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Matrices & determinants · Lesson 2 of 2

Determinants and inverses

The determinant ad − bc of a 2 × 2 matrix and what it measures, singular matrices, the inverse of a 2 × 2 matrix, finding a matrix from its inverse, solving simultaneous equations with an inverse, and 3 × 3 determinants.

16 minYou should already know: Linear & simultaneous equations
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The determinant

The determinant of a 2×22 \times 2 matrix is one number, written with straight bars:

∣abcd∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

Multiply down the leading diagonal, then take away the product of the other diagonal.

abcdad − bc
DeterminantLeading diagonal minus the other diagonal: ad − bc

Try it

The determinantChange the entries with the sliders
−6−5−4−3−2−1123456−6−5−4−3−2−1123456xy
2113matrix2 × 3 − 1 × 1 = 5ad − bc3/5−1/5−1/52/5inverse
The columns (2, 1) and (1, 3) are the sides of the shaded shape. The unit square (area 1) becomes a shape of area |ad − bc| = 5. Its inverse is 1/5 times d−b−ca.

A 2×22 \times 2 matrix moves every point of the plane. The determinant tells you what happens to areas: the unit square becomes a shape whose area is the size of the determinant. Press “Make it singular” to see the determinant become 0 and the square flatten into a line.

The inverse of a 2 × 2 matrix

The inverse A−1A^{-1} undoes AA: AA−1=A−1A=IAA^{-1} = A^{-1}A = I. For A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}:

  1. swap aa and dd;
  2. change the signs of bb and cc;
  3. divide by the determinant.
A−1=1ad−bc(d−b−ca)A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

If ad−bc=0ad - bc = 0 there is no inverse: the matrix is singular.

A⁻¹ =1ad − bcd−b−caswap a, d · change signs of b, c
InverseSwap a and d, change the signs of b and c, divide by ad − bc

A matrix TT with ST=IST = I (or TS=ITS = I) is just the inverse of SS: T=S−1T = S^{-1}. Find it with the same rule.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q13 (b)

Given the matrices A=(25−1−3)\mathbf A = \begin{pmatrix} 2 & 5 \\ -1 & -3 \end{pmatrix} and B=(3−241)\mathbf B = \begin{pmatrix} 3 & -2 \\ 4 & 1 \end{pmatrix}, find: (i) BA\mathbf{BA}; (ii) the determinant of BA\mathbf{BA}.

  1. (i) Which way round?

    BA\mathbf{BA} means rows of B\mathbf B times columns of A\mathbf A.

    Think first. BA: whose rows and whose columns?

  2. Multiply

    • Row 1: 3(2)+(−2)(−1)=8{3(2) + (-2)(-1) = 8} and 3(5)+(−2)(−3)=21{3(5) + (-2)(-3) = 21}.
    • Row 2: 4(2)+1(−1)=7{4(2) + 1(-1) = 7} and 4(5)+1(−3)=17{4(5) + 1(-3) = 17}.
    • So BA=(821717){\mathbf{BA} = \begin{pmatrix} 8 & 21 \\ 7 & 17 \end{pmatrix}}.

    Think first. Work out each entry.

  3. (ii) The determinant

    • ∣BA∣=8×17−21×7{|\mathbf{BA}| = 8 \times 17 - 21 \times 7}.
    • =136−147=−11{= 136 - 147 = -11}.

    Think first. ad − bc for the answer to (i).

Finding a matrix from its inverse

The inverse of A−1A^{-1} is AA itself. So to find PP from P−1P^{-1}, find the inverse of P−1P^{-1} with the same rule.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q7 (b)

Given that P−1=(−114−3)P^{-1} = \begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix}, find the matrix PP.

  1. The determinant of the given matrix

    • ∣P−1∣=(−1)(−3)−(1)(4){|P^{-1}| = (-1)(-3) - (1)(4)}.
    • =3−4=−1{= 3 - 4 = -1}.

    Think first. ad − bc for P⁻¹.

  2. Swap and change signs

    (−3−1−4−1)\begin{pmatrix} -3 & -1 \\ -4 & -1 \end{pmatrix}.

    Think first. Swap −1 and −3; change the signs of 1 and 4.

  3. Divide by the determinant

    • P=1−1(−3−1−4−1){P = \frac{1}{-1}\begin{pmatrix} -3 & -1 \\ -4 & -1 \end{pmatrix}}.
    • P=(3141){P = \begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}}.

    Think first. Divide every entry by −1.

  4. Check

    (3141)(−114−3)=(1001)\begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ✓.

    Think first. What should P × P⁻¹ give?

Solving simultaneous equations with an inverse

Two equations such as 3x+y=73x + y = 7 and 2x+y=52x + y = 5 can be written as one matrix equation:

(3121)(xy)=(75)\begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 5 \end{pmatrix}

Multiplying both sides on the left by the inverse gives (xy)\begin{pmatrix} x \\ y \end{pmatrix}.

Each equation is a straight line, and the solution is the point where the two lines cross. Make the determinant 0 and see why there is then no inverse: the lines never cross, or they are the same line.

Solving with an inverse matrixSet the coefficients and the right-hand sides
−8−6−4−22468−8−6−4−22468xy(2, 3)
2x + y = 7first line3x + 4y = 18second line5determinant ad − bc(2, 3)solution (x, y)
|A| = (2)(4) − (1)(3) = 5. A⁻¹ = (1/5) × (4, −1 / −3, 2). Then x = (4 × 7 − 1 × 18) ÷ 5 = 2 and y = ((−3) × 7 + 2 × 18) ÷ 5 = 3. Check: 2 × 2 + 1 × 3 = 7 ✓. On the graph, (2, 3) is where the two lines cross.

3 × 3 determinants

For a 3×33 \times 3 determinant, expand along the top row. Each top-row entry multiplies the 2×22 \times 2 determinant left when you cover its row and column. The signs go +,−,++, -, +:

abcdefghi+−+a(ei − fh)− b(di − fg)+ c(dh − eg)
3 × 3 determinantTop row, signs + − +, each times the 2 × 2 left over

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.

When an entry holds an unknown, expand in the same way. The determinant becomes an expression: set it equal to the value you are given and solve. (The 3×33 \times 3 identity I3I_3 has determinant 1: along the top row, only 1×1×11 \times 1 \times 1 is left.)

Worked example · JAMB 2000

JAMB 2000 · UME · Q18

Find the values of tt for which the determinant of the matrix (t−400−1t+1134t−2)\begin{pmatrix} t - 4 & 0 & 0 \\ -1 & t + 1 & 1 \\ 3 & 4 & t - 2 \end{pmatrix} is zero.

  1. Expand along the top row

    • The top row is t−4{t - 4}, 0{0}, 0{0}, so only the first term is left.
    • (t−4)((t+1)(t−2)−1×4){(t - 4)\big((t + 1)(t - 2) - 1 \times 4\big)}.

    Think first. Two of the top entries are 0.

  2. Simplify the bracket

    • (t+1)(t−2)−4=t2−t−6{(t + 1)(t - 2) - 4 = t^2 - t - 6}.
    • t2−t−6=(t−3)(t+2){t^2 - t - 6 = (t - 3)(t + 2)}.

    Think first. Expand, then factorise.

  3. Set it to zero

    • (t−4)(t−3)(t+2)=0{(t - 4)(t - 3)(t + 2) = 0}.
    • So t=4{t = 4}, t=3{t = 3} or t=−2{t = -2}: option D.

    Think first. When is a product zero?

Your turn

WAEC 2020 · Paper 2 · Q6 (b)

  1. (b)

    Given that X=(2486)X = \begin{pmatrix} 2 & 4 \\ 8 & 6 \end{pmatrix}, Y=(abcd)Y = \begin{pmatrix} a & b \\ c & d \end{pmatrix} and XY=(1216810)XY = \begin{pmatrix} 12 & 16 \\ 8 & 10 \end{pmatrix}, find: (i) matrix YY; (ii) the determinant of YY.

Worked solution (try it first)

(b)(i)

  1. Multiply row by column: XY=(2a+4c2b+4d8a+6c8b+6d)XY = \begin{pmatrix} 2a + 4c & 2b + 4d \\ 8a + 6c & 8b + 6d \end{pmatrix}
    =(1216810)= \begin{pmatrix} 12 & 16 \\ 8 & 10 \end{pmatrix}.
  2. The first column gives 2a+4c=122a + 4c = 12 and 8a+6c=88a + 6c = 8.
  3. Multiply the first by 4: 8a+16c=488a + 16c = 48.
  4. Take away the second: 10c=4010c = 40, so c=4c = 4, and 2a=12−16=−42a = 12 - 16 = -4, so a=−2a = -2.
  5. The second column gives 2b+4d=162b + 4d = 16 and 8b+6d=108b + 6d = 10.
  6. Multiply the first by 4: 8b+16d=648b + 16d = 64.
  7. Take away the second: 10d=5410d = 54, so d=275d = \frac{27}{5}, and 2b=16−1085=−2852b = 16 - \frac{108}{5} = -\frac{28}{5}, so b=−145b = -\frac{14}{5}.
  8. So Y=(−2−2454525)Y = \begin{pmatrix} -2 & -2\frac45 \\ 4 & 5\frac25 \end{pmatrix}.

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