Manipulating Determinants Revision Notes for AQA A-Level Further Maths

Manipulating Determinants

Calculating determinants using any row or column

When working with a $3 \times 3$ matrix, the determinant can be calculated using any row or any column. This flexibility is particularly useful when simplifying calculations.

For a matrix $\mathbf{A} = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ , the determinant is defined as:

$\det(\mathbf{A}) = a\begin{vmatrix} e & f \\ h & i \end{vmatrix} - b\begin{vmatrix} d & f \\ g & i \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix}$

This formula uses the first row (top row) to expand the determinant. The signs alternate according to a pattern.

The sign matrix

The pattern of signs for expanding a determinant follows this sign matrix:

$\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$

chatImportant

This checkerboard pattern starts with a positive sign in the top-left corner and alternates across rows and columns. When you expand along any row or column, you must apply these signs to each term.

Memory aid: Think "plus minus plus, minus plus minus, plus minus plus" - the pattern starts with + in the top-left corner and alternates like a checkerboard.

Expanding using different rows or columns

You can choose any row or column to calculate the determinant, and you will always get the same answer. The choice of row or column should be strategic - look for rows or columns containing zeros, as these will simplify your calculation significantly.

For example, using the second row to expand the determinant:

$\det(\mathbf{A}) = -d\begin{vmatrix} b & c \\ h & i \end{vmatrix} + e\begin{vmatrix} a & c \\ g & i \end{vmatrix} - f\begin{vmatrix} a & b \\ g & h \end{vmatrix}$

Similarly, using the third column:

$\det(\mathbf{A}) = c\begin{vmatrix} d & e \\ g & h \end{vmatrix} - f\begin{vmatrix} a & b \\ g & h \end{vmatrix} + i\begin{vmatrix} a & b \\ d & e \end{vmatrix}$

All three methods produce the same result: $aei - afh - bdi + bfg + cdh - ceg$ .

infoNote

Exam tip: Always check for rows or columns with zeros before choosing how to expand. This will save you time and reduce calculation errors.

The transpose property

A particularly useful property relates the determinant of a matrix to the determinant of its transpose.

The transpose of a matrix, denoted $\mathbf{M}^T$ , is formed by swapping its rows and columns. The element in row $i$ , column $j$ becomes the element in row $j$ , column $i$ .

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Key property:

$|\mathbf{M}| = |\mathbf{M}^T|$

This means the determinant of a matrix equals the determinant of its transpose. This property confirms why calculating a determinant using the second row gives the same answer as using the second column.

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This property is particularly useful in proofs and when checking your work. If you calculate the determinant using a row and want to verify, you could calculate using the corresponding column of the transpose.

Row and column operations

Strategic manipulation of matrices using row and column operations can significantly simplify determinant calculations. However, different operations have different effects on the determinant value.

Operations that preserve the determinant value

The following operations can be performed without affecting the value of the determinant:

Adding or subtracting any multiple of a row to another row
- Example: $R_1 \rightarrow R_1 + 3R_2$ leaves the determinant unchanged
Adding or subtracting any multiple of a column to another column
- Example: $C_3 \rightarrow C_3 - 2C_1$ leaves the determinant unchanged

These operations are extremely useful for creating zeros in the matrix, which simplifies the calculation when you expand along that row or column.

infoNote

Exam strategy: Use these operations to create a row or column with multiple zeros before expanding the determinant. This dramatically reduces the amount of calculation required.

Operations that change the sign of the determinant

The following operations change the sign of the determinant:

Swapping two rows
- If you swap any two rows, the determinant changes sign: $\det(\mathbf{A}) \rightarrow -\det(\mathbf{A})$
Swapping two columns
- If you swap any two columns, the determinant changes sign: $\det(\mathbf{A}) \rightarrow -\det(\mathbf{A})$

chatImportant

Remember: If you perform an even number of swaps, the signs cancel out and the determinant remains positive. Always keep track of your swap operations!

Scalar multiplication and division

Operations involving scalars have predictable effects on the determinant:

Multiplying a row or column by a scalar
- Multiplying any row or column by a scalar $k$ multiplies the entire determinant by $k$
- If $R_i \rightarrow kR_i$ , then $\det(\mathbf{A}) \rightarrow k\det(\mathbf{A})$
Dividing a row or column by a scalar
- Dividing any row or column by a scalar $k$ divides the entire determinant by $k$
- If $R_i \rightarrow \frac{1}{k}R_i$ , then $\det(\mathbf{A}) \rightarrow \frac{1}{k}\det(\mathbf{A})$

infoNote

Exam tip: When you factor out a scalar from a row or column, remember to account for this in your final answer by multiplying the calculated determinant by that scalar.

Worked example: Simplifying determinants using operations

Let's work through a complete example showing how to use row and column operations to simplify a determinant calculation.

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Worked Example: Using Operations to Simplify Determinants

Problem: Use row and column operations to show that $\begin{vmatrix} 24 & 20 & -1 \\ 9 & 4 & -2 \\ 3 & 6 & 2 \end{vmatrix} = 6\begin{vmatrix} 0 & 0 & -1 \\ 2 & 5 & 0 \\ 1 & 6 & 2 \end{vmatrix}$ , then evaluate this determinant.

Solution steps:

Step 1: Add row 3 to row 2

$\begin{vmatrix} 24 & 20 & -1 \\ 9 & 4 & -2 \\ 3 & 6 & 2 \end{vmatrix} = \begin{vmatrix} 24 & 20 & -1 \\ 12 & 10 & 0 \\ 3 & 6 & 2 \end{vmatrix}$

This creates a zero in the $(2,3)$ position.

Step 2: Divide row 2 by 2

$= 2\begin{vmatrix} 24 & 20 & -1 \\ 6 & 5 & 0 \\ 3 & 6 & 2 \end{vmatrix}$

Remember to factor out the 2, as dividing by 2 divides the determinant by 2.

Step 3: Subtract $4 \times$ row 2 from row 1

$= 2\begin{vmatrix} 0 & 0 & -1 \\ 6 & 5 & 0 \\ 3 & 6 & 2 \end{vmatrix}$

This creates two zeros in the first row.

Step 4: Divide column 1 by 3

$= 6\begin{vmatrix} 0 & 0 & -1 \\ 2 & 5 & 0 \\ 1 & 6 & 2 \end{vmatrix}$

We've now shown the required equivalence. The factor is $2 \times 3 = 6$ .

Step 5: Evaluate the determinant using the top row

$6\begin{vmatrix} 0 & 0 & -1 \\ 2 & 5 & 0 \\ 1 & 6 & 2 \end{vmatrix} = 6 \times (-1) \times \begin{vmatrix} 2 & 5 \\ 1 & 6 \end{vmatrix}$

$= 6 \times (-1) \times (12 - 5) = 6 \times (-1) \times 7 = -42$

Key insight: By strategically creating zeros, we reduced the calculation to evaluating just one $2 \times 2$ determinant instead of three.

Factorising determinants

Row and column operations can also be used to factorise determinants, which is particularly useful when determinants contain algebraic expressions.

Strategy for factorising determinants

Follow these three steps:

Step 1: Take out common factors from rows or columns

Look for common factors in each row or column
Factor these out in front of the determinant

Step 2: Perform row and column operations to create rows or columns where the elements have a common factor

Use addition/subtraction operations strategically
Aim to create patterns that reveal common factors

Step 3: Multiply out the determinant

Expand along a convenient row or column
Simplify the result

Worked example: Factorising with algebraic expressions

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Worked Example: Factorising Determinants with Algebraic Expressions

Problem: Factorise $\begin{vmatrix} a & b & 1 \\ a^2 & b^2 & 1 \\ a^3 & b^3 & 1 \end{vmatrix}$ and determine when the system of equations $ax + by + z = 4$ , $a^2x + b^2y + z = -2$ , $a^3x + b^3y + z = 0$ has a unique solution.

Solution:

Part a) Factorising the determinant:

Starting with:

$\begin{vmatrix} a & b & 1 \\ a^2 & b^2 & 1 \\ a^3 & b^3 & 1 \end{vmatrix}$

Step 1: Factor out $a$ from column 1 and $b$ from column 2:

$= ab\begin{vmatrix} 1 & 1 & 1 \\ a & b & 1 \\ a^2 & b^2 & 1 \end{vmatrix}$

Step 2: Subtract column 3 from column 1, then from column 2:

$= ab\begin{vmatrix} 0 & 0 & 1 \\ a-1 & b-1 & 1 \\ a^2-1 & b^2-1 & 1 \end{vmatrix}$

Step 2 (continued): Factor out $(a-1)$ from column 1:

Note that $a^2 - 1 = (a+1)(a-1)$ , so:

$= ab(a-1)\begin{vmatrix} 0 & 0 & 1 \\ 1 & b-1 & 1 \\ a+1 & b^2-1 & 1 \end{vmatrix}$

Step 2 (continued): Factor out $(b-1)$ from column 2:

Note that $b^2 - 1 = (b+1)(b-1)$ , so:

$= ab(a-1)(b-1)\begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & 1 \\ a+1 & b+1 & 1 \end{vmatrix}$

Step 3: Multiply out using the first row:

$= ab(a-1)(b-1) \times 1 \times \begin{vmatrix} 1 & 1 \\ a+1 & b+1 \end{vmatrix}$

$= ab(a-1)(b-1)[(b+1)-(a+1)]$

$= ab(a-1)(b-1)(b-a)$

Part b) Conditions for a unique solution:

The system can be written as:

$\begin{pmatrix} a & b & 1 \\ a^2 & b^2 & 1 \\ a^3 & b^3 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 0 \end{pmatrix}$

From part a, we know:

$\det\begin{pmatrix} a & b & 1 \\ a^2 & b^2 & 1 \\ a^3 & b^3 & 1 \end{pmatrix} = ab(a-1)(b-1)(b-a)$

For a unique solution, the determinant must be non-zero:

$ab(a-1)(b-1)(b-a) \neq 0$

This means: $a \neq 0$ , $b \neq 0$ , $a \neq 1$ , $b \neq 1$ , and $a \neq b$

Therefore, the system has a unique solution if these conditions are met: $a \neq 0, b \neq 0, a \neq 1, b \neq 1, a \neq b$ .

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Exam insight: This example demonstrates how factorising determinants is crucial when working with systems of equations. The determinant being zero indicates no unique solution exists.

Remember!

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Key Points to Remember:

Any row or column can be used to calculate a determinant - choose strategically by looking for zeros to simplify calculations.
The transpose property $|\mathbf{M}| = |\mathbf{M}^T|$ means the determinant value is unchanged when rows and columns are swapped.
Preserve determinant value by adding or subtracting multiples of rows/columns; change sign by swapping rows or columns; multiply/divide the determinant when multiplying/dividing a row or column by a scalar.
Create zeros before expanding - use row and column operations to simplify the matrix before calculating the determinant, making the expansion much easier.
Factorise systematically by extracting common factors from rows/columns first, then using operations to reveal more common factors, before finally multiplying out the simplified determinant.

Manipulating Determinants (AQA A-Level Further Maths): Revision Notes