Systems of Linear Equations Revision Notes for AQA A-Level Further Maths

Systems of Linear Equations

Introduction to determinants

When working with square matrices, you can calculate a special scalar value called the determinant. This value is particularly important because it allows you to find the inverse of a matrix and determine whether the matrix has certain properties.

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Key definition: For a 2×2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the determinant of $A$ is calculated using the formula:

$\det(A) = ad - bc$

The determinant can be written as $|A|$ or $\det(A)$ .

Calculating determinants

To find the determinant of a matrix, you multiply the elements on the leading diagonal (top-left to bottom-right) and subtract the product of the other diagonal.

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Worked Example: Finding a Determinant

Find the determinant of $A = \begin{pmatrix} 5 & 2 \\ -1 & -3 \end{pmatrix}$

Using the formula $\det(A) = ad - bc$ :

$\det(A) = (5 \times -3) - (2 \times -1)$ $= -15 - (-2)$ $= -15 + 2$ $= -13$

Singular matrices

A matrix is classified based on the value of its determinant.

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Key definition: If $\det(A) = 0$ , then $A$ is called a singular matrix.

Singular matrices have special properties that distinguish them from non-singular matrices (matrices with non-zero determinants).

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Critical Property: Singular matrices do not have an inverse. Only non-singular matrices (with $\det(A) \neq 0$ ) can have inverses.

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Worked Example: Finding the Value for a Singular Matrix

Find the value of $k$ for which the matrix $\begin{pmatrix} 3 & -1 \\ -2 & k \end{pmatrix}$ is singular.

For the matrix to be singular, its determinant must equal zero:

$\det\begin{pmatrix} 3 & -1 \\ -2 & k \end{pmatrix} = (3 \times k) - (-1 \times -2) = 0$

$3k - 2 = 0$

$k = \frac{2}{3}$

Therefore, when $k = \frac{2}{3}$ , the matrix is singular.

Properties of determinants

Determinants have useful properties when working with matrix multiplication.

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Key property: For any square matrices $A$ and $B$ :

$\det(AB) = \det(A) \times \det(B)$

This multiplicative property is extremely useful for working with combined matrices.

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Worked Example: Using the Multiplicative Property

The 2×2 matrices $A$ , $B$ , and $C$ are such that $C = AB$ . Given that $C = \begin{pmatrix} 2 & -4 \\ 3 & -5 \end{pmatrix}$ and $\det(A) = \frac{1}{4}$ , calculate the determinant of $B$ .

First, calculate $\det(C)$ using the formula $ad - bc$ :

$\det(C) = 2 \times (-5) - (-4) \times 3$ $= -10 + 12$ $= 2$

Now use the property $\det(AB) = \det(A) \times \det(B)$ :

$\frac{1}{4} \times \det(B) = 2$

$\det(B) = 2 \times 4 = 8$

Matrix inverses

The identity matrix

Before defining the inverse, we need to understand the identity matrix.

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Key definition: The identity matrix $I$ is the matrix where $AI = IA = A$ for any compatible matrix $A$ .

For 2×2 matrices, the identity matrix is:

$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Notice that the leading diagonal contains ones and all other elements are zero. The determinant of the identity matrix is $\det(I) = 1 - 0 = 1$ .

Defining the inverse matrix

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Key definition: The inverse of a matrix $A$ is $A^{-1}$ where:

$AA^{-1} = A^{-1}A = I$

This means that when you multiply a matrix by its inverse (in either order), you get the identity matrix.

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Singular matrices do not have an inverse. Only non-singular matrices (with $\det(A) \neq 0$ ) have inverses.

Formula for the inverse of a 2×2 matrix

For a 2×2 matrix, there is a formula to find the inverse directly.

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Key formula: If $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , then the inverse matrix is:

$A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

The process involves:

Swap the elements on the leading diagonal ( $a$ and $d$ )
Change the signs of the off-diagonal elements ( $b$ and $c$ )
Multiply by $\frac{1}{\det(A)}$

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Worked Example: Finding the Inverse of a Matrix

Find the inverse of the matrix $B = \begin{pmatrix} 2 & 5 \\ -1 & 4 \end{pmatrix}$

Step 1: Calculate the determinant using $ad - bc$ :

$\det(B) = (2 \times 4) - (5 \times -1) = 8 - (-5) = 13$

Step 2: Apply the inverse formula:

$B^{-1} = \frac{1}{13} \begin{pmatrix} 4 & -5 \\ 1 & 2 \end{pmatrix}$

Step 3: Verify by checking that $BB^{-1} = I$ :

$BB^{-1} = \begin{pmatrix} 2 & 5 \\ -1 & 4 \end{pmatrix} \times \frac{1}{13} \begin{pmatrix} 4 & -5 \\ 1 & 2 \end{pmatrix}$

$= \frac{1}{13} \begin{pmatrix} 2 & 5 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} 4 & -5 \\ 1 & 2 \end{pmatrix}$

$= \frac{1}{13} \begin{pmatrix} 13 & 0 \\ 0 & 13 \end{pmatrix}$

$= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$

This confirms our inverse is correct.

Determinant of an inverse matrix

Using the property that $\det(AB) = \det(A) \times \det(B)$ with $B = A^{-1}$ , you can derive an important result.

Since $AA^{-1} = I$ , we have:

$\det(AA^{-1}) = \det(A) \times \det(A^{-1})$

But $AA^{-1} = I$ , so:

$\det(I) = \det(A) \times \det(A^{-1})$

The determinant of the identity matrix equals 1, therefore:

$1 = \det(A) \times \det(A^{-1})$

$\det(A^{-1}) = \frac{1}{\det(A)}$

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Key property: For any square, non-singular matrix $A$ :

$\det(A^{-1}) = \frac{1}{\det(A)}$

Self-inverse matrices

Some special matrices are their own inverse.

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Key definition: A matrix $C$ is self-inverse if $C^{-1} = C$ .

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Worked Example: Determinant of a Self-Inverse Matrix

Given that the matrix $C$ is self-inverse, determine the possible values of $\det(C)$ .

Since $C$ is self-inverse, we know that $C^{-1} = C$ .

Therefore:

$\det(C^{-1}) = \det(C)$

Using the property $\det(A^{-1}) = \frac{1}{\det(A)}$ :

$\frac{1}{\det(C)} = \det(C)$

$[\det(C)]^2 = 1$

$\det(C) = \pm 1$

So a self-inverse matrix must have a determinant of either +1 or -1.

Solving systems of linear equations using matrices

Matrices provide a powerful method for solving systems of simultaneous linear equations. The approach involves rewriting the system in matrix form and then using the inverse matrix to find the solution.

Matrix representation of systems

Consider the system of equations:

$ax + by = x'$
$cx + dy = y'$

This can be written in matrix form as:

$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix}$

Or more compactly as $M\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix}$

Strategy for solving systems of linear equations

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Strategy for solving systems of linear equations:

Rewrite the system using matrices - Express the equations in the form $M\mathbf{x} = \mathbf{b}$ where $M$ is the coefficient matrix, $\mathbf{x}$ is the variable vector, and $\mathbf{b}$ is the constant vector.
Pre-multiply or post-multiply by the inverse - Multiply both sides by $M^{-1}$ to get $M^{-1}M\mathbf{x} = M^{-1}\mathbf{b}$ .
Use the fact that $AA^{-1} = A^{-1}A = I$ - Simplify the left side to get $\mathbf{x} = M^{-1}\mathbf{b}$ , then calculate the values.

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Worked Example: Solving a System of Equations

Use matrices to solve the simultaneous equations:

$2x + 3y = 11$
$4x - y = -6$

Step 1: Write the system in matrix form:

$\begin{pmatrix} 2 & 3 \\ 4 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 11 \\ -6 \end{pmatrix}$

Step 2: Find the inverse of the coefficient matrix $A = \begin{pmatrix} 2 & 3 \\ 4 & -1 \end{pmatrix}$ :

$\det(A) = (2 \times -1) - (3 \times 4) = -2 - 12 = -14$

$A^{-1} = \frac{1}{-14} \begin{pmatrix} -1 & -3 \\ -4 & 2 \end{pmatrix} = \frac{1}{14} \begin{pmatrix} 1 & 3 \\ 4 & -2 \end{pmatrix}$

Step 3: Pre-multiply both sides by $A^{-1}$ :

$A^{-1}A\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\begin{pmatrix} 11 \\ -6 \end{pmatrix}$

Since $A^{-1}A = I$ :

$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{14} \begin{pmatrix} 1 & 3 \\ 4 & -2 \end{pmatrix} \begin{pmatrix} 11 \\ -6 \end{pmatrix}$

$= \frac{1}{14} \begin{pmatrix} 11 - 18 \\ 44 + 12 \end{pmatrix} = \frac{1}{14} \begin{pmatrix} -7 \\ 56 \end{pmatrix}$

$= \begin{pmatrix} -\frac{1}{2} \\ 4 \end{pmatrix}$

Therefore, $x = -\frac{1}{2}$ and $y = 4$ .

Matrix proof techniques

You can construct matrix proofs by pre-multiplying or post-multiplying both sides of an equation. This technique is useful for proving identities involving inverses.

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Worked Example: Proving the Inverse of a Product

Prove that $(AB)^{-1} = B^{-1}A^{-1}$

Start with the definition that $(AB)^{-1}(AB) = I$ :

Post-multiply both sides by $B^{-1}$ : $(AB)^{-1}(AB)B^{-1} = IB^{-1}$

Since $BB^{-1} = I$ and $IB^{-1} = B^{-1}$ : $(AB)^{-1}A = B^{-1}$

Post-multiply both sides by $A^{-1}$ : $(AB)^{-1}AA^{-1} = B^{-1}A^{-1}$

Since $AA^{-1} = I$ : $(AB)^{-1} = B^{-1}A^{-1}$

This proves the required result.

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Key property: If $A$ and $B$ are non-singular matrices, then:

$(AB)^{-1} = B^{-1}A^{-1}$

Notice how the order reverses when taking the inverse of a product.

Exam tips

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Important Exam Tips:

Always check that $\det(A) \neq 0$ before attempting to find an inverse - if the determinant is zero, the matrix has no inverse.
When solving systems of equations, clearly show the matrix form before finding the inverse.
Verify your inverse by checking that $AA^{-1} = I$ - this is good practice in exams.
Remember that matrix multiplication is not commutative: $(AB)^{-1} \neq A^{-1}B^{-1}$ , but $(AB)^{-1} = B^{-1}A^{-1}$ .
For proofs involving inverses, use the property $AA^{-1} = A^{-1}A = I$ by pre-multiplying or post-multiplying appropriately.
Scientific calculators can find determinants and inverses, but you must show your working in exams.

Remember!

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Key Points to Remember:

The determinant of a 2×2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\det(A) = ad - bc$
A singular matrix has $\det(A) = 0$ and has no inverse; only non-singular matrices (with $\det(A) \neq 0$ ) have inverses
The inverse of $A$ is found using $A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ and satisfies $AA^{-1} = A^{-1}A = I$
Important determinant properties: $\det(AB) = \det(A) \times \det(B)$ and $\det(A^{-1}) = \frac{1}{\det(A)}$
To solve systems of linear equations, write in matrix form $M\mathbf{x} = \mathbf{b}$ , then pre-multiply by $M^{-1}$ to get $\mathbf{x} = M^{-1}\mathbf{b}$

Systems of Linear Equations (AQA A-Level Further Maths): Revision Notes