Determinants and inverses of 2×2 matrices | Year 11

01 / LEARN

The idea, step by step

The determinant is a scalar

For $A = (\begin{array}{cccccccccc} a & b \\ c & d \end{array})$ , the determinant is $a d - b c$ . Keep the order in the subtraction. It is one number, not another matrix.

A non-zero determinant means the two columns give independent directions, so the corresponding linear transformation can be undone uniquely.

Construct and verify the inverse

When $a d - b c \neq 0$ , interchange a and d, change the signs of b and c, then divide every entry by the determinant. The inverse satisfies $A^{- 1} A = A A^{- 1} = I$ .

If the determinant is zero, stop: there is no inverse. Dividing by zero is not a way to produce an inverse. Check a calculated inverse with one of the products, particularly when negative signs or fractions appear.

A^{- 1} = \frac{1}{a d - b c} (\begin{array}{cccccccccc} d & - b \\ - c & a \end{array})

A common trap

Using $a d + b c$ or forgetting to divide the whole adjugate.

Instead: Calculate $a d - b c$ separately first, then apply the scalar to all four entries.

02 / WORKED EXAMPLES

See the thinking

Example 1

A determinant-one inverse

Find the inverse of $A = (\begin{array}{cccccccccc} 2 & 1 \\ 3 & 2 \end{array})$ .

The determinant is $2 (2) - 1 (3) = 1$ .
Swap the diagonal entries and negate the off-diagonal entries.
Multiplication with A gives $(\begin{array}{cccccccccc} 1 & 0 \\ 0 & 1 \end{array})$ .

A^{- 1} = (\begin{array}{cccccccccc} 2 & - 1 \\ - 3 & 2 \end{array})

Example 2

An inverse with fractions

Find $B^{- 1}$ for $B = (\begin{array}{cccccccccc} 4 & 0 \\ 1 & 2 \end{array})$ .

The determinant is 8.
The adjugate is $(\begin{array}{cccccccccc} 2 & 0 \\ - 1 & 4 \end{array})$ .
Divide every entry by 8.

B^{- 1} = (\begin{array}{cccccccccc} \frac{1}{4} & 0 \\ - \frac{1}{8} & \frac{1}{2} \end{array})

Check your understanding

For which value of $k$ is $(\begin{array}{cccccccccc} 1 & 2 \\ 3 & k \end{array})$ non-invertible?

Show answer

$k = 6$ .

The determinant is $k - 6$ , which is zero at 6.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find $det (\begin{array}{cccccccccc} 3 & 2 \\ 1 & 4 \end{array})$ .
2 marks
Find the inverse of $(\begin{array}{cccccccccc} 1 & 2 \\ 0 & 1 \end{array})$ .
2 marks
Find the inverse of $(\begin{array}{cccccccccc} 2 & 0 \\ 0 & 5 \end{array})$ .
2 marks
Find $t$ if $det (\begin{array}{cccccccccc} t & 1 \\ 2 & 3 \end{array}) = 7$ .
2 marks
Find the inverse of $(\begin{array}{cccccccccc} 0 & - 1 \\ 1 & 0 \end{array})$ .
2 marks
Show that $(\begin{array}{cccccccccc} 2 & 4 \\ 1 & 2 \end{array})$ has no inverse.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $det (\begin{array}{cccccccccc} 3 & 2 \\ 1 & 4 \end{array})$ .
2 marks
$3 (4) - 2 (1) = 10$ .
$10$ .
Find the inverse of $(\begin{array}{cccccccccc} 1 & 2 \\ 0 & 1 \end{array})$ .
2 marks
The determinant is 1.
Swap diagonal entries and negate the off-diagonal entries.
$(\begin{array}{cccccccccc} 1 & - 2 \\ 0 & 1 \end{array})$ .
Find the inverse of $(\begin{array}{cccccccccc} 2 & 0 \\ 0 & 5 \end{array})$ .
2 marks
Each diagonal scale is undone by its reciprocal.
The determinant is 10, so the inverse exists.
$(\begin{array}{cccccccccc} \frac{1}{2} & 0 \\ 0 & \frac{1}{5} \end{array})$ .
Find $t$ if $det (\begin{array}{cccccccccc} t & 1 \\ 2 & 3 \end{array}) = 7$ .
2 marks
$3 t - 2 = 7$ .
Thus $3 t = 9$ .
$t = 3$ .
Find the inverse of $(\begin{array}{cccccccccc} 0 & - 1 \\ 1 & 0 \end{array})$ .
2 marks
The determinant is 1.
The inverse reverses the quarter-turn.
$(\begin{array}{cccccccccc} 0 & 1 \\ - 1 & 0 \end{array})$ .
Show that $(\begin{array}{cccccccccc} 2 & 4 \\ 1 & 2 \end{array})$ has no inverse.
2 marks
$2 (2) - 4 (1) = 0$ .
The inverse formula would require division by zero.
Its determinant is zero.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: matrices and transformations; original lessons follow the approved school-sized chapter structure.

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