Properties and Arithmetic (AQA A-Level Further Maths): Revision Notes

Understanding Matrix Order

For example:

• The matrix $\begin{pmatrix} 5 & 3 & -1 \\ -4 & 0 & 5 \end{pmatrix}$ has order $2 \times 3$ (2 rows, 3 columns)
• The matrix $\begin{pmatrix} 2 \\ 7 \\ -5 \end{pmatrix}$ has order $3 \times 1$ (3 rows, 1 column)

Worked Example: Matrix Addition and Subtraction

Given three matrices:

• $A = \begin{pmatrix} 2 & 5 \\ 0 & 4 \\ -1 & -3 \end{pmatrix}$
• $B = \begin{pmatrix} 2 & -3 \\ 0 & 5 \end{pmatrix}$
• $C = \begin{pmatrix} 6 & 15 \\ 0 & 12 \\ -3 & -9 \end{pmatrix}$

We can analyse various operations:

Part a: Finding A + B

This is not possible because $A$ and $B$ do not have the same order. Matrix $A$ is $3 \times 2$ while matrix $B$ is $2 \times 2$.

Part b: Finding A + C

Since both $A$ and $C$ have order $3 \times 2$, we can add them:

$A + C = \begin{pmatrix} 2 & 5 \\ 0 & 4 \\ -1 & -3 \end{pmatrix} + \begin{pmatrix} 6 & 15 \\ 0 & 12 \\ -3 & -9 \end{pmatrix}$

Add the corresponding elements:

$= \begin{pmatrix} 2+6 & 5+15 \\ 0+0 & 4+12 \\ -1+(-3) & -3+(-9) \end{pmatrix}$

$= \begin{pmatrix} 8 & 20 \\ 0 & 16 \\ -4 & -12 \end{pmatrix}$

Part c: Finding 2B

This involves scalar multiplication (covered in the next section).

Part d: Finding C - 3A

First, multiply matrix $A$ by 3, then subtract from $C$:

$C - 3A = \begin{pmatrix} 6 & 15 \\ 0 & 12 \\ -3 & -9 \end{pmatrix} - 3\begin{pmatrix} 2 & 5 \\ 0 & 4 \\ -1 & -3 \end{pmatrix}$

$= \begin{pmatrix} 6-3 \times 2 & 15-3 \times 5 \\ 0-3 \times 0 & 12-3 \times 4 \\ -3-3 \times(-1) & -9-3 \times(-3) \end{pmatrix}$

$= \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}$

This result is called a zero matrix.

Example: Scalar Multiplication

If $B = \begin{pmatrix} 2 & -3 \\ 0 & 5 \end{pmatrix}$, then:

$2B = 2\begin{pmatrix} 2 & -3 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 2 \times 2 & 2 \times(-3) \\ 2 \times 0 & 2 \times 5 \end{pmatrix} = \begin{pmatrix} 4 & -6 \\ 0 & 10 \end{pmatrix}$

Every element in the original matrix is multiplied by 2.

Conformability Examples

• A $3 \times 2$ matrix can be multiplied by a $2 \times 4$ matrix (the middle numbers match)
• A $3 \times 2$ matrix cannot be multiplied by a $3 \times 2$ matrix (the middle numbers don't match)

Worked Example: Matrix Multiplication

Given:

• $A = \begin{pmatrix} 3 & 0 \\ -1 & 2 \\ 7 & -4 \end{pmatrix}$ (order $3 \times 2$)
• $B = \begin{pmatrix} 5 & 1 \\ -3 & 0 \end{pmatrix}$ (order $2 \times 2$)

Part a: Finding AB

First, check conformability: $A$ has 2 columns and $B$ has 2 rows, so they are conformable.

The product will have order $3 \times 2$.

$AB = \begin{pmatrix} 3 & 0 \\ -1 & 2 \\ 7 & -4 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ -3 & 0 \end{pmatrix}$

Calculate each element:

For the element in row 1, column 1: $(3 \times 5) + (0 \times (-3)) = 15 + 0 = 15$

For the element in row 1, column 2: $(3 \times 1) + (0 \times 0) = 3 + 0 = 3$

For the element in row 2, column 1: $((-1) \times 5) + (2 \times (-3)) = -5 + (-6) = -11$

For the element in row 2, column 2: $((-1) \times 1) + (2 \times 0) = -1 + 0 = -1$

For the element in row 3, column 1: $(7 \times 5) + ((-4) \times (-3)) = 35 + 12 = 47$

For the element in row 3, column 2: $(7 \times 1) + ((-4) \times 0) = 7 + 0 = 7$

Therefore: $AB = \begin{pmatrix} 15 & 3 \\ -11 & -1 \\ 47 & 7 \end{pmatrix}$

Notice that the product of a $3 \times 2$ matrix and a $2 \times 2$ matrix gives a $3 \times 2$ matrix.

Part b: Finding BA

Matrix $B$ has 2 columns but matrix $A$ has 3 rows, so they are not conformable for multiplication in this order. Therefore, $BA$ is not possible.

This demonstrates an important point: even if $AB$ exists, $BA$ may not exist.

Part c: Finding B²

$B^2$ means $B \times B$. This is only possible if $B$ is a square matrix (same number of rows and columns).

$B^2 = \begin{pmatrix} 5 & 1 \\ -3 & 0 \end{pmatrix}\begin{pmatrix} 5 & 1 \\ -3 & 0 \end{pmatrix}$

$= \begin{pmatrix} (5 \times 5) + (1 \times (-3)) & (5 \times 1) + (1 \times 0) \\ ((-3) \times 5) + (0 \times (-3)) & ((-3) \times 1) + (0 \times 0) \end{pmatrix}$

$= \begin{pmatrix} 22 & 5 \\ -15 & -3 \end{pmatrix}$

Example: Matrix Transpose

The transpose of $\begin{pmatrix} 1 & -2 & 4 \\ 0 & 5 & -6 \end{pmatrix}$ is $\begin{pmatrix} 1 & 0 \\ -2 & 5 \\ 4 & -6 \end{pmatrix}$.

The first row of the original matrix becomes the first column of the transpose, and the second row becomes the second column.

Worked Example: Transpose Properties

Given:

• $A = \begin{pmatrix} 1 & 0 & -3 \\ k & 2 & 4 \\ 5 & -6 & 0 \end{pmatrix}$
• $B = \begin{pmatrix} 2 & k & 0 \\ -3 & 1 & -5 \\ 0 & 1 & -1 \end{pmatrix}$

Find $(AB)^T$ without first finding $AB$.

Solution:

Step 1: Write down the transpose of each matrix.

$A^T = \begin{pmatrix} 1 & k & 5 \\ 0 & 2 & -6 \\ -3 & 4 & 0 \end{pmatrix}$

$B^T = \begin{pmatrix} 2 & -3 & 0 \\ k & 1 & 1 \\ 0 & -5 & -1 \end{pmatrix}$

Step 2: Use the property $(AB)^T = B^T A^T$.

$(\mathbf{AB})^T = \begin{pmatrix} 2 & -3 & 0 \\ k & 1 & 1 \\ 0 & -5 & -1 \end{pmatrix}\begin{pmatrix} 1 & k & 5 \\ 0 & 2 & -6 \\ -3 & 4 & 0 \end{pmatrix}$

Step 3: Multiply the matrices.

$= \begin{pmatrix} 2 & 2k-6 & 28 \\ k-3 & k^2+6 & 5k-6 \\ 3 & -14 & 30 \end{pmatrix}$

Worked Example: Proving Associativity

To prove the associative property $(AB)C = A(BC)$ for $2 \times 2$ matrices, we use general matrices with subscript notation:

Let:

• $A = \begin{pmatrix} a_1 & a_2 \\ a_3 & a_4 \end{pmatrix}$
• $B = \begin{pmatrix} b_1 & b_2 \\ b_3 & b_4 \end{pmatrix}$
• $C = \begin{pmatrix} c_1 & c_2 \\ c_3 & c_4 \end{pmatrix}$

The proof involves:

1. First calculating $AB$
2. Then multiplying by $C$ to get $(AB)C$
3. Separately calculating $BC$
4. Then multiplying $A$ by the result to get $A(BC)$
5. Showing both expressions are equal

This systematic approach works by expanding all the bracket multiplications and simplifying.

Worked Example: Finding Unknown Elements

Given that $\begin{pmatrix} x & 5 \\ 3 & 3y \end{pmatrix}\begin{pmatrix} 1 \\ 5 \end{pmatrix} = \begin{pmatrix} y \\ x \end{pmatrix}$, find the values of $x$ and $y$.

Solution:

Step 1: Multiply the matrices on the left-hand side.

$\begin{pmatrix} x & 5 \\ 3 & 3y \end{pmatrix}\begin{pmatrix} 1 \\ 5 \end{pmatrix} = \begin{pmatrix} x \times 1 + 5 \times 5 \\ 3 \times 1 + 3y \times 5 \end{pmatrix} = \begin{pmatrix} x+25 \\ 3+15y \end{pmatrix}$

Step 2: Since the matrices are equal, equate corresponding elements.

$\begin{pmatrix} x+25 \\ 3+15y \end{pmatrix} = \begin{pmatrix} y \\ x \end{pmatrix}$

This gives us two equations:

• $x + 25 = y$
• $3 + 15y = x$

Step 3: Solve the equations simultaneously.

From the first equation: $y = x + 25$

Substitute into the second equation:

• $3 + 15(x + 25) = x$
• $3 + 15x + 375 = x$
• $378 = x - 15x$
• $378 = -14x$
• $x = -27$

Therefore: $y = -27 + 25 = -2$

Worked Example: Writing Systems as Matrix Equations

Part a: Write the system $x + 7y = -5$ and $3x - y = 29$ as a matrix equation.

The coefficient matrix contains the coefficients of $x$ and $y$:

$\begin{pmatrix} 1 & 7 \\ 3 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -5 \\ 29 \end{pmatrix}$

Part b: Write the system $3x - 2y - 5z = -3$, $x + 7y + 4z = 30$, and $5x - 9z = 13$ as a matrix equation.

Notice that the number of variables determines the number of columns in the coefficient matrix, and the number of equations determines the number of rows:

$\begin{pmatrix} 3 & -2 & -5 \\ 1 & 7 & 4 \\ 5 & 0 & -9 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -3 \\ 30 \\ 13 \end{pmatrix}$

Where $y$ is missing in the third equation, we use 0 as its coefficient.

Worked Example: Matrix Induction Proof

Prove by induction that $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & 0 \\ 2n & 1 \end{pmatrix}$ for all positive integers $n$.

Solution:

Step 1: Check the base case ($n = 1$)

When $n = 1$:

LHS: $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^1 = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$

RHS: $\begin{pmatrix} 1 & 0 \\ 2(1) & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$

The statement is true for $n = 1$.

Step 2: Assume the statement is true for $n = k$

Assume $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^k = \begin{pmatrix} 1 & 0 \\ 2k & 1 \end{pmatrix}$

Step 3: Prove it's true for $n = k + 1$

We need to show that $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 0 \\ 2(k+1) & 1 \end{pmatrix}$

Starting with the left-hand side:

$\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}^k \times \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$

Using our assumption:

$= \begin{pmatrix} 1 & 0 \\ 2k & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$

Multiply the matrices:

$= \begin{pmatrix} (1 \times 1) + (0 \times 2) & (1 \times 0) + (0 \times 1) \\ (2k \times 1) + (1 \times 2) & (2k \times 0) + (1 \times 1) \end{pmatrix}$

$= \begin{pmatrix} 1 & 0 \\ 2k + 2 & 1 \end{pmatrix}$

$= \begin{pmatrix} 1 & 0 \\ 2(k+1) & 1 \end{pmatrix}$

This is what we needed to show.

Step 4: Write the conclusion

Since the statement is true for $n = 1$, and assuming it's true for $n = k$ implies it's true for $n = k + 1$, the statement is true for all positive integers $n$ by mathematical induction.

Worked Example: UK Cities Rainfall Data

The table shows the probabilities of a spring day being rainy in four UK cities:

Part a: Write a matrix $A$ to represent the table and a vector $\mathbf{x}$ to represent the number of days in each month.

The matrix $A$ has cities as rows and months as columns:

$A = \begin{pmatrix} 0.32 & 0.30 & 0.29 \\ 0.38 & 0.33 & 0.37 \\ 0.42 & 0.37 & 0.36 \\ 0.45 & 0.38 & 0.38 \end{pmatrix}$

The vector $\mathbf{x}$ contains the number of days in each month (March has 31 days, April has 30, May has 31):

$\mathbf{x} = \begin{pmatrix} 31 \\ 30 \\ 31 \end{pmatrix}$

Part b: Show how to calculate the total number of rainy days expected in each city over the three months.

We multiply the matrix by the vector:

$A\mathbf{x} = \begin{pmatrix} 0.32 & 0.30 & 0.29 \\ 0.38 & 0.33 & 0.37 \\ 0.42 & 0.37 & 0.36 \\ 0.45 & 0.38 & 0.38 \end{pmatrix}\begin{pmatrix} 31 \\ 30 \\ 31 \end{pmatrix}$

Calculate each element:

For London: $0.32 \times 31 + 0.30 \times 30 + 0.29 \times 31 = 27.91$

For Edinburgh: $0.38 \times 31 + 0.33 \times 30 + 0.37 \times 31 = 33.15$

For Cardiff: $0.42 \times 31 + 0.37 \times 30 + 0.36 \times 31 = 35.28$

For Belfast: $0.45 \times 31 + 0.38 \times 30 + 0.38 \times 31 = 37.13$

Therefore:

$A\mathbf{x} = \begin{pmatrix} 27.91 \\ 33.15 \\ 35.28 \\ 37.13 \end{pmatrix}$

Conclusion: We expect a total of 28 rainy days in London, 33 in Edinburgh, 35 in Cardiff, and 37 in Belfast (answers to the nearest day).