3 × 3 determinants and Cramer's rule · Matrices & linear transformations

Evaluating a 3 × 3 determinant

Expand along the top row. Each entry multiplies the $2 \times 2$ determinant left when you cover its row and column, with the signs $+\ -\ +$ :

Along the top rowa(ei − fh) − b(di − fg) + c(dh − eg)

Step through one, term by term. Watch the sign of the middle term: it is always taken away.

Expanding a 3 × 3 determinantPress Next to take each top entry in turn

0running total

Expand along the top row. Each top entry is multiplied by the 2 × 2 determinant left when you cover its row and column, and the signs go + − + along the top, from the chessboard of signs. Press Next.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q9

Given that $F = \begin{pmatrix} 3 & 4 & 2 \\ 0 & -1 & 2 \\ 2 & 1 & -2 \end{pmatrix}$ :

Find $|F|$ , the determinant of $F$ .

Use the determinant of $F$ to solve the system of equations $3x + 4y + 2z = -1$ , $-y + 2z = -8$ , $2x + y - 2z = 6$ simultaneously.

The determinant
- First term: ${3\big((-1)(-2) - 2(1)\big) = 3(0) = 0}$ .
- Second term: ${-4\big(0(-2) - 2(2)\big) = -4(-4) = 16}$ .
- Third term: ${2\big(0(1) - (-1)(2)\big) = 2(2) = 4}$ .
- So ${|F| = 0 + 16 + 4 = 20}$ .
Think first. Cover the row and column of each top entry.
Cramer's rule
- ${\Delta_x = -20}$ , ${\Delta_y = 40}$ and ${\Delta_z = -60}$ .
- ${x = \frac{-20}{20} = -1}$ , ${y = \frac{40}{20} = 2}$ and ${z = \frac{-60}{20} = -3}$ .
Think first. Replace each column in turn by −1, −8, 6.
Check in the equations
- ${3(-1) + 4(2) + 2(-3) = -1}$ ✓, ${-2 + 2(-3) = -8}$ ✓, ${2(-1) + 2 - 2(-3) = 6}$ ✓.

Unknowns in a determinant

Expand as usual; the determinant becomes an expression in the unknown. Set it equal to the given value and solve.

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q11 (a)

Find the positive value of $y$ for which $\begin{vmatrix} y & 1 & 0 \\ 1 & y & 1 \\ 0 & 1 & y \end{vmatrix} = 0$ (4 d.p.).

Expand
- ${y(y^2 - 1) - 1(y - 0) + 0 = y^3 - 2y}$ .
Solve
- ${y^3 - 2y = 0}$ , so ${y(y^2 - 2) = 0}$ .
- The positive value is ${y = \sqrt2 \approx 1.4142}$ .

More: unknowns in a determinant

WAEC 2023 · Paper 1 · Q24Given that $P = \begin{pmatrix} x & 4 \\ 3 & 7 \end{pmatrix}$ , $Q = \begin{pmatrix} x & 3 \\ 1 & 2x \end{pmatrix}$ and the …

Cramer’s rule

For three equations with coefficient determinant $\Delta \ne 0$ , each unknown is a ratio of determinants. $\Delta_x$ is $\Delta$ with the $x$ column replaced by the right-hand sides; $\Delta_y$ and $\Delta_z$ likewise:

Cramer's rulex = Δx ÷ Δ, y = Δy ÷ Δ, z = Δz ÷ Δ

Cramer's rulePick a system, then a letter

x + y + z = 6
2x − y + z = 3
x + 2y − z = 2

7Δ7Δx1x

The orange column of Δ holds the coefficients of x. Take it out and put the gold right-hand sides 6, 3, 2 in its place: Δx = 6(−1) − 1(−5) + 1(8) = 7. So x = 7 ÷ 7 = 1.

Always check the answers in all three equations: a sign slip in one determinant shows up at once.

More: 3 × 3 determinants

Your turn

WAEC 2019 · Paper 2 · Q11

Matrices & linear transformations

(a)
Using determinants, solve the following equations simultaneously:

$5x - 6y + 4z = 15$ $7x + 4y - 3z = 19$ $2x + y + 6z = 46$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$\Delta = \begin{vmatrix} 5 & -6 & 4 \\ 7 & 4 & -3 \\ 2 & 1 & 6 \end{vmatrix}$
$= 5(24 + 3) + 6(42 + 6) + 4(7 - 8)$
$= 135 + 288 - 4$
$= 419$ .
Replace the $x$ column by $15, 19, 46$ : $\Delta_x = 15(27) + 6(114 + 138) + 4(19 - 184)$
$= 1257$ .
Replace the $y$ column: $\Delta_y = 5(114 + 138) - 15(48) + 4(322 - 38)$
$= 1676$ .
Replace the $z$ column: $\Delta_z = 5(184 - 19) + 6(322 - 38) + 15(7 - 8)$
$= 2514$ .
So $x = \dfrac{1257}{419} = 3$ , $y = \dfrac{1676}{419} = 4$ and $z = \dfrac{2514}{419} = 6$ .
Check: $15 - 24 + 24 = 15$ ✓, $21 + 16 - 18 = 19$ ✓, $6 + 4 + 36 = 46$ ✓.

Report a problem with this question

2	1	3
1	−2	1
4	0	2

1	1	1
2	−1	1
1	2	−1

6	1	1
3	−1	1
2	2	−1