Date: 02/01/2025
Class: #24
Syllabus Topic: Simultaneous Equations and Matrix Transformation
Title: Solving Simultaneous Equations using the Matrix Method, Transformation Matrices
Solving matrices using Simultaneous Equations
Recall: In Algebra, there are 2 methods used to solve simultaneous equations:
•
Elimination
•
Substitution
We are now going to learn the Matrix method.
Changing simultaneous equations from algebraic form to matrix form
Example 1:
Algebraic form:
2𝑥 + 𝑦 = 0
𝑥 − 3𝑦 = 7
Change to matrix form:
𝑥
2 1
0
) (𝑦) = ( )
1 −3
7
(
Example 2:
Algebraic form:
𝑥 + 4𝑦 = 1
2𝑥 − 3𝑦 = −9
Change to matrix form:
𝑥
1 4
1
) (𝑦) = ( )
2 −3
−9
(
Example 3:
Algebraic form:
4𝑚 − 𝑛 = 3
2𝑚 − 4𝑛 = 5
Change to matrix form:
3
4 −1 𝑚
)( ) = ( )
5
2 −4 𝑛
(
Note
Consider
𝑥
2 1
0
) (𝑦) = ( )
1 −3
7
(
in the form
2
1
1
)
−3
where 𝐴 = (
Now,
𝐴𝑋 = 𝐵
[coefficient matrix]
𝑥
𝑋 = (𝑦)
[variable matrix]
0
𝐵=( )
7
[column matrix]
𝐴𝑋 = 𝐵
𝐴−1 𝐴𝑋 = 𝐴−1 𝐵
𝐼𝑋 = 𝐴−1 𝐵
𝑿 = 𝑨−𝟏 𝑩
Worked Example 1
Solve 2𝑥 + 𝑦 = 0
𝑥 − 3𝑦 = 7
Solution:
Change to matrix form:
(
𝑥
2 1
0
) (𝑦) = ( )
1 −3
7
in the form
𝐴𝑋 = 𝐵
∴ 𝑋 = 𝐴−1 𝐵
Note that 𝑨−𝟏 is placed to the left of 𝑩 and not to the right. Remember that matrix
multiplication is NOT commutative.
Find 𝐴−1 :
det (𝐴) = 𝑎𝑑 − 𝑏𝑐
= (2)(−3) − (1)(1)
= −6 − 1
= −7
𝐴−1 =
1
(𝑎𝑑𝑗)
𝑑ⅇ𝑡
−3 −1
)
−1 2
1
= −7 (
3
1
= (71
7
7
2)
−7
𝑑
−𝑐
𝑎𝑑𝑗(𝐴) = (
−𝑏
)
𝑎
−3 −1
)
−1 2
=(
So, we have:
3
1
𝑥
0
(𝑦) = (71 7 2) ( )
−7 7
7
3
1
(7 × 0) + (7 × 7)
)
=( 1
2
(7 × 0) + (− 7 × 7)
0+1
)
=(
0 + (−2)
1
)
−2
=(
∴
𝑥 = 1 and 𝑦 = −2
Worked Example 2
Solve 𝑥 + 4𝑦 = 1
2𝑥 − 3𝑦 = −9
Solution:
Change to matrix form:
𝑥
1 4
1
) (𝑦) = ( )
2 −3
−9
(
in the form
𝐴𝑋 = 𝐵
∴ 𝑋 = 𝐴−1 𝐵
1
2
where 𝐴 = (
Find 𝐴−1 :
4
).
−3
𝑑
−𝑐
det (𝐴) = 𝑎𝑑 − 𝑏𝑐
𝑎𝑑𝑗(𝐴) = (
−3 −4
)
−2 1
= (1)(−3) − (2)(4)
=(
= −3 − 8
= −11
1
𝐴−1 = 𝑑ⅇ𝑡 (𝑎𝑑𝑗)
−3 −4
)
−2 1
1
= −11 (
3
= (11
2
11
4
11
1)
− 11
So, by rearranging, we have:
(
𝑥
1 4
1
) (𝑦 ) = ( )
2 −3
−9
3
4
𝑥
1
(𝑦) = (11 11 ) ( )
2
1
−9
−
11
11
3
4
(11 × 1) + (11 × −9)
)
=( 2
1
(11 × 1) + (− 11 × −9)
3
−𝑏
)
𝑎
36
− 11
9)
+
11
11
= (11
2
−3
)
1
=(
OR (“shortcut” – a way to avoid the fractions)
1
𝑥
1
3 −4
(𝑦 ) =
(
)( )
−9
−11 −2 1
1
−3 + 36
)
−2 − 9
1
33
)
−11
= − 11 (
= − 11 (
−3
)
1
=(
∴
𝑥 = −3 and 𝑦 = 1
Worked Example 3
Solve 4𝑥 − 𝑦 = 3
2𝑥 − 4𝑦 = 5
Solution:
Change to matrix form:
3
4 −1 𝑥
)( ) = ( )
5
2 −4 𝑦
(
in the form
𝐴𝑋 = 𝐵
∴ 𝑋 = 𝐴−1 𝐵
4
2
where 𝐴 = (
−1
)
−4
Find 𝐴−1 :
det (𝐴) = 𝑎𝑑 − 𝑏𝑐
= (4)(−4) − (2)(−1)
= −16 − (−2)
= −14
𝑑
−𝑐
𝑎𝑑𝑗(𝐴) = (
−𝑏
)
𝑎
−4 1
)
−2 4
=(
𝐴−1 =
1
(𝑎𝑑𝑗)
𝑑ⅇ𝑡
−4 1
)
−2 4
1
= −14 (
2
1
− 14
2)
−7
= (71
7
So, by rearranging, we have:
(
3
4 −1 𝑥
) (𝑦) = ( )
5
2 −4
2
𝑥
(𝑦) = (71
7
1
− 14 3
2 ) (5)
−7
2
1
( × 3) + (− 14 × 5)
)
= ( 71
2
(7 × 3) + (− 7 × 5)
6
5
− 14
10)
−
7
7
= (73
1
=( 2 )
−1
OR (“shortcut” – a way to avoid the fractions)
1 −4 1 3
𝑥
(𝑦 ) =
(
)( )
−14 −2 4 5
−12 + 5
)
−6 + 20
1
= −14 (
1
=( 2 )
−1
∴
1
𝑥 = 2 and 𝑦 = −1
Worked Example 4
Solve 2𝑥 + 𝑦 = 5
2𝑥 + 3𝑦 = 7
Solution:
Change to matrix form:
2 1 𝑥
5
)( ) = ( )
2 3 𝑦
7
(
in the form
𝐴𝑋 = 𝐵
∴ 𝑋 = 𝐴−1 𝐵
2
2
where 𝐴 = (
1
).
3
Find 𝐴−1 :
det (𝐴) = 𝑎𝑑 − 𝑏𝑐
= (2)(3) − (2)(1)
=6−2
=4
1
𝐴−1 = 𝑑ⅇ𝑡 (𝑎𝑑𝑗)
1
= 4(
3 −1
)
−2 2
𝑑
−𝑐
𝑎𝑑𝑗(𝐴) = (
−𝑏
)
𝑎
3 −1
)
−2 2
=(
So, by rearranging, we have:
(
2 1 𝑥
5
) (𝑦 ) = ( )
2 3
7
𝑥
1 3
(𝑦 ) = 4 (
−2
−1 5
)( )
2
7
𝑥
1
15 − 7
(𝑦 ) = 4 (
)
−10 + 14
𝑥
1 8
(𝑦 ) = 4 ( )
4
𝑥
2
(𝑦 ) = ( )
1
∴
𝑥 = 2 and 𝑦 = 1
Matrix Transformation - Introduction
2 4
Consider the matrix (
).
1 0
Size/Order of the matrix is 2 × 2.
Note: The size or order of a matrix is
stated in the form: rows by column
3
Consider the matrix ( ).
4
Size/Order of the matrix is 2 × 1.
This matrix is called a column matrix.
Identity Matrix
The 2 × 2 identity matrix is (
Recall:
The additive identity is 0.
That is, 𝑎 + 0 = 0 + 𝑎 = 𝑎.
1 0
).
0 1
The multiplicative identity is 1.
That is, 𝑎 × 1 = 1 × 𝑎 = 𝑎.
5 −2
Consider the matrix (
).
1 0
When the identity matrix is multiplied to this matrix, the matrix remains unchanged.
(
1 0 5
)(
0 1 1
(1 × 5) + (0 × 1)
−2
)=(
(0 × 5) + (1 × 1)
0
=(
5
1
(1 × −2) + (0 × 0)
)
(0 × −2) + (1 × 0)
−2
)
0
Transformation Matrices
Consider the point 𝐴(1,4).
1
Now, (
0
0
1
)𝐴 = (
1
0
0 1
)( )
1 4
1
=( )
4
𝑦
𝐴(1,4)
𝑥
The identity matrix leaves the point 𝐴 unchanged.
How about we make one of the “1” negative?
Consider the point 𝐴(1,4). Let 𝐴′ be the image of 𝐴.
Now,
−1 0
𝐴′ = (
)𝐴
0 1
−1 0 1
=(
)( )
0 1 4
−1 + 0
)
0+4
=(
−1
=( )
4
where −1 is the 𝑥-coordinate and 4 is the 𝑦-coordinate
𝑦
𝐴′(−1,4)
𝐴(1,4)
𝑥
The matrix (
−1 0
) represents a reflection in the 𝒚-axis.
0 1
Consider the point 𝐴(1,4).
Let 𝐴′ be the image of 𝐴.
Now,
1 0
𝐴′ = (
)𝐴
0 −1
1 0
1
=(
)( )
0 −1 4
1+0
=(
)
0−4
1
=( )
−4
where 1 is the 𝑥-coordinate and −4 is the 𝑦-coordinate
𝑦
𝐴(1,4)
𝑥
𝐴′(1, −4)
The matrix (
1 0
) represents a reflection in the 𝒙-axis.
0 −1
Exercise
What do these transformation matrices represent?
−1 0
)
0 1
A) (
B) (
C) (
1 0
)
0 −1
0 1
)
1 0
2 0
)
0 2
D) (
Solution:
−1 0
)
0 1
→ reflection in the 𝑦-axis
1 0
)
0 −1
→ reflection in the 𝑥-axis
0 1
)
1 0
→ reflection in the line 𝑦 = 𝑥
2 0
)
0 2
→ enlargement of scale factor 2
A) (
B) (
C) (
D) (
Enlargement
Consider the points below which connect to form a triangle.
𝐴 = (1, 5)
𝐵 = (1, 1)
𝐶 = (3, 1)
The graph is shown below.
𝑦
𝐴(1, 5)
𝐵(1, 1)
𝐶(3, 1)
𝑥
Suppose that the points undergo an enlargement of scale factor 2.
2
0
The matrix is (
0
).
2
The image points are as follows:
2 0 1
𝐴′ = (
)( )
0 2 5
2
𝐴′ = ( )
10
2 0 1
𝐵′ = (
)( )
0 2 1
2
𝐵′ = ( )
2
2 0 3
𝐶′ = (
)( )
0 2 1
6
𝐶′ = ( )
2
𝐴′(2, 10)
𝑦
𝐴(1, 5)
𝐵′(2, 2)
𝐵(1, 1)
𝐶′(6, 2)
𝐶(3, 1)
𝑥
Examples of Transformation Matrices
Reflection or Flip
𝑥-axis
𝑦-axis
1 0
)
0 −1
(
The line 𝑦 = 𝑥
The line 𝑦 = −𝑥
(
(
0
1
1
)
0
−1 0
)
0 1
(
0 −1
)
−1 0
Translation or Slide
𝑥
Use the vector: (𝑦)
where 𝑥 represents the movement in the horizontal
and 𝑦 represents the vertical movement.
Common Rotations
Rotation 90 degrees clockwise (or 270 degrees anticlockwise):
(
0 1
)
−1 0
Rotation 180 degrees:
(
−1 0
)
0 −1
Rotation 270 degrees clockwise (or 90 degrees anticlockwise):
(
0
1
−1
)
0
Rotation of Transformation Matrices
Sometimes we label the transformation matrix as “ T “.
𝑻𝑨 = 𝑨′
−1 0 1
−1
)( ) = ( )
0 1 4
4
For example, (
If we want to go backwards,
Note: 𝐴′ is called 𝐴 prime.
𝑨 = 𝑻−𝟏 𝑨′
Recall:
2𝑥 + 3𝑦 = 1
𝑥 + 4𝑦 = 0
In matrix form,
(
1
2 3 𝑥
) (𝑦 ) = ( )
0
1 4
𝑥
2
(𝑦) = (
1
3 −1 1
) ( )
0
4
Past Paper Question - January 2021 – Question 10(b)
A right-angled triangle, 𝑀, has vertices 𝑋(1, 1), 𝑌(3, 1) and 𝑍(3, 4). When 𝑀 is
transformed by the matrix 𝑁 = (
0
1
1
), the image is 𝑀′.
0
Find the coordinates of the vertices of 𝑀′.
[2]
Solution:
(
0 1
) represents a reflection in the line 𝑦 = 𝑥.
1 0
So, we need to interchange coordinates to find 𝑀′.
Therefore,
𝑋′(1, 1)
𝑌′(1, 3)
𝑍′(4, 3)
Another method to obtain the points:
0 1 1
𝑋′ = (
)( )
1 0 1
(0 × 1) + (1 × 1)
)
𝑋′ = (
(1 × 1) + (0 × 1)
0+1
𝑋′ = (
)
1+0
1
𝑋′ = ( )
1
Similarly,
0 1 3
𝑌′ = (
)( )
1 0 1
1
𝑌′ = ( )
3
And,
0 1 3
𝑍′ = (
)( )
1 0 4
4
𝑍′ = ( )
3
Visually, here is the ∆𝑋𝑌𝑍 reflecting in the line 𝑦 = 𝑥 to obtain ∆𝑋′𝑌′𝑍′.
𝑦
𝑍(3, 4)
𝑌(1, 3)
𝑍′(4, 3)
𝑀
𝑀′
𝑋(1,1)
𝑋′(1,1)
𝑌′(3,1)
𝑥
Past Paper Question - January 2020 – Question 10(a)
The transformation 𝑀 = (
(i)
0
𝑞
𝑝
) maps the point 𝑅 onto 𝑅′ as shown in the diagram below.
0
Determine the values of 𝑝 and 𝑞.
Solution:
(i)
From the graph, 𝑅(2, −5) and 𝑅′(5, 2).
Now,
(
0
𝑞
𝑝
2
5
)( ) = ( )
0 −5
2
(0 × 2) + (𝑝 × −5)
5
(
)=( )
(𝑞 × 2) + (0 × −5)
2
−5𝑝
5
(
)=( )
2𝑞
2
[2]
By comparing the equivalent matrices, we have
−5𝑝 = 5
2𝑞 = 2
5
2
𝑝 = −5
𝑞=2
𝑝 = −1
𝑞=1
Past Paper Question - January 2014 – Question 11(a)
The matrix, 𝑇, is such that
2
1
𝑇=(
−1
)
3
(i)
Determine, 𝑇 −1 , the inverse of 𝑇.
(ii)
The matrix 𝑇 maps the point (𝑎, 𝑏) onto the point (4, 9). Determine the values
[3]
of 𝑎 and 𝑏.
[4]
Solution:
(i)
2 −1
𝑎
) in the form (
1 3
𝑐
𝑇=(
𝑏
)
𝑑
Finding the 𝑇 −1 :
𝑑ⅇ𝑡 = 𝑎𝑑 − 𝑏𝑐
= (2)(3) − (−1)(1)
=6+1
=7
𝑎𝑑𝑗(𝑇) = (
𝑑
−𝑐
−𝑏
)
𝑎
3 1
)
−1 2
=(
1
𝑇 −1 = 𝑑ⅇ𝑡 (𝑎𝑑𝑗)
1
= 7(
3 1
)
−1 2
3
7
= (−1
7
(ii)
1
7
2)
7
𝑎
4
𝑇( ) = ( )
𝑏
9
𝑎
4
∴ ( ) = 𝑇 −1 ( )
𝑏
9
𝑎
1 3
1 4
( ) = 7(
)( )
𝑏
−1 2 9
𝑎
1 12 + 9
( )= (
)
7 −4 + 18
𝑏
𝑎
1 21
( ) = 7( )
𝑏
14
𝑎
3
( )=( )
𝑏
2
Past Paper Question - January 2013 – Question 11(b)
0 −1
) represents a single transformation.
1 0
The matrix 𝐽 = (
The image of the point 𝑃 under the transformation 𝐽 is (5, 4).
Determine the coordinates of 𝑃.
Solution:
[3]
5
𝐽𝑃 = ( )
4
(
0 −1
5
)𝑃 = ( )
1 0
4
𝑃=(
0 −1 −1 5
) ( )
1 0
4
𝑑ⅇ𝑡 = 𝑎𝑑 − 𝑏𝑐
= (0)(0) − (−1)(1)
=1
𝐽−1 =
1
(𝑎𝑑𝑗)
𝑑ⅇ𝑡
1 0
1
= 1(
)
−1 0
0 1
)
−1 0
=(
Therefore,
0 1 5
𝑃=(
)( )
−1 0 4
(0 × 5) + (1 × 4)
)
𝑃=(
(−1 × 5) + (0 × 4)
4
𝑃=( )
−5
𝑎𝑑𝑗(𝐽) = (
𝑑
−𝑐
−𝑏
)
𝑎
=(
0 1
)
−1 0
Past Paper Question - January 2009 – Question 11(b)
−1 0
).
0 1
The transformation 𝑅 is represented by the matrix (
0 1
).
−1 0
The transformation 𝑆 is represented by the matrix (
(i)
Write a single matrix, in the form (
𝑎
𝑐
𝑏
) to represent the combined
𝑑
transformation 𝑆 followed by 𝑅.
(ii)
[2]
Calculate the image of the point (5, −2) under the combined transformation in
(b) (i) above.
[3]
Solution:
(i)
0 1
)
−1 0
𝑆=(
and
𝑅=(
−1
0
𝑆 followed by 𝑅.
This means 𝑅𝑆.
−1 0
0 1
)(
)
0 1 −1 0
𝑅𝑆 = (
=(
(ii)
0 −1
)
−1 0
(0 × 5) + (−1 × −2)
0 −1
5
)
)( ) = (
(−1 × 5) + (0 × −2)
−1 0
−2
(
0+2
)
−5 + 0
=(
2
)
−5
=(
0
)
1