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Matrices & linear transformations · Lesson 2 of 3

3 × 3 determinants and Cramer's rule

Evaluating a 3 × 3 determinant along the top row, solving equations for an unknown in a determinant, and solving three simultaneous equations by Cramer's rule.

18 minYou should already know: Matrices & determinants
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  3. 3

Evaluating a 3 × 3 determinant

Expand along the top row. Each entry multiplies the 2×22 \times 2 determinant left when you cover its row and column, with the signs + − ++\ -\ +:

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

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q9

Given that F=(3420−1221−2)F = \begin{pmatrix} 3 & 4 & 2 \\ 0 & -1 & 2 \\ 2 & 1 & -2 \end{pmatrix}:

Find ∣F∣|F|, the determinant of FF.

Use the determinant of FF to solve the system of equations 3x+4y+2z=−13x + 4y + 2z = -1, −y+2z=−8-y + 2z = -8, 2x+y−2z=62x + y - 2z = 6 simultaneously.

  1. The determinant

    • First term: 3((−1)(−2)−2(1))=3(0)=0{3\big((-1)(-2) - 2(1)\big) = 3(0) = 0}.
    • Second term: −4(0(−2)−2(2))=−4(−4)=16{-4\big(0(-2) - 2(2)\big) = -4(-4) = 16}.
    • Third term: 2(0(1)−(−1)(2))=2(2)=4{2\big(0(1) - (-1)(2)\big) = 2(2) = 4}.
    • So ∣F∣=0+16+4=20{|F| = 0 + 16 + 4 = 20}.

    Think first. Cover the row and column of each top entry.

  2. Cramer's rule

    • Δx=−20{\Delta_x = -20}, Δy=40{\Delta_y = 40} and Δz=−60{\Delta_z = -60}.
    • x=−2020=−1{x = \frac{-20}{20} = -1}, y=4020=2{y = \frac{40}{20} = 2} and z=−6020=−3{z = \frac{-60}{20} = -3}.

    Think first. Replace each column in turn by −1, −8, 6.

  3. Check in the equations

    • 3(−1)+4(2)+2(−3)=−1{3(-1) + 4(2) + 2(-3) = -1} ✓, −2+2(−3)=−8{-2 + 2(-3) = -8} ✓, 2(−1)+2−2(−3)=6{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 yy for which ∣y101y101y∣=0\begin{vmatrix} y & 1 & 0 \\ 1 & y & 1 \\ 0 & 1 & y \end{vmatrix} = 0 (4 d.p.).

  1. Expand

    • y(y2−1)−1(y−0)+0=y3−2y{y(y^2 - 1) - 1(y - 0) + 0 = y^3 - 2y}.

  2. Solve

    • y3−2y=0{y^3 - 2y = 0}, so y(y2−2)=0{y(y^2 - 2) = 0}.
    • The positive value is y=2≈1.4142{y = \sqrt2 \approx 1.4142}.

More: unknowns in a determinant

Cramer’s rule

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

Δxpbcqefrhix = Δx ÷ Δp, q, r goin column x
Cramer's rulex = Δx ÷ Δ, y = Δy ÷ Δ, z = Δz ÷ Δ
Cramer's rulePick a system, then a letter
7Δ7Δx1x
x + y + z = 6; 2x − y + z = 3; x + 2y − z = 2. Replace the x column by 6, 3, 2: 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

  1. (a)

    Using determinants, solve the following equations simultaneously:

    5x−6y+4z=155x - 6y + 4z = 15 7x+4y−3z=197x + 4y - 3z = 19 2x+y+6z=462x + y + 6z = 46

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)
  1. Δ=∣5−6474−3216∣\Delta = \begin{vmatrix} 5 & -6 & 4 \\ 7 & 4 & -3 \\ 2 & 1 & 6 \end{vmatrix}
    =5(24+3)+6(42+6)+4(7−8)= 5(24 + 3) + 6(42 + 6) + 4(7 - 8)
    =135+288−4= 135 + 288 - 4
    =419= 419.
  2. Replace the xx column by 15,19,4615, 19, 46: Δx=15(27)+6(114+138)+4(19−184)\Delta_x = 15(27) + 6(114 + 138) + 4(19 - 184)
    =1257= 1257.
  3. Replace the yy column: Δy=5(114+138)−15(48)+4(322−38)\Delta_y = 5(114 + 138) - 15(48) + 4(322 - 38)
    =1676= 1676.
  4. Replace the zz column: Δz=5(184−19)+6(322−38)+15(7−8)\Delta_z = 5(184 - 19) + 6(322 - 38) + 15(7 - 8)
    =2514= 2514.
  5. So x=1257419=3x = \dfrac{1257}{419} = 3, y=1676419=4y = \dfrac{1676}{419} = 4 and z=2514419=6z = \dfrac{2514}{419} = 6.
  6. Check: 15−24+24=1515 - 24 + 24 = 15 ✓, 21+16−18=1921 + 16 - 18 = 19 ✓, 6+4+36=466 + 4 + 36 = 46 ✓.

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