Further Pure 1
Revision Topic 2: Introduction to Matrices
The OCR syllabus says that candidates should:
(a) carry out operations of matrix addition, subtraction and multiplication, and recognise the terms
null (or zero) matrix and identity (or unit) matrix;
(b) recall the meaning of the terms singular and non-singular as applied to square matrices, and,
for 2 × 2 matrices evaluate determinants and find inverses of non-singular matrices;
(c) understand and use the result, for non-singular matrices, that (AB)-1 = B-1 A-1.
(d) formulate a problem involving the solution of 2 linear simultaneous equations in 2 unknowns
as a problem involving the solution of a matrix equation.
(e) understand the cases that may arise concerning the consistency or inconsistency of 2 linear
simultaneous equations, relate them to the singularity or otherwise of the corresponding square
matrix, and solve consistent systems.
Section 1: Introduction to matrices



 2 3 0 
A matrix is a rectangular array of numbers arranged in rows and columns, e.g. 
.
 1 1 7 
A matrix is usually denoted by a bold capital letter, e.g. A (or A when hand-written).
 2 3 0 
The matrix 
 has 2 rows and 3 columns and so is described as a 2  3 matrix. The
 1 1 7 
order of this particular matrix is said to be 2  3.
In general a m  n matrix will have m rows and n columns.
1.1


Special matrices
5
A matrix with a single column, e.g.  2  , is called a column vector. Likewise a matrix with
6
 
just one row, e.g. (1 -4), is called a row vector. We usually label column vectors and row
vectors by lower case bold letters, e.g.
1
 2
a   2 1 9  , b    .
6
 
 3
If a matrix has an equal number of rows and columns, it is called a square matrix, e.g.
 0 7
 2 4  is an example of a 2  2 square matrix.



A zero matrix (or null matrix) has all its entries as 0. For example, a 4  2 zero matrix is
0
0
0
0

0
0
0
.
0

0

A diagonal matrix has zeros everywhere except on the leading diagonal, e.g.
9 0

 2 0 0


0 4
 0  1 0  or 
0 0
 0 0 6



0 0


0 

0 0 
.
7 0 

0  2 
0
An identity matrix is a diagonal matrix that has 1’s on the leading diagonal,
1 0 0


1 0
 and the 3  3 identity matrix is I   0 1 0  .
e.g. the 2  2 identity matrix is I  
0 1
0 0 1


1.2
General notation
 a11

 a 21
A general m  n matrix can be written as A   a 31

 

 a m1
a12
a13
a 22
a 23
a 32
a 33


am2
a m3
 a1n 

 a 2n 
 a 3n  .

  

 a mn 
The entry (or element) in the ith row and jth column is denoted a ij .
1.3
Addition and subtraction of matrices
Two matrices can only be added or subtracted if they have the same size (i.e they have the same
number of rows and columns).
To add (or subtract) two matrices, we add (or subtract) the corresponding entries.
Example:
 3 2 1 3 
 2 3 7 6 


If A  2 4 3 0 and B   4 5 0 2  then
 6 0 1 5 
 2 4 3 5 




 5 1 8 3 
A + B =  2 9 3 2  and
 8 4 4 10 


 1 5 6 9 
A – B = 6 1 3 2 .
 4 4 2 0 


1.4
Multiplication by a scalar
The matrix kA, where k is a scalar (i.e. a single number), is obtained by multiplying all of the
entries in matrix A by k.
Example:
  2 3


If A =  0 1  then 2A =
 4 2


  4 6


 0 2  and -3A =
 8 4


 9
 6


 3 .
 0
  12  6 


1.5
Multiplication of matrices
For two matrices A and B, the product AB is only defined if the number of columns of A is the
same as the number of rows in B.
To calculate AB we multiply each row of A by each column of B.
We can illustrate this process by an example:
1 2 


Suppose A =  3  1 and B =
2 5 


 5 3  2


4 1 2 
then
1 3  2 1
1  (2)  2  2   13 5 2 
1 2 
 1 5  2  4

 5 3  2  
 

   3  5  (1)  4 3  3  (1)  1 3  (2)  (1)  2    11 8  8  .
AB =  3  1
 2 5  4 1 2   2  5  5  4
2  3  5 1
2  (2)  5  2   30 11 6 



General result
If A is of size m  n and B is of size n  p then AB is of size m  p,
i.e. (m  n)  (n  p) = (m  p)
Example:
1 3 2
2 0
 4




If A = 4 2 5 , B  1 5 and C   1  then
 1 2 3 
 2 3 
 2




 



 1 3 2  2 0   2  3  4 0  15  6 
AB =  4 2 5  1 5    8  2  10 0 10 15  =
 1 2 3  2 3   2  2  6 0  10  9 


 

9 
 3
 20 25  .
 10 19 


 1 3 2  4   4  3  4   11 
AC =  4 2 5  1  = 16  2  10  =  24  .
 1 2 3  2   4  2  6   8 

  
  
BC cannot be calculated as the number of columns in B does not equal the number of rows in C.
Note:
AB will not in general equal BA. If AB exists, then BA does not necessarily exist. This means that
matrix multiplication is not commutative.
Note 2:
Pre- or post-multiplying a matrix by the identity matrix (of appropriate size) will not alter it.
 4  1
 4  1

 1 0 

   2 5  and
E.g.  2 5  
 3 1  0 1  3 1 





 1 0 0  4  1  4  1


 

 0 1 0  2 5    2 5  .
 0 0 1  3 1   3 1 


 

So we have the result that for a general matrix A,
IA = AI = A
1.6
The transpose of a matrix
To find the transpose of a matrix A you swap the rows and columns over.
Example:
 2 0
 2 1 2 
If A   1 3  then the transpose of A, written AT, is AT  
.
 2 4 
0 3 4 


Example 2:
 4 4 2
 4 5 7 
T 


If B = 5 9 6 , then B  4 9 3  .
 7 3 1 
2 6 1 




Examination Question 1 (AQA 2005)
The matrices A and B are defined by
 3 4
A

 4 3
(a)
(b)
 0 2
B

 2 0
Calculate the matrices:
(i)
A + B;
(ii)
AB.
Show that A + B – AB = kI, where k is an integer and I is the 2 × 2 identity matrix.
Examination Question 2: AQA January 2003
The matrices A and B are defined by
 1 1
A
,
 1 2 
 1 0 3
B
.
 0 1 1
Find the matrices:
(a)
AB;
(b)
BTAT.
Section 2: Determinants and inverses of 2 by 2 matrices
2.1
Determinant of a 2 by 2 matrix
a b
The determinant of the 2 × 2 matrix A  
 is defined as the number
c d
det(A) = ad – bc.
Example 1:
 2 3
If A  
 , then det(A) = (2×5) – (3×1) = 10 – 3 = 7.
 1 5
Example 2:
 2 1
If B  
 , then det(B) = (2×3) – (-1×4) = 6 + 4 = 10.
4 3 
2.2
Singular matrices
A matrix is called singular if its determinant is 0.
Example:
 3 2 
A= 
 is a singular matrix since det(A) = 12 – 12 = 0.
 6 4 
Worked examination style question
3 
k
Find the restrictions needed on the value of k if the matrix A  
is to be non-singular.
4
k

4 

Solution: A is non-singular if det(A) ≠ 0.
But
det(A) = k(k – 4) – 12 = k2 – 4k – 12 = (k + 2)(k – 6).
So we require:
(k + 2)(k – 6) ≠ 0
i.e.
k ≠ -2 and k ≠ 6.
Examination style question
 2c 2 
Find the values of c for which the matrix 
 is singular.
 4 c 
Examination style question
2
Simplify x
2x
2 .
x4
Note: |A| is another way of writing det(A).
2.3
The inverse of a 2  2 matrix
Suppose that two (square) matrices A and B are such that
AB = BA = I.
A is then called the inverse of B (and B is the inverse of A). We write A  B 1 and B  A 1 .
Example:
 3 1  2  1  1 0 

  
.
 5 2   5 3   0 1 
Note that 
 3 1   2  1
 is 
 and the inverse of
5 2   5 3 
So the inverse of 
 2  1  3 1 

 is 
 .
  5 3  5 2
General result
a b
There are two steps in order to find the inverse of a general matrix A = 
:
c d 
Step 1: Swap the two elements on the leading diagonal and change the sign of the other two
entries, giving the matrix:
 d b 
 c a  .


Step 2: Divide each of the entries in this new matrix by the determinant of A.
a b
So we have that the inverse of 
 is:
c d 
a b 
c d 


1

1  d b 

.
ad  bc  c a 
Note: When the determinant of a matrix is zero, the inverse matrix is does not exist. So we can
only find inverses for non-singular matrices.
Example:
 4 1 
Find the inverse of B = 
.
2 2 
Solution:
The determinant of B is det B = (4  2) – (1  2) = 10.
So the inverse of B is
B 1 
1  2 1

.
10  2 4 
Note: As a check we could work out BB1 and make sure this product gives the identity matrix.
Example 2:
1 0 
Find the inverse of C = 
.
 6 5 
As det C = 5, the inverse matrix is
C1 
1  5 0  1  5 0  1  5 0 

 
 
.
5  6 1  5  6 1  5  6 1
2.3
Finding the inverse of a product of matrices
You should be aware of the result that
(AB)-1 = B-1A-1.
We can show that this result is true, since
( AB)B 1A 1  ABB 1A 1
 AIA 1
as B-1 is the inverse of B
 AA 1
I
as I is the identity matrix
So the inverse of AB is B-1A-1.
Examination style question:
Let A and B be the matrices:
 2 3
A

 1 1 
3 4
B
.
1 1
a) Calculate ABA1 and AB1A1 . Verify that (ABA1)1 = AB1A1 .
b) Prove that the result (ABA1)1 = AB1A1 is true for any non-singular matrices A and B.
Solution:
1 1 3 
 2 3
A
 A 1  


5 1 2 
 1 1 
1  1 4   1 4 
3 4
B
 B 1  



1
1
1  1 3   1 3 


So
 2 3  3 4   1 1 3  
ABA 1  

 

 1 1  1 1   5 1 2  
1  9 11  1 3 
 


5  2 3  1 2 
(working out AB first)
1  20 5 
 

5  5 0 
 4 1


 1 0 
Likewise
 2 3  1 4   1 1 3  
AB1A 1  

 

 1 1  1 3   5 1 2  
=
1  1 1  1 3 



5  2 7  1 2 
So
AB1A1 =
1  0 5   0 1 


.
5  5 20   1 4 
 4 1
(ABA ) =  

 1 0 
1 1
1

1  0 1   0 1 



1  1 4   1 4 
Therefore (ABA1)1 = AB1A1 .
Finally we need to prove the result (ABA1)1 = AB1A1 more generally.
Using the result (ABC)1  C1B1A1 we see that
 
(ABA1)1 = A1
1
B1A1  AB1A1 .
Examination question:
1 3
a c
The matrices M and U are given by M  
 and U   0 b  , where ab ≠ 0.
0
2




-1
(i)
Find M .
(ii)
a d 
Show that MUM-1 is of the form 
 and find an expression for d in terms of a, b and c.
0 b
(iii)
Given that D = MUM 1 , express D-1 similarly in terms of M and U.
2.4
Solving matrix equations
Solving the equation AX = B
Consider an equation of the form AX = B, where A and B are known matrices and X is the matrix
we want to find.
This matrix equation can be solved for X provided that A is a non-singular matrix, i.e. provided that
A-1 exists.
To find X, we multiply the equation AX = B through by A-1:
A-1AX = A-1B
i.e.
IX = A-1B
So
X = A-1B
Solving the equation XA = B
This matrix equation can be solved for X provided that the inverse of A exists. This time to find X
we need to multiply through by A-1 on the right hand side:
XAA-1= B A-1
i.e.
XI = B A-1
so
X = B A-1
Example
Find the matrix X such that
 2 3
 10 4 
 1 4  X   5 9  .




Solution:
 2 3
Note that 

 1 4 
1

1  4 3 

.
11  1 2 
1
 2 3
To find X, we need to multiply through by 
 on the left hand side:
 1 4 
1
 2 3   10
X
 
 1 4   5
1  4 3  10
X 
i.e.

11  1 2  5
So the solution is
 5 1 
X
.
0 2 
4
9 
4  1  55 11  5 1


.
9  11  0 22   0 2 
Example 2:
7 
8 1   5

Find X such that X 


 1 2   9 16 
Solution:
1  2 1
8 1 
is
The inverse of 


 . Therefore
17  1 8 
 1 2 
7  2 1 1  17
51 
7   1  2 1  1  5
5
 
X
=









 9 16   17  1 8   17  9 16  1 8  17  34 119 
 1 3 
=
.
 2 7 
Examination style question
 4 2   8 4 
Find the matrix X such that X 

.
 1 2   1 7 
3
Solving simultaneous equations in two variables using matrices
3.1
Introduction to solving equations using matrices
The equations
ax  by  
cx  dy  
can be represented in matrix form:
 a b  x    
 c d  y      .

   
 x
From the work in the previous section, we can find the solution vector   by multiplying by the
 y
a b
inverse of the matrix 
:
c d 
1
 x   a b   
 y  c d    .
  
  
a b
Note: A unique solution exists provided that the matrix 
 is non-singular.
c d 
Example:
Using a matrix method, solve:
3x – y = 5
-4x + 2y = -9.
Solution:
 3 1  x   5 
 4 2   y    9  .

   
The equations can be written in matrix form:
1
1  2 1
 3 1
 
We find the inverse of the matrix: 

.
2  4 3
 4 2 
To find the solutions, we multiply through by this inverse matrix:
 x  1  2 1  5  1  1   0.5 
 y    4 3  9    7    3.5  .
  2
  2   

So the solution to the equations are: x = 0.5 and y = -3.5.
Examination style question:
Using a matrix method, solve the equations:
2x + 5y = 2
-3x – 8y = -2
3.2
Consistent and inconsistent equations
The equations
ax + by =  
cx + dy =  
can be interpreted geometrically as two straight lines in two dimensional space.
a b
The equations have a unique solution provided that the determinant of the matrix 
 is nonc d 
zero. The lines then intersect at a single point.
a b
If det 
  0 , the equations either:
c d
a)
have no solutions (this would occur if the lines represented by equations  and  were
parallel to each other)
b)
have an infinite number of solutions (this would occur if equations  and  represented
the same line).
Example:
Solve the equations
2x  3 y  5
kx  6 y  2
for all possible values of k.
Solution
 2 3 
There are two cases to explore determined by the determinant of the matrix 
.
k 6 
 2 3 
 2 3 
 12  (3k )  12  3k . So det 
Notice that det 

  0 if k = -4.
k 6 
k 6 
Case 1: If k = -4, the equations are
2x  3 y  5
(1)
.
4 x  6 y  2
(2)
These equations will not have a unique solution. If we try eliminating x from these equations by
finding 2 ×(1) + (2) we get the equation
0 = 12.
This clearly is a contradiction. We say that the equations are inconsistent and so the equations
have no solutions.
Case 2: k ≠ -4.
 2 3   x   5 
 k 6  y    2

   
The inverse of the matrix exists when k ≠ -4. So the solution vector is:
We can represent the equations in matrix form:
1
1  6
 x   2 3   5 
 y    k 6   2 

  
   12  3k  k
3  5 
.
2 
 2 
Therefore:
1  36 
 x
 y 


  12  3k  4  5k 
36
12
x

,
So the equations have a unique solution:
12  3k 4  k
y
4  5k
.
12  3k
Example 2:
Solve the equations
2x – 5y = 6
-3x + ky = -9
for all possible values of k.
 2 5 
Solution: The solution will depend upon the determinant of the matrix 
.
 3 k 
 2 5 
Since det 
  2k  15 , the equations will have a unique solution provided that 2k – 15 ≠ 0,
 3 k 
i.e. provided that k ≠ 7.5.
Case 1: k = 7.5
The solution will not be unique. The equations are
2x – 5y = 6 
-3x + 7.5y = -9 
These equations are equivalent to each other as  = -1.5×. So in this case, the equations are
consistent but yield an infinite number of solutions.
We can get expressions for the solutions by introducing a parameter, y = t, and solving the
equations in terms of t.
Then from equation :
2x = 5y + 6
i.e.
x = 2.5t + 3
(as y = t).
So the solutions in this case are
x = 2.5t + 3, y = t.
Case 2: k ≠ 7.5
The equations can be expressed as
 2 5  x   6 
 3 k  y    9  .

   
1 k 5
 2 5 
In this case, the inverse of 
will exist and is


.
2k  15  3 2 
 3 k 
So
1  k 5  6 
1  6k  45 
 x
 y 
 3 2  9  

.
  2k  15 
  2k  15  0 
Therefore the solutions are
6k  45 3(2k  15)
x

 3 and y = 0.
2k  15
2k  15
Examination style question
Find the value of k for which the following equations do not have a unique solution. Solve the
equations for this value of k.
6x – y = 7
kx + 2y = -14.
Self-review:
Syllabus
Carry out operations of matrix addition,
subtraction and multiplication, and recognise the
terms null (or zero) matrix and identity (or unit)
matrix.
Recall the meaning of the terms singular and
non-singular as applied to square matrices, and,
for 2 × 2 matrices evaluate determinants and
find inverses of non-singular matrices.
Understand and use the result, for nonsingular matrices, that (AB)-1 = B-1 A-1.
Formulate a problem involving the solution of 2
linear simultaneous equations in 2 unknowns as
a problem involving the solution of a matrix
equation.
Understand the cases that may arise concerning
the consistency or inconsistency of 2 linear
simultaneous equations, relate them to the
singularity or otherwise of the corresponding
square matrix, and solve consistent systems.
Self-review   