
λ1 = 5
λ2 = 2
λ3 = 3
Solving for the λ1 eigenvector X1 ,

−2 0
 0 0
−5 0

1 0

→ −5 0
0 0

1 0
→ 0 0
0 0
4

4 0
0 0 
7 0

−2 0
7 0 
0 0

−2 0
3 0 
0 0

1 0 0 0
→ 0 0 1 0 
0 0 0 0
 
0

X1 = t 1 
0

In a similar fashion we find that λ2 and λ3 have as eigenvectors,
 4 
 
1
5
X2 = t  0 
X3 = t  0 
1
1
So in the end we have that

 

0 45 1
5 0 0
P −1 A  1 0 0  =  0 2 0 
0 1 1
0 0 3
3.1 #6, 3.2 #10 c.) d.) f.) & 12 originally typed up by Olga Krylova. Edited
by Geoff Williams.
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