1MA01: Mathematical Methods Tutorial Sheet 4
1
1. Matrix inverses using Gauss-Jordan elimination
Determine the inverse of the matrices:
A=
2 3
3 5
!
3 4
5 7
; B=
2 3 1 0
3 5 0 1
!
!

1 3/2 1/2 0
3 5 0 1
r1/2
−→

−2 2 1

2 0 
; C= 1

0 −1 0
!
−→
1 3/2 1/2 0
0 1 −3 2
r2∗2
−→
r1−3/2r2
−→
A
3 4 1 0
5 7 0 1
!
=
5 −3
−3 2
(2)
−→
!
!
1 4/3 1/3 0
5 7 0 1
r1/3
!
1 0 5 −3
0 1 −3 2
Then
−1
!
1 3/2 1/2 0
0 1/2 −3/2 1
r2−3r1
!
−→
1 4/3 1/3 0
0 1 −5 3
r2∗3
−→
r1−4/3r2
−→
B
1
=
!
1 0 7 −4
0 1 −5 3
Then
−1
!
1 4/3 1/3 0
0 1/3 −5/3 1
r2−5r1
7 −4
−5 3
(1)
!
!
(3)
Sinéad Ryan, ryan@maths.tcd.ie, see also http://www.maths.tcd.ie/˜ryan/123.html






−2 2 1 1 0 0
r2↔r3 
−→  0 1 0 0 0 −1 

1 2 0 0 1 0
−2 2 1 1 0 0

2 0 0 1 0 

 1
0 −1 0 0 0 1
−1 4 1 1 1 0
r1+r3 

−→  0 1 0 0 0 −1 
1 2 0 0 1 0








1 −4 −1 −1 −1 0
r1∗(−1) 
0 −1 
0 0
−→  0 1

1
0
1 2
0 0
1 −4 −1 −1 −1 0
r3−r1) 
0 0
0 −1 
−→  0 1

0 6
1 1
2
0
1 0 −1 −1 −1 −4
r1+4r2) 
0 −1 
−→  0 1 0 0

0 6 1
1
2
0
1 0 −1 −1 −1 −4
r3−6r2) 
0 −1 
−→  0 1 0 0

0 0 1
1
2
6


1 0 0 0 1 2
r1+r3) 
−→  0 1 0 0 0 −1 

0 0 1 1 2 6
Then

C −1

0 1 2

=  0 0 −1 

1 2 6
(4)
2. Solving systems of equations with matrix inverses
Use the inverse of the matrix A in Question 1 to solve the following 2
different systems of linear equations by matrix multiplication.
2x + 3y = 10
3x + 5y = 15
and
2x + 3y = 4
3x + 5y = 7
Hint: first express these systems of equations in matrix form.
From Question 1 we know that the inverse of the matrix
2 3
3 5
A=
!
−1
is A
5 −3
−3 2
=
!
(5)
Now recall that a system of linear equations, such as given here, can
be expressed in matrix form as Ax = b. If you premultiply both sides
of this equation by the inverse of A then you get
Ax
A Ax
Ix
x
−1
=
=
=
=
b
A−1 b
A−1 b,
A−1 b,
using the property of matrix inverse
using the property of the identity matrix
So you see that if we know A−1 we can solve the system of equations
by just multiplying it with b.
For the particular question here,
2x + 3y = 10
3x + 5y = 15
2 3
3 5
is
!
x
y
!
=
10
15
!
.
(6)
Then by the algebra above
x
y
!
10
15
!
=
5 −3
−3 2
!
!
=
5(10) − 3(15)
−3(10) + 2(15)
=
5
0
!
yielding x = 5 and y = 0.
Similarly for the second set of equations, we can write
2x + 3y = 4
3x + 5y = 7
is
2 3
3 5
!
x
y
!
=
4
7
!
.
(7)
and now we can reuse the same A−1 as in the first set of equations to
get
x
y
!
!
=
5 −3
−3 2
=
5(4) − 3(7)
−3(4) + 2(7)
=
−1
2
4
7
!
!
!
yielding x = −1 and y = 2.
Notice that once you know A−1 solving these systems of equations by
matrix multiplication is very easy - a lot less work than using GaussJordan elimination. Also, it is sometimes possible to reuse the same
A−1 to solve a new problem, if only the right-hand side of the equations
changes.
3. Determinants and invertibility
Compute the determinant of the following matrices and comment on
whether or not they are invertible.
A=
2 1
3 6
!
; B=
2 3
6 9
!


3 1 6


; C =  −5 0 −2 
4 6 −1
(8)
det(A) = 2(6) − 1(3) = 9
det(B) = 2(9) − 3(6) = 0
det(C) = 3(0(−1) − (−2)(6)) − (1(−5(−1) − (−2)(4))) + 6((−5)(6) − 0(4)) = −157
Note A and C are invertible since their determinants are nonzero. B
is singular ie not invertible.
4. Properties of matrix inverse
Verify that
• (AB)−1 = B −1 A−1
• det(AB) = det(A)det(B)
for the matrices A and B in question 1 above.
First,
2 3
3 5
(AB) =
−1
(AB)
!
3 4
5 7
47 −29
−34 21
=
!
!
You should check this using Gaussian elimination.
Now, from question 1
−1
−1
B A
!
5 −3
−3 2
!
=
7 −4
−5 3
=
7(5) + (−4)(−3) 7(−3) + (−4)(2)
(−5)(5) + 3(−3) −5(−3) + 3(2)
=
47 −29
−34 21
!
!
as required.
Secondly,
!
21 29
det(AB) = det
34 47
= (21)(47) − (29)(34)
= 987 − 986 = 1
and calculating determinants.
det(A) = 2(5) − 3(3) = 1
(9)
det(B) = 3(7) − 4(5) = 1
(10)
and
and so det(A)det(B) = 1 ∗ 1 = 1 as required.