Course 111: Algebra, 21st Feb 2007
To be handed in at tutorials on Feb 26th and 27th.
1. Prove the following:
(i) if a matrix A is invertible and symmetric then A−1 is symmetric.
Note that this proof uses the result that for AT invertible then (AT )−1 =
(A−1 )T . To prove this you should show that AT (A−1 )T = AT (AT )−1 =
I, where I is the identity matrix.
To see this, consider AT (A−1 )T = (A−1 A)T = I T = I. Similarly,
(A−1 )T AT = (A(A−1 ))T = I T = I.
The proof required then follows.
Proof
Write (A−1 )T = (AT )−1 = (A)−1 = A−1 .
(ii) if a matrix A is invertible then AAT and AT A are both invertible.
Proof
If A is invertible then AT is invertible since the inverse of AT is just
(A−1 )T . We also have that the product of invertible matrices is invertible and so both AAT and AT A are invertible.
2. Use LU decomposition to find a solution to the following system of
linear equations
−2x1 + 4x2 − 3x3 = −1
3x1 − 2x2 + x3 = 17
−4x2 + 3x3 = −9
Reducing the coefficient matrix to row-echelon form:


−2 4 −3


 3 −2 1 
0 −4 3
1 −2 23
1/4R2 

→  0 1 − 87 
0 −4 3


Then
1 −2 23
1 −2 23

 R2 −3R1 
7 
 2 −2 1  →  0 4 − 2 
0 −4 3
0 −4 3
−1/2R1

R3 +4R2

→
→


1 −2 23
1 −2 32

7  −R3 
7 
 0 1 −8  →  0 1 −8  = U
0 0 − 21
0 0
1




−2 0
0

4
0 
L= 3
.
1
0 −4 − 2
Now, to solve the system, first solve





−2 0
0
y1
−1




4
0   y2  =  17 
 3
.
1
0 −4 − 2
y3
−9
to get
1
2
31
3y1 + 4y2 = 17 ⇒ y2 =
8
1
−4y2 − y3 = −9 ⇒ y3 = −13
2
−2y1 = 1 ⇒ y1 =
Then solve
1
1 −2 23
x1
2

 31 

7 
 0 1 − 8   x2  =  8 
0 0
1
x3
−13







to get
x3 = −13 ⇒ x3 = −13
31
15
7
⇒ x2 = −
x2 − x3 =
8
8
2
1
3
⇒ x1 = 5.
x1 − 2x2 + x3 =
2
2
3. Prove that for A and B, both n×n matrices, det (AB) = det(A)det(B).
(see lecture notes for a hint on how to start.)
Proof
A complete proof is given in Dr Colm Ó Dúnlaing’s online notes for
this course (from last year). Page 45 of
http://www.maths.tcd.ie/~odunlain/111/book.pdf