Academic Skills Advice
Matrices Summary
To transpose a matrix write the rows as columns.
For example:
2 1 5
A = [1 5 0
2 1 4
0
9]
2
2
1
5
0
AT =
1
5
0
9
2
1
4
2
To add or subtract matrices they must be the same dimensions. Just add or subtract the
corresponding numbers.
For example:
2 5 −1
[4 1 2] +
−3 3 4
3
[7
10
−2 3
5
4
2 ] = [11
5 −5
7
3 2
5 4]
8 −1
To multiply matrices the number of columns in the 1st must equal the number of rows in
the 2nd because you multiply across the row and down the column.
For example:
(6 + 10) (−9 + 14)
3 2
2 −3 7
[4 6 ] × [
] = [(8 + 30) (−12 + 42)
5 7 8
(2 − 25) (−3 − 35)
1 −5
16
= [ 38
−23
(21 + 16)
(28 + 48)]
(7 − 40)
5
37
30
76 ]
−38 −33
Transformations
When setting up the multiplication for a transformation the transformation matrix always
goes first.
Rotations:
Clockwise: [
𝑐𝑜𝑠𝜃
−𝑠𝑖𝑛𝜃
𝑠𝑖𝑛𝜃
]
𝑐𝑜𝑠𝜃
Reflections:
0
[
1
1
] : swaps 𝑥 𝑎𝑛𝑑 𝑦.
0
𝑐𝑜𝑠𝜃
Anticlockwise: [
𝑠𝑖𝑛𝜃
−𝑠𝑖𝑛𝜃
]
𝑐𝑜𝑠𝜃
Scaling / magnification:
[
3 0
] : makes 𝑥 3x bigger and 𝑦 2x bigger
0 2
Composite:
If a matrix A is reflected by T1 then magnified by T2 to give matrix B then the single
transformation that would transform A into B is T 2T1 (in that order).
© H Jackson 2010 /2013/ Academic Skills
1
The Inverse (A-1):
AA = I
-1
1
(the identity matrix) e.g. [0
0
0 0
1 0]
0 1
or [
1 0
]
0 1
Finding the Inverse:
1
𝑎
If: A = [
𝑐
A-1 = |𝐴| x adj
𝑏
]
𝑑
|𝐴| = (𝑎𝑑) − (𝑏𝑐)
Where: |𝐴| = determinant
adj = adjoint
adj = [
𝑑
−𝑐
−𝑏
]
𝑎
n.b. If det=0 matrix is singular and there is no inverse. If det≠0, matrix is non-singular.
Solving Simultaneous Equations:
If :
then:
𝐴𝑥 = 𝐵
𝑥 = 𝐴−1 𝐵
where :
𝐴 is the matrix
𝑥 is the 𝑥, 𝑦, 𝑧 (𝑒𝑡𝑐) values
𝐵 is the answers
All you need to remember to find the values of 𝑥, 𝑦, 𝑧 etc is to do: inverse x answers
Example:
We have:
5𝑥 − 6𝑦 = −8
−3𝑥 + 4𝑦 = 6
Solve
and
[
−8
5 −6 𝑥
] [𝑦] = [ ]
6
−3 4
𝐴
So:
simultaneously
𝑥 =
𝐵
𝑥
−8
[𝑦] = 𝐴−1 [ ]
6
𝑥
[𝑦 ] =
4
[
2 3
1
© H Jackson 2010 /2013/ Academic Skills
1 4
6 −8
] [ ]= [ ]
2 6
5 6
(Use the method shown above to find the
inverse then multiply by the answers.)
=
2
[ ]
3
2
3 X 3 Matrices
Most operations are done exactly the same for a 3x3 matrix as for a 2x2. The formula for
finding the inverse is the same but finding the determinant and the adjoint are a bit trickier.
The Minor:
The minor of each element is the determinant of everything that is not in the same row or
column as the element. It is used when finding the determinant and the adjoint.
e.g. Find the minor of each element in the top row of the matrix:
𝒂
[𝒅
𝒈
𝒃 𝒄
𝒆 𝒇]
𝒉 𝒊
To find the minor of element ‘𝑎’ cover up everything in the same row or column and the
minor is the determinant of what is left:
𝑎
𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓
𝑖
the minor of a is:
𝒆 𝒇
|
| = 𝒆𝒊 − 𝒇𝒉
𝒉 𝒊
To find the minor of element ‘𝑏’ cover up everything in the same row or column and the
minor is the determinant of what is left:
𝑎
𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓
𝑖
the minor of b is:
𝒅 𝒇
|
| = 𝒅𝒊 − 𝒇𝒈
𝒈 𝒊
To find the minor of element ‘𝑐’ cover up everything in the same row or column and the
minor is the determinant of what is left:
𝑎
𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓
𝑖
the minor of c is:
𝒅 𝒆
|
| = 𝒅𝒉 − 𝒆𝒈
𝒈 𝒉
The signs:
When finding the determinant or the adjoint we use alternate signs as follows:
+
[−
+
− +
+ −]
− +
© H Jackson 2010 /2013/ Academic Skills
3
Finding the Inverse
Remember:
1
Where: |𝐴| = determinant
adj = adjoint
A-1 = |𝐴| x adj
Determinant:
For the top row only, multiply each number by its minor and apply the alternate signs:
The minor of each
element in the top row
𝑎
If: A = [𝑑
𝑔
𝑏
𝑒
ℎ
𝑐
𝑓]
𝑖
|𝐴| = +𝑎 |
𝑒
ℎ
𝑑
𝑓
|−𝑏|
𝑔
𝑖
𝑓
𝑑
|+𝑐|
𝑔
𝑖
𝑒
|
ℎ
Adjoint:
The adjoint is the transpose of the cofactors (𝑐 𝑇 )
To find the cofactors you find the ‘minor’ of each element (and use the alternate signs as
shown above).
𝑎
If: A = [𝑑
𝑔
𝑏
𝑒
ℎ
The minors of
each element
𝑐
𝑓]
𝑖
Matrix of cofactors =
𝑒
+|
ℎ
𝑑
𝑓
|−|
𝑔
𝑖
𝑓
𝑑
|+|
𝑔
𝑖
𝑒
|
ℎ
𝑏
−|
ℎ
𝑎
𝑐
| + |𝑔
𝑖
𝑐
𝑎
|
−
|
𝑖
𝑔
𝑏
|
ℎ
𝑏
𝑒
𝑎
𝑐
| − |𝑑
𝑓
𝑐
𝑎
𝑓 | + |𝑑
𝑏
|
𝑒
+|
Nb. To find the adjoint we then need to transpose this matrix.
© H Jackson 2010 /2013/ Academic Skills
4
𝟐
e.g. Find the inverse of the following matrix: A = [𝟑
𝟒
|𝐴|
Determinant:
Adjoint
(𝑪𝑻 ):
3
+|
1
1
−|
1
1
[+ |3
Inverse:
=
A-1 =
x adj
© H Jackson 2010 /2013/ Academic Skills
= −38
2
3 2
3 3
| −|
| +|
|
1
4 1
4 1
1
5
2 1
5
2 5
| +|
| −|
| =[ 4
−18
4 1
1
4 1
−13
11
2 1
5
2 5
| −|
| +|
|]
3 3
2
3 2
𝟏
=−
[𝟓
𝟑𝟖
−𝟗
𝟏
−9
2]
3
We have transposed
the matrix of cofactors
to get the adjoint.
1
4
−13
[ 5 −18 11 ]
−9
2
3
1
|𝐴|
𝟓
𝟐]
𝟏
3 2
3 2
3 3
= +2 |
| − 1|
| + 5|
|
1 1
4 1
4 1
= 2(1) − 1(−5) + 5(−9)
Cofactors =
𝟏
𝟑
𝟏
𝟒
−𝟏𝟖
𝟐
−𝟏𝟑
𝟏𝟏 ]
𝟑
5
Solving Simultaneous Equations Using Cramer’s Rule:
Cramers method is a way to solve simultaneous equations using matrices.
𝑎
e.g. Solve [
𝑐
𝑒
𝑏 𝑥
] [𝑦] = [𝑓 ] to find 𝑥 𝑎𝑛𝑑 𝑦
𝑑
Replace 1st column with
answer column
To find 𝒙:
𝑥=
𝑒
|
𝑓
𝑎
|
𝑐
𝑏
|
𝑑
𝑏
|
𝑑
Determinant of matrix with 𝟏𝐬𝐭 column replaced with the answer column.
𝑦=
𝑎
|𝑐
𝑎
|
𝑐
𝑒
𝑓|
𝑏
|
𝑑
Determinant of matrix with 𝟐𝐧𝐝 column replaced with the answer column.
Determinant of original matrix
To find 𝒚:
Determinant of original matrix
Notice that the bottom of the fraction is always the determinant of the original
matrix.
𝟑𝒙 − 𝟒𝒚 = 𝟏𝟎
𝒙 + 𝟕𝒚 = −𝟓
e.g. Solve the simultaneous equations:
Think of these as the answers.
3 −4 𝑥
10
] [𝑦 ] = [ ]
−5
1 7
Set up a matrix:
[
Determinant of matrix:
3 −4
|
| = 𝟐𝟓
1 7
For help with finding the
determinant, see Determinant
Summary sheet.
answers
To find 𝑥:
|
10 −4
|
−5 7
25
=
50
25
= 2
answers
To find 𝑦:
|
© H Jackson 2010 /2013/ Academic Skills
3 10
|
1 −5
25
=
−25
25
= -1
6
𝟐𝒙 + 𝒚 − 𝟑𝒛 = 𝟕
𝒙 + 𝟑𝒚 + 𝒛 = 𝟔
𝒙 − 𝟒𝒚 − 𝟓𝒛 = −𝟐
e.g. Solve the simultaneous equations:
Think of these as the answers.
Set up the matrix:
2 1 −3 𝑥
7
𝑦
[1 3
]
[
]
=
[
1
6]
1 −4 −5 𝑧
−2
Determinant of matrix:
2
|1
1
To find 𝑥
answers
7
|6
−2
To find 𝑦
1 −3
3
1 |=5
−4 −5
1
3
−4
5
−3
1|
−5
5
= 1
=
10
5
= 2
answers
2 1
7
|1 3
6|
1 −4 −2
5
© H Jackson 2010 /2013/ Academic Skills
5
answers
2 7 −3
|1 6
1|
1 −2 −5
5
To find 𝑧
=
=
−5
5
= -1
7