MATH 10260 Problem Sheet 8 Solutions
1. (a)


1 2 x+1
3 
D= 1 x
1 3
3
Notice that when we compute the determinant of this matrix, we
get terms which are a product containing one entry from each
row. The highest power of x in |D| comes from the product of x
and x + 1. Thus the highest power of x in the determinant is 2.
That means that |D| = 0 is a quadratic equation, which has at
most two solutions for x. We can just read off the solutions by
looking at the matrix. If x = 3, the second row is equal to the
third, and by property 5 of the class notes this means |D| = 0.
Similary, if we let x = 2, the first and second rows are the same.
So in all, the values of x we are looking for are 2 and 3.
(b)


1 1 1
F = 2 3 x 
4 9 x2
Again, we can figure out how many answers we could possibly
find. x and x2 are in the same column here so they never appear
multiplied together in the determinant. This means again the
highest power of x that we have in |F | is 2. So we’re looking for
at most two values of x. If x = 2, column 1 = column 2 and
if x = 3, column 2 = column 3. In both cases property 5 of
determinants tells us that |F | = 0
(c) Writing out DF is a bit annoying. We will get a lot of entries
which are polynomials in x. We don’t have to do this at all
though. Property 9 of determinants tells us that
|DF | = |D| |F |
. So |DF | = 0 when either |D| = 0 or |F | = 0. From parts (a)
and (b) we know that this is when x = 2 or x = 3.
2.


a1 a2 a3
A =  b1 b2 b3  ,
c1 c2 c3


a1 + 2b1 − 3c1 a2 + 2b2 − 3c2 a3 + 2b3 − 3c3

−c1
−c2
−c3
B=
b1
b2
b3
1
The property of determinants that we want to use first is number 6 in
the class notes. Determinants are additive in rows and columns. Since
the first row is
a1 + 2b1 − 3c1 a2 + 2b2 − 3c2 a3 + 2b3 − 3c3 =
a1 a2 a3 + 2b1 2b2 2b3 − 3c1 3c2 3c3
That means we have
a1
a2
a3
|B| = −c1 −c2 −c3
b1
b2
b3
2b1 2b2 2b3 −3c1 −3c2 −3c3 + −c1 −c2 −c3 + −c1 −c2 −c3 b1
b2
b3 b2
b3 b1
Now by property 2,
a1
a2
a3
|B| = −c1 −c2 −c3
b1
b2
b3
b1
b2
b3 −c1 −c2 −c3 +2 −c1 −c2 −c3 +3 −c1 −c2 −c3 b1
b2
b3 b1
b2
b3 The second and third terms have repeated rows, so they are nothing
other than zero. In the first term, we can interchange row 2 and 3 if
we multiply in front by -1, from property 3.
a1
a2
a3 b2
b3 |B| = − b1
−c1 −c2 −c3 By property 2 again
a1 a2 a3 |B| = b1 b2 b3 = |A| = 3
c1 c2 c3 We just use the property about constant multiples of rows and columns
for C:
a1
a1 a2 a3 a1
a
a
3a
a
2
3
2
3
b2
b3 = 3·(−2) b1 b2 b3 3b2
b3 = 3 b1
|C| = b1
c1 c2 c3 −2c1 −2c2 −2c3 −2c1 −6c2 −2c3 det(C) = −6 · det(A) = −18
2
Note that by property 4 about the determinant of the transpose of a
matrix
a2 a1 a3 det(D) = det(DT ) = b2 b1 b3 = −det(A)
c2 c1 c3 = -3 since the matrix above can be gotten by swapping two columns
of A.
3.
det(A−1 B T ) = det(A−1 )det(B T ) =
1
det(B) = 1
det(A)
det(CD) = det(C)det(D) = (−18)(−3) = 54
4. (a)
2
3
2
11
0 1
4 4
2 −4 −2 3
=
3 −1 0 2
8 −4 6 11
3 0
4 2 −4 −2 3 −1 0 8 −4 6 (Added the third row to the first one)
4 3 0
4 3 0
4
4
3 2 −4 −2 3 2 −4 −2 =
= 2 2 3 −1 0 = 0
2 3 −1 0 8 6 0
4 3 0
8 4 (Subtracted the second row from the last one). The final determinant is 0 because of the repeated row.
(b)
4
2 3 −4 −2 0 1 −3 3 −2 1 5 4
2 3 −4 =
−2 0 1 −3 3 −2 1 5 8 −2 6 4 8 −2 6 4 (Swap the second and third rows, then swap the first and the new
second row - overall the sign doesn’t change)
−2 0
−2 0
1 −3 1 −3 0
0
2
5 −10 2
5 −10 =
= 2 3 −2 1
10 5 6 −4 2
0 −2 10 −8 0 −2 10 −8 3
We multiply the third row by 12 so that we can add three times
the first row to it. This gives us a factor of 21 in front of the
determinant.
−2 0
−2 0 1 −3 1 −3 1 0
2
5 −10 1 0 2 5 −10 =
= 1 2 0 0 15 −19 2 0 −4 5
0 −2 10 −8 0 0 15 −18 −2 0 1 −3 1 0 2 5 −10 = = −2 · 15 = −30
2 0 0 15 −19 0 0 0
1 (c) This matrix is lower triangular, so we just multiply the diagonal
terms to get −120.
5. (a) Expand
2 0
3 2
2 3
11 8
about the first row:
1
4 2 −4 −2 3 2 −2 3 2 −4 −4 −2 = 2 3 −1 0 −0 + 2 3 0 −4 2 3 −1 −1 0 8 −4 6 11 8 6 11 8 −4 −4 6 −4 −2 2 −2 = 2 −3 + (−1) −0
−4 6 8 6 3 −2 2 −2 −0
+ 3
+ −2 11 6 8 6 3 −1 2 −1 2 3 − 2
−4 3 11 −4 + (−4) 11 8 8 −4 =2(96 − 28) + (−56 + 120) − 4(−12 − 6 + 68) = 0
(b) Expand about
4
2 3
3 −2 1
−2 0 1
8 −2 6
the third row:
−4 4 2 3
2 3 −4 4 2 −4 5 = −2 −2 1 5 −0 + 3 −2 5 −(−3) 3 −2 1
−3 8 −2 6
8 −2 4 −2 6 4 4 4
= −2 2 + 4 +3 4 −2 1
5 − 4 −2 6
4
3 −2
5 − 4 4
8 −2
3 −2
1 + 3 8 −2
6
−2
1 5 − 3 −2
6 4
3
−2 5 − 2 −2 4
8
3
−2 1 − 2 8
−2 6
= − 2(−52 − 6 + 40) + (8 + 56 − 40) + 3(−40 − 20 + 30) = 36 + 24 − 90 = −30
(c)
2
−5
3
4
0
3
2
2
0 0
0 0
4 0
1 −5
3 0 0
= 2 2 4 0
2 1 −5
= 6 4 0
1 −5
= −120
6. To find the adjoint, we just need to get the cofactor matrix and take
the transpose. Remember, the (i, j)th cofactor is the determinant of
the matrix we get when we delete the ith row and the j th column.
(a) e.g. the (1, 1)st cofactor in (a) is
2 −4 −2 3 −1 0 = −3(−32) − 1(28) = 68
8 −4 6 


68 −24 64 −50

 34 −12 32 −25 


cofactors = 
 68 −24 64 −50  , adjoint = 
−34 12 −32 25

68
34
68 −34
−24 −12 −24 12 

64
32
64 −32 
−50 −25 −50 25
(b)


−18 60
24
30
 −28 50

34
30 
 , adjoint = 
cofactors = 
 −54 120 72

60 
26 −55 −38 −30

−18
60
24
30

28 −54 26
50 120 −55 

34 72 −38 
30 60 −30
(c)



−60 −100 95 −69
−60
0
0
0
 0

 −100 −40
−40
20
−12
0
0
 , adjoint = 
cofactors = 
 0
 95
0
−30 −6 
20 −30 0
0
0
0
24
−69 −12 −6 24
5




where an is is the leading term of V (x) - the coefficient of xn−1 .
We can see exactly what the term in front of xn−1 is by calculating
the determinant in a different way. Using cofactor expansion about
the final row, the term in that expansion involving xn−1 is
1 α1
α12 . . . α1n−2 1 α2
α22 . . . α2n−2 ...
n−2 1 αn−1 α2
n−1 . . . αn−1
which is the Vandermonde determinant of one smaller size.
That gives us a way to use an induction proof. For the 2 × 2 Vandermonde matrix,
1 α1 1 α2 = α2 − α1
Which agrees with the formula we believe is true.
Assume the formula is true for all Vandermonde matrices of size up to
(n − 1) × (n − 1). Then we have just shown that the determinant of
the n × n Vandermonde matrix is
n−2 2
1 α1
α
.
.
.
α
1
1
1 α2
α22 . . . α2n−2 Y
(αn − αi )
...
i<n
n−2 1 αn−1 α2
n−1 . . . αn−1
And the determinant above is, by our induction hypothesis,
Y
Y
(αj − αi )
(αn − αi )
i<j<n
i<n
=
Y
i<j≤n
As was to be shown.
8
(αj − αi )