1MA01: Mathematical Methods Tutorial Sheet 3
1
1
Sinéad Ryan, ryan@maths.tcd.ie, see also http://www.maths.tcd.ie/ ryan/123.html
1. Write down the 3 × 3 identity matrix, denoted I.
Given the 3 × 3 matrix,


1 2 5

A= 6 0 1 
,
3 2 7
(1)
show, by calculation, that AI = IA = A.


1 0 0


I= 0 1 0 
0 0 1


(2)

1 0 0
1 2 5



AI =  6 0 1   0 1 0 
0 0 1
3 2 7


1(1) + 2(0) + 5(0) 1(0) + 2(1) + 5(0) 1(0) + 2(0) + 5(1)

=  6(1) + 0(0) + 1(0) 6(0) + 0(1) + 1(0) 6(0) + 0(0) + 1(1) 

3(1) + 2(0) + 7(0) 3(0) + 2(1) + 7(0) 3(0) + 2(0) + 7(1)




1 2 5


=  6 0 1 
3 2 7
= A
and

1 0 0
1 2 5


IA =  0 1 0   6 0 1 

0 0 1
3 2 7


1(1) + 0(6) + 0(3) 1(2) + 2(1) + 0(0) + 0(2) 1(5) + 0(1) + 0(7)

0(2) + 1(0) + 0(2)
0(5) + 1(1) + 0(7) 
=  0(1) + 1(6) + 0(3)

0(5) + 0(1) + 1(7)


1 2 5

= 
 6 0 1 
3 2 7
= A
as required, by matrix multiplication.
2. Sketch the lines represented by the two equations below.
Solve for a and b, by writing in matrix form and using Gauss-Jordan
elimination, the following system of equations.
5a − b = 14
2a + 3b = 26
Check your answer is correct.
20
15
10
2a+3b=26
b
5
0
-b
-5
4
=1
5a
-10
-15
-20
0
2
5 −1 14
2 3 26
4
!
a
r1−2r2
−→
r2−2r1
−→
r2/17
−→
6
8
10
−→
Giving the solutions, a=4 and b=6.
Check this is correct by substitution.
!
!
!
1 −7 −38
0 17 102
1 −7 −38
0 1 6
r1+7r2
1 −7 −38
2 3 26
1 0 4
0 1 6
!
3. Use Gauss-Jordan elimination to solve the following equations for x, y, z.
x + y + 2z = 5
x + y + z = −10
2x + 3y + 4z = 2
Check your answer is correct.
The augmented matrix is


1 1 2 5


 1 1 1 −10 
2 3 4 2


1 1 2 5
r2−r1 
−→  0 0 −1 −15 

2 3 4 2


1 1 2 5
(−1)r2↔r3 

−→  2 3 4 2 
0 0 1 15





1 1 2 5
r2−2r1 
−→  0 1 0 −8 

0 0 1
15
1 0 2 13
r1−r2 

−→  0 1 0 −8 
0 0 1 15

1 0 0 −17
r1−2r3 
−→  0 1 0 −8 

0 0 1
15
giving the solutions x = −17, y = −8, z = 15.
4. Consider the system of equations
2x − y = −5
−2x + y = −2
Sketch the two lines represented by these equations. What do you
notice?
Can you solve for x and y using Gauss-Jordan elimination? What
happens?
20
15
y
-5
2x-y=
10
=-2
-2x+y
5
0
0
2
x
4
6
The lines are parallel, non-intersecting. There is no solution.
Trying Gauss-Jordan:
2 −1 −5
−2 1 −2
!
r2+r1
−→
2 −1 −5
0 0 −7
!
No further progress can be made with E.R.Os ie no solution via GaussJordan.