Matrices:
CONTENTS
MATRIX ADDITION
Matrices are a way of storing numbers in a grid. Addition and subtraction can be done in the obvious way, providing the matrices are of the same size.
e.g.1
\[\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
+
\begin{bmatrix}
e & f \\
g & h
\end{bmatrix}\
=
\begin{bmatrix}
a+e & b+f \\
c+g & d+h
\end{bmatrix}\]
e.g.2
\[\begin{bmatrix}
6 & 1 & 0 \\
5 & 2 & 3
\end{bmatrix}
+
\begin{bmatrix}
4 & 3 & 2 \\
3 & 1 & 4
\end{bmatrix}\
=
\begin{bmatrix}
10 & 4 & 2 \\
8 & 3 & 7
\end{bmatrix}\]
MATRIX MULTIPLICATION
Multiplication is done by multiplying one row by one column pairwise.
The output goes in the position where the row and column intersect.
Multiplication can only be done if the number of columns in the first matrix is the same as the number of rows in the second.
e.g.1
\[\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\times
\begin{bmatrix}
e & f \\
g & h
\end{bmatrix}\
=
\begin{bmatrix}
ae+bg & af+bh \\
ce+dg & cf+dh
\end{bmatrix}\]
e.g.2
\[\begin{align}
\begin{bmatrix}
6 & 1 & 0 \\
5 & 2 & 3
\end{bmatrix}
\times
\begin{bmatrix}
4 & 5 \\
3 & 4 \\
1 & 7
\end{bmatrix}\
& =
\begin{bmatrix}
(6 \times 4 + 1 \times 3 + 0 \times 1) & (6 \times 5 + 1 \times 4 + 0 \times 7) \\
(5 \times 4 + 2 \times 3 + 3 \times 1) & (5 \times 5 + 2 \times 4 + 3 \times 7)
\end{bmatrix} \\
& =
\begin{bmatrix}
27 & 34 \\
29 & 54
\end{bmatrix}
\end{align}\]
If the matrices are recording data, one matrix may need to be transposed to make the multiplication valid.
The transpose of a matrix `A` is denoted `A^T` and is formed by making the rows into the columns of the matrix.
e.g.3 Alice, Bob and Charlie are selling pens and pencils. The number that they have sold of each is recorded in the matrix A, where the columns represent Alice, Bob and Charlie and the rows represent pens and pencils respectively. The prices of a pen and pencil in pence are recorded in the respective rows of matrix B. Calculate the revenue each of them makes.
\[A =
\begin{bmatrix}
4 & 5 & 2 \\
2 & 4 & 4
\end{bmatrix},
B =
\begin{bmatrix}
20 \\
15
\end{bmatrix}\]
\[A^T \times B =
\begin{bmatrix}
4 & 2 \\
5 & 4 \\
2 & 4
\end{bmatrix}
\times
\begin{bmatrix}
20 \\
15
\end{bmatrix}
=
\begin{bmatrix}
110 \\
160 \\
100
\end{bmatrix}\]
∴ Alice makes 110p (£1.10), Bob makes 160p (£1.60) and Charlie makes 100p (£1.00)
e.g.4 Solve the matrix equation
\(\begin{bmatrix}
4 & 2 \\
3 & 5
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
10 \\
18
\end{bmatrix}\)
\[\begin{align}
\begin{bmatrix}
4 & 2 \\
3 & 5
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
& =
\begin{bmatrix}
10 \\
18
\end{bmatrix} \\
\begin{bmatrix}
4x + 2y \\
3x + 5y
\end{bmatrix}
& =
\begin{bmatrix}
10 \\
18
\end{bmatrix}
\end{align}\]
`4x + 2y = 10 => y = 5 - 2x`
`3x + 5y = 18`
\[\begin{align}
3x + 5(5 - 2x) & = 18 \\
3x + 25 - 10x & = 18 \\
7x & = 7 \\
x & = 1
\end{align}\]
\[\begin{align}
x = 1 ⇒ y & = 5 - 2(1) \\
& = 3
\end{align}\]
∴ `x = 1, y = 3`
Matrix multiplication is not commutative, but is associative. i.e.
`AB != BA`
`A xx (B xx C) = (A xx B) xx C`
FINDING THE INVERSE AND DETERMINANT OF A 2x2 MATRIX
The identity matrix, `I_n` of size n is defined as the n x n matrix with 1s on the main diagonal (from top left to bottom right) and 0s everywhere else.
\[I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\]
The inverse of a matrix `A` is defined as the matrix which multiplies with `A` to give the identity matrix.
Division can only be done by multiplying by the inverse
e.g.1 Find the inverse of the matrix
\(\begin{bmatrix}
2 & 7 \\
1 & 4
\end{bmatrix}\)
\[\begin{align}
\begin{bmatrix}
2 & 7 \\
1 & 4
\end{bmatrix}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
& =
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\
\begin{bmatrix}
2a + 7c & 2b + 7d \\
a + 4c & b + 4d
\end{bmatrix}
& =
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\end{align}\]
`2a + 7c = 1`
`2b + 7d = 0`
`a + 4c = 0`
`b + 4d = 1`
Solving simultaneously, `a = 4, b = -7, c = -1, d = 2`
\[∴
\begin{bmatrix}
2 & 7 \\
1 & 4
\end{bmatrix}^{-1}
=
\begin{bmatrix}
4 & -7 \\
-1 & 2
\end{bmatrix}\]
We can find a general solution to the inverse of a 2 x 2 matrix `A`.
\[A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\] \[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\]
The expression `ad - bc` is called the determinant and is denoted `|A|` or `det(A)`. If it is 0, the matrix is said to be singular, and has no inverse.
e.g.2 Find the determinant of the matrix
\(A =
\begin{bmatrix}
4 & -1 \\
-8 & 2
\end{bmatrix}\).
Hence determine if there are unique solutions to the simultaneous equations `4x - y = 3` and `-8x + 2y = 4`.
\[\begin{bmatrix}
4 & -1 \\
-8 & 2
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
3 \\
4
\end{bmatrix}\]
\[\begin{align}
det(A) & =
\begin{vmatrix}
4 & -1 \\
-8 & 2
\end{vmatrix} \\
& = 4 \times 2 - (-1 \times -8) \\
& = 0
\end{align}\]
∴ There are no unique solutions to the simultaneous equations.
e.g.3 Find the inverse of the matrix
\(A =
\begin{bmatrix}
4 & -1 \\
-8 & 2
\end{bmatrix}\).
Hence solve the simultaneous equations `4x + y = 3` and `7x + 2y = 4`.
\[\begin{bmatrix}
4 & 1 \\
7 & 2
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
3 \\
4
\end{bmatrix}\]
\[\begin{align}
A^{-1} & =
\frac{1}{4 \times 2 - 1 \times 7}
\begin{bmatrix}
2 & -1 \\
-7 & 4
\end{bmatrix} \\
& =
\begin{bmatrix}
2 & -1 \\
-7 & 4
\end{bmatrix}
\end{align}\]
\[\begin{align}
\begin{bmatrix}
2 & -1 \\
-7 & 4
\end{bmatrix}
\begin{bmatrix}
4 & 1 \\
7 & 2
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
& =
\begin{bmatrix}
2 & -1 \\
-7 & 4
\end{bmatrix}
\begin{bmatrix}
3 \\
4
\end{bmatrix} \\
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
& =
\begin{bmatrix}
2 \\
-5
\end{bmatrix} \\
\begin{bmatrix}
x \\
y
\end{bmatrix}
& =
\begin{bmatrix}
2 \\
-5
\end{bmatrix}
\end{align}\]
When taking the inverse of a product, multiply the inverses of the two matrices in reverse order.
`(AB)^(-1) = B^(-1) A^(-1)`
e.g.4 Find `(AB^(-1)A^(-1))^(-1)` when \(A =
\begin{bmatrix}
4 & 1 \\
7 & 2
\end{bmatrix}\)
and \(B =
\begin{bmatrix}
3 & 5 \\
-2 & -4
\end{bmatrix}\)
\[\begin{align}
(A B^{-1} A^{-1})^{-1} & = (A^{-1})^{-1} (B^{-1})^{-1} A^{-1} \\
& = ABA^{-1} \\
& = \begin{bmatrix}
4 & 1 \\
7 & 2
\end{bmatrix}
\begin{bmatrix}
3 & 5 \\
-2 & -4
\end{bmatrix}
\begin{bmatrix}
2 & -1 \\
-7 & 4
\end{bmatrix} \\
& = \begin{bmatrix}
10 & 16 \\
17 & 27
\end{bmatrix}
\begin{bmatrix}
2 & -1 \\
-7 & 4
\end{bmatrix} \\
& = \begin{bmatrix}
-92 & 54 \\
-155 & 91
\end{bmatrix}
\end{align}\]
FINDING THE INVERSE AND DETERMINANT OF A 3x3 MATRIX
One method of finding the determinant of a 3 x 3 matrix is to multiply each diagonal together (including going off the side of the matrix and rejoining on the other side) and adding the answers if the diagonal is top left to bottom right and subtracting it if it is top right to bottom left.
\[\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = aei + bfg + cdh - ceg - fha - ibd\]
e.g.1 Find the determinant of the matrix
\(A = \begin{bmatrix}
4 & 2 & 3 \\
3 & 1 & 2 \\
3 & -1 & 4
\end{bmatrix}\)
\[\begin{align}
|A| & = (4 \times 1 \times 4) + (2 \times 2 \times 3) + (3 \times 3 \times -1) - (3 \times 1 \times 3) - (2 \times -1 \times 4) - (4 \times 2 \times 3) \\
& = -6
\end{align}\]
Each number in a 3 x 3 matrix has a cofactor. To find it, cross out the row and column containing the number to leave a 2 x 2 matrix. Then, find the determinant of this matrix and change sign based on the position of the number in the original matrix using this pattern:
\[\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\]
Another way to find the determinant of a 3 x 3 matrix is to take a row (or column) and multiply each number by its cofactor and add them up. This method also works to find the determinant of higher dimension matrices.
\[\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}\]
To find the inverse matrix, first find the matrix cofactors. Then transpose the matrix. Then divide by the determinant.
e.g.2 Find the inverse of the matrix \(A =
\begin{bmatrix}
1 & 3 & 0 \\
2 & -1 & -2 \\
3 & 1 & 3
\end{bmatrix}\)
\[\begin{align}
|A|
& = 1 \times \begin{vmatrix}
-1 & -2 \\
1 & 3
\end{vmatrix}
- 3 \times \begin{vmatrix}
2 & -2 \\
3 & 3
\end{vmatrix}
+ 0 \times \begin{vmatrix}
2 & -1 \\
3 & 1
\end{vmatrix} \\
& = 1 \times -1 - 3 \times 12 + 0 \times 5 \\
& = -37
\end{align}\]
Matrix of Cofactors, \(C =
\begin{bmatrix}
-1 & -12 & 5 \\
-9 & 3 & 8 \\
-6 & 2 & -7
\end{bmatrix}\)
\[\begin{align}
A^{-1} & = - \frac{1}{37} C^T \\
& = - \frac{1}{37}
\begin{bmatrix}
-1 & -9 & -6 \\
-12 & 3 & 2 \\
5 & 8 & -7
\end{bmatrix} \\
& = \frac{1}{37}
\begin{bmatrix}
1 & 9 & 6 \\
12 & -3 & -2 \\
-5 & -8 & 7
\end{bmatrix}
\end{align}\]
2D MATRIX TRANSFORMATIONS
A matrix can be used to describe a transformation. The coordinates of a point are written as a column vector and to perform the transformation, left multiply by the transformation matrix.
A matrix can be used to describe the following transformations:
- Rotation about the origin of any angle
- Reflection in any line through the origin
- Enlargement of any scale factor centered at the origin
- Stretch (normally parallel to the x- or y-axis)
- Shear (normally in direction parallel to x- or y-axis)
To identify the transformation represented by a matrix, find the image of the unit square. The unit square is the square starting from the origin, in the first quadrant, with side length 1. i.e. the square with coordinates (0, 0), (1, 0), (1, 1) and (0, 1).
e.g.1 Find the image of the unit square under transformation by the matrix
\(\begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}\).
Hence describe the transformation.
\[\begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}
\begin{bmatrix}
0 \\
0
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0
\end{bmatrix}\]
\[\begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix}
=
\begin{bmatrix}
-1 \\
0
\end{bmatrix}\]
\[\begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
-1 \\
-1
\end{bmatrix}\]
\[\begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}
\begin{bmatrix}
0 \\
1
\end{bmatrix}
=
\begin{bmatrix}
0 \\
-1
\end{bmatrix}\]
This is a rotation of 180° about the origin.
The images of the points (1, 0) and (0, 1) form the columns of the transformational matrix.
A combination of transformations, `A` followed by `B`, is the matrix product `BA`.
e.g.2 Write down the matrix `A` representing an enlargement about the origin by a scale factor of 3 and the matrix `B` representing a rotation about the origin 90° clockwise. Write down the matrix for A followed by B.
Under an enlargement scale factor 3 about the origin, (1, 0) is mapped to (3, 0) and (0, 1) is mapped to (0, 3)
\(∴ A =
\begin{bmatrix}
3 & 0 \\
0 & 3
\end{bmatrix}\)
Under a rotation 90° clockwise about the origin, (1, 0) is mapped to (0, -1) and (0, 1) is mapped to (1, 0)
\(∴ B =
\begin{bmatrix}
0 & 1 \\
-1 & 0
\end{bmatrix}\)
\[\begin{align}
A \text{ then } B & = BA \\
& = \begin{bmatrix}
0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
3 & 0 \\
0 & 3
\end{bmatrix} \\
& = \begin{bmatrix}
0 & 3 \\
-3 & 0
\end{bmatrix}
\end{align}\]
e.g.3 Describe the transformation given by the matrix \(\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}\)
Enlargement scale factor 2 about the origin and rotation 90° anticlockwise about the origin (in either order).
A shear is a transformation which has a line which remains fixed. The other points move in a direction parallel to the fixed line by a distance proportional to the distance the point is from the fixed line
e.g.4 Describe the transformation given the matrix \(\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\)
Shear
y-axis invariant
(1, 0) mapped to (1, 2)
The determinant of a 2 x 2 matrix represents the area scale factor of the transformation. A negative determinant means that the image is a reflection of the object.
e.g.5 Find the area of the image of the unit square under the transformation given by the matrix `A` =
\(\begin{bmatrix}
3 & 2 \\
1 & 5
\end{bmatrix}\)
\[\begin{align}
\text{Area of Image} & = \text{Area of Unit Square} \times |A| \\
& = 1 \times (3 \times 5 - 1 \times 2) \\
& = 13
\end{align}\]
Summary of 2D Transformation Matrices:
| Rotation anticlockwise about the origin by angle \(\theta\): | \(\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\) |
| Reflection in `x`-axis: | \(\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\) |
| Reflection in `y`-axis: | \(\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\) |
| Reflection in `y=x`: | \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) |
| Reflection in `y=-x`: | \(\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}\) |
| Enlargement by scale factor k centered at the origin: | \(\begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}\) |
| Stretch by scale factor k parallel to the x-axis (y-axis invariant): | \(\begin{bmatrix} k & 0 \\ 0 & 1 \end{bmatrix}\) |
| Stretch by scale factor k parallel to the y-axis (x-axis invariant): | \(\begin{bmatrix} 1 & 0 \\ 0 & k \end{bmatrix}\) |
| Shear, x-axis invariant, (0, 1) mapped to (k, 1): | \(\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}\) |
| Shear, y-axis invariant, (1, 0) mapped to (1, k): | \(\begin{bmatrix} 1 & 0 \\ k & 1 \end{bmatrix}\) |
INVARIANT LINES AND LINES OF INVARIANT POINTS
An invariant point `x` under a transformation `A` has `Ax = x`. Note that (0, 0) is always invariant. A line of invariant points is a line on which every point is invariant (i.e. maps to itself under the transformation `A`).
e.g.1 Show that (1, 1) is invariant under the matrix `A` =
\(\begin{bmatrix}
0 & 1 \\
-1 & 2
\end{bmatrix}\). Show further that any point on the line `y=x` is invariant under `A`.
\[\begin{bmatrix}
0 & 1 \\
-1 & 2
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
1 \\
1
\end{bmatrix}\]
∴ (1, 1) is invariant
\[\begin{bmatrix}
0 & 1 \\
-1 & 2
\end{bmatrix}
\begin{bmatrix}
x \\
x
\end{bmatrix}
=
\begin{bmatrix}
x \\
x
\end{bmatrix}\]
∴ Any point on `y=x` is invariant
A line is invariant under a transformation `A` if every point on the line is mapped to another point on the line.
e.g.2 Show that `y=x` is invariant under the matrix `A` =
\(\begin{bmatrix}
7 & -1 \\
-2 & 8
\end{bmatrix}\). Find the other line through the origin that is invariant under `A`.
\[\begin{bmatrix}
7 & -1 \\
-2 & 8
\end{bmatrix}
\begin{bmatrix}
x \\
x
\end{bmatrix}
=
\begin{bmatrix}
6x \\
6x
\end{bmatrix}\]
`(6x, 6x)` satisfies `y=x`
∴ `y=x` is an invariant line
For a point on `y=kx`:
\[\begin{bmatrix}
7 & -1 \\
-2 & 8
\end{bmatrix}
\begin{bmatrix}
x \\
kx
\end{bmatrix}
=
\begin{bmatrix}
7x - kx \\
-2x + 8kx
\end{bmatrix}\]
\[\begin{align}
\text{Invariant} ⇒ -2x + 8kx & = k(7x - kx) \\
-2 + 8k & = k(7 - k) \\
k^2 + k -2 & = 0 \\
(k - 1)(k + 2) & = 0 \\
k & = 1, -2
\end{align}\]
∴ `y=-2x` is also invariant
e.g.3 Show that any line of the form `3y + 4x = k` is invariant under the matrix
\(\begin{bmatrix}
0.28 & 0.96 \\
0.96 & -0.28
\end{bmatrix}\)
\[y = \frac{k-4x}{3}\]
\[\begin{bmatrix}
0.28 & 0.96 \\
0.96 & -0.28
\end{bmatrix}
\begin{bmatrix}
x \\
\frac{k-4x}{3}
\end{bmatrix}
=
\begin{bmatrix}
0.28x + 0.96\left(\frac{k-4x}{3}\right) \\
0.96x - 0.28\left(\frac{k-4x}{3}\right)
\end{bmatrix}\]
\[\begin{align}
& 3\left(0.96x - 0.28\left(\frac{k-4x}{3}\right)\right) + 4\left(0.28x + 0.96\left(\frac{k-4x}{3}\right)\right) \\
& = 2.88x - 0.28(k - 4x) + 1.12x + 1.28(k - 4x) \\
& = k
\end{align}\]
∴ `3y + 4x = k` is invariant
3D MATRIX TRANSFORMATIONS
As with 2D transformations, the columns represent the images of the basic unit vectors - (1, 0, 0), (0, 1, 0) and (0, 0, 1).
The determinant is the volume scale factor.
| Rotation about the `x`-axis by angle \(\theta\): | \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{bmatrix}\) |
| Rotation about the `y`-axis by angle \(\theta\): | \(\begin{bmatrix} \cos \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta \end{bmatrix}\) |
| Rotation about the `z`-axis by angle \(\theta\): | \(\begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\) |
| Reflection in `x=0` (`y`–`z` plane) : | \(\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) |
| Reflection in `y=0` (`x`–`z` plane) : | \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) |
| Reflection in `z=0` (`x`–`y` plane) : | \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}\) |
e.g.1 Let `A` be a reflection in `x=0` (the y-z plane) and `B` be a rotation of 60° about the y-axis (x = z = 0). Find the image of (1, 4, 7) after `A` followed by `B`.
\[A = \begin{bmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}\]
\[B = \begin{bmatrix}
\cos(60) & 0 & \sin(60) \\
0 & 1 & 0 \\
-\sin(60) & 0 & \cos(60)
\end{bmatrix}
= \begin{bmatrix}
\frac{1}{2} & 0 & \frac{\sqrt{3}}{2} \\
0 & 1 & 0 \\
-\frac{\sqrt{3}}{2} & 0 & \frac{1}{2}
\end{bmatrix}\]
\[\begin{align}
\text{A then B} & = BA \\
& = \begin{bmatrix}
\frac{1}{2} & 0 & \frac{\sqrt{3}}{2} \\
0 & 1 & 0 \\
-\frac{\sqrt{3}}{2} & 0 & \frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} \\
& = \begin{bmatrix}
-\frac{1}{2} & 0 & \frac{\sqrt{3}}{2} \\
0 & 1 & 0 \\
\frac{\sqrt{3}}{2} & 0 & \frac{1}{2}
\end{bmatrix}
\end{align}\]
\[\begin{align}
\text{Image of (1, 4, 7)} & = \begin{bmatrix}
-\frac{1}{2} & 0 & \frac{\sqrt{3}}{2} \\
0 & 1 & 0 \\
\frac{\sqrt{3}}{2} & 0 & \frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
1 \\
4 \\
7
\end{bmatrix} \\
& =
\begin{bmatrix}
\frac{7\sqrt{3}-1}{2} \\
4 \\
\frac{7+\sqrt{3}}{2}
\end{bmatrix}
\end{align}\]
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