SolutionDeterminants

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Solutions - Determinants
V
det(A) = 2
∣∣∣∣∣ 6 50 7
∣∣∣∣∣− 1
∣∣∣∣∣ 3 5−4 7
∣∣∣∣∣− 2
∣∣∣∣∣ 3 6−4 0
∣∣∣∣∣
= 2(6× 7− 0) + 1(3× 7− 5× (−4))− 2(0− 6× (−4))
= 2× 42 + 1× 41− 2× 24 = 77.
VI
We can write
(ABC)T = ((AB)C)T
because the matrix multiplication is associative. Now using the property (AB)T = BT AT , then we
have:
(ABC)T = ((AB)C)T = CT (AB)T
Again using this property
(ABC)T = ((AB)C)T = CT (AB)T = CT BT AT
So to prove that (ABC)T = CT BT AT , we use the associativity rule and the property (AB)T = BT AT .
VIII
∣∣∣∣∣∣∣∣
1 2 3
4 5 6
8 8 9
∣∣∣∣∣∣∣∣ = 1×
∣∣∣∣∣ 5 68 9
∣∣∣∣∣− 2×
∣∣∣∣∣ 4 68 9
∣∣∣∣∣ + 3×
∣∣∣∣∣ 4 58 8
∣∣∣∣∣ = 1× (45− 48)− 2× (36− 48) + 3× (32− 40) = −3
∣∣∣∣∣∣∣∣∣∣
1 1 2 1
3 1 4 5
7 6 1 2
1 1 3 4
∣∣∣∣∣∣∣∣∣∣
= 60
IX
1. We multiply out each of the bracketed terms in turn, and then re-assemble the calculation.
Firstly, we have
(A−1B2)−1 = (B2)
−1
(A−1)
−1
= (BB)−1(A−1)
−1
= B−1B−1A
since (A−1)−1 = A. Next, we get
(ABA−1)−1 = (A−1)
−1
B−1A−1 = AB−1A−1,
2
so
(A−1B2)−1(BA−1)(ABA−1)−1(BA)(A−1B)
= (B−1B−1A)(BA−1)(AB−1A−1)(BA)(A−1B)
= (B−1B−1A)B(A−1A)(B−1A−1)B(AA−1)B
= (B−1B−1A)B(I)B−1A−1)B(I)B
= (B−1B−1A)BB−1A−1BB
= (B−1B−1A)IA−1BB = B−1B−1(AA−1)BB
= B−1B−1(I)BB = B−1(B−1B)B = B−1IB
= B−1B = I.
2. Proof by induction:
• we �rst verify that what is to demonstrate is true for the �rst values of n, so lets consider
n = 2:
A =
[
a11 0
a21 a22
]
then
det(A) = a11a22 − a21 × 0 = a11a22
So for n = 2, it is true that if aij = 0 for i < j then
det(A) = a11a22 · · · ann =
n∏
i=1
aii
for A = [aij ].
• we verify again (just to be fully convinced) for n = 3:
det(A) =
∣∣∣∣∣∣∣∣
a11 0 0
a21 a22 0
a31 a32 a33
∣∣∣∣∣∣∣∣ = a11 ×
∣∣∣∣∣ a22 0a32 a33
∣∣∣∣∣− 0×
∣∣∣∣∣ a21 0a31 a33
∣∣∣∣∣ + 0
∣∣∣∣∣ a21 a22a31 a32
∣∣∣∣∣ = a11a22a33
So for n = 3, it is true that if aij = 0 for i < j then
det(A) = a11a22 · · · ann =
n∏
i=1
aii
for A = [aij ].
• now we cannot check for every n that it is true (because we would need to check for an
in�nite number of values n). So now let assume it is true for a matrix of size n× n, and
use this assumption to show it is true for a matrix of size (n + 1)× (n + 1):
� so we assume ∣∣∣∣∣∣∣∣∣∣∣
a11 0 0 · · · 0
a21 a22 0 · · · 0
...
. . .
...
an1 · · · ann
∣∣∣∣∣∣∣∣∣∣∣
= a11a22 · · · ann
3
� Now lets calculate:∣∣∣∣∣∣∣∣∣∣∣∣∣
a11 0 0 · · · 0
a21 a22 0 · · · 0
...
. . .
...
an1 · · · ann 0
an+1,1 · · · an+1,n+1
∣∣∣∣∣∣∣∣∣∣∣∣∣
= a11
∣∣∣∣∣∣∣∣∣∣∣
a22 0 · · · 0
. . .
...
an,2 ann 0
an+1,2 · · · an+1,n+1
∣∣∣∣∣∣∣∣∣∣∣
The other terms of the determinant are multiplied by 0 (and therefore not even
written). So now the determinant of the (n + 1) × (n + 1) matrix depends of the
determinant of a n× n matrix having all its coef�cients above the diagonal being 0.
According to the assumption we made, we know this determinant to be equal to the
product of the elements in this diagonal so:∣∣∣∣∣∣∣∣∣∣∣∣∣
a11 0 0 · · · 0
a21 a22 0 · · · 0
...
. . .
...
an1 · · · ann 0
an+1,1 · · · an+1,n+1
∣∣∣∣∣∣∣∣∣∣∣∣∣
= a11(a22 · · · annan+1,n+1) =
n+1∏
i=1
aii
• So now we now that:
if aij = 0 for i < j then
det(A) = a11a22 · · · ann =
n∏
i=1
aii
for A = [aij ].
is true for n = 2, 3 and if it is true for n then it is true for n + 1. So by induction, we can
say it is true for any n.