The determinant
The determinant of a 2 × 2 2 \times 2 2 × 2
∣ a b c d ∣ = a d − b c \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc a c b d = a d − b c Multiply down the leading diagonal, then take away the product of the other diagonal.
a b c d ad − bc Determinant Leading diagonal minus the other diagonal: ad − bc
Try it
The determinant Change the entries with the sliders
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 x y (2 1 | 1 3) matrix 2 × 3 − 1 × 1 = 5 ad − bc (3/5 −1/5 | −1/5 2/5) inverse
a = 2 b = 1 c = 1 d = 3 Make it singular
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 | −c a). A 2 × 2 2 \times 2 2 × 2
The inverse of a 2 × 2 matrix
The inverse A − 1 A^{-1} A − 1 A A A A A − 1 = A − 1 A = I AA^{-1} = A^{-1}A = I A A − 1 = A − 1 A = I A = ( a b c d ) A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} A = ( a c b d ) a a a d d d b b b c c c
A − 1 = 1 a d − b c ( d − b − c a ) A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} A − 1 = a d − b c 1 ( d − c − b a ) If a d − b c = 0 ad - bc = 0 a d − b c = 0 singular .
A⁻¹ = 1 ad − bc d −b −c a swap a, d · change signs of b, c Inverse Swap a and d, change the signs of b and c, divide by ad − bc Worked example · WAEC 2019
WAEC 2019 · Paper 2 · Q13 (b)
Given the matrices A = ( 2 5 − 1 − 3 ) \mathbf A = \begin{pmatrix} 2 & 5 \\ -1 & -3 \end{pmatrix} A = ( 2 − 1 5 − 3 ) B = ( 3 − 2 4 1 ) \mathbf B = \begin{pmatrix} 3 & -2 \\ 4 & 1 \end{pmatrix} B = ( 3 4 − 2 1 ) B A \mathbf{BA} BA B A \mathbf{BA} BA
(i) Which way round?
B A \mathbf{BA} BA B \mathbf B B A \mathbf A A
Think first. BA: whose rows and whose columns?
Multiply
B A = ( 3 ( 2 ) + ( − 2 ) ( − 1 ) 3 ( 5 ) + ( − 2 ) ( − 3 ) 4 ( 2 ) + 1 ( − 1 ) 4 ( 5 ) + 1 ( − 3 ) ) = ( 8 21 7 17 ) \mathbf{BA} = \begin{pmatrix} 3(2) + (-2)(-1) & 3(5) + (-2)(-3) \\ 4(2) + 1(-1) & 4(5) + 1(-3) \end{pmatrix} = \begin{pmatrix} 8 & 21 \\ 7 & 17 \end{pmatrix} BA = ( 3 ( 2 ) + ( − 2 ) ( − 1 ) 4 ( 2 ) + 1 ( − 1 ) 3 ( 5 ) + ( − 2 ) ( − 3 ) 4 ( 5 ) + 1 ( − 3 ) ) = ( 8 7 21 17 )
Think first. Work out each entry.
(ii) The determinant
∣ B A ∣ = 8 × 17 − 21 × 7 = 136 − 147 = − 11 |\mathbf{BA}| = 8 \times 17 - 21 \times 7 = 136 - 147 = -11 ∣ BA ∣ = 8 × 17 − 21 × 7 = 136 − 147 = − 11
Think first. ad − bc for the answer to (i).
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Finding a matrix from its inverse
The inverse of A − 1 A^{-1} A − 1 A A A P P P P − 1 P^{-1} P − 1 P − 1 P^{-1} P − 1
Worked example · WAEC 2018
WAEC 2018 · Paper 2 · Q7 (b)
Given that P − 1 = ( − 1 1 4 − 3 ) P^{-1} = \begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix} P − 1 = ( − 1 4 1 − 3 ) P P P
The determinant of the given matrix
∣ P − 1 ∣ = ( − 1 ) ( − 3 ) − ( 1 ) ( 4 ) = 3 − 4 = − 1 |P^{-1}| = (-1)(-3) - (1)(4) = 3 - 4 = -1 ∣ P − 1 ∣ = ( − 1 ) ( − 3 ) − ( 1 ) ( 4 ) = 3 − 4 = − 1
Think first. ad − bc for P⁻¹.
Swap and change signs
( − 3 − 1 − 4 − 1 ) \begin{pmatrix} -3 & -1 \\ -4 & -1 \end{pmatrix} ( − 3 − 4 − 1 − 1 )
Think first. Swap −1 and −3; change the signs of 1 and 4.
Divide by the determinant
P = 1 − 1 ( − 3 − 1 − 4 − 1 ) = ( 3 1 4 1 ) P = \frac{1}{-1}\begin{pmatrix} -3 & -1 \\ -4 & -1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix} P = − 1 1 ( − 3 − 4 − 1 − 1 ) = ( 3 4 1 1 )
Think first. Divide every entry by −1.
Check
( 3 1 4 1 ) ( − 1 1 4 − 3 ) = ( 1 0 0 1 ) \begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ( 3 4 1 1 ) ( − 1 4 1 − 3 ) = ( 1 0 0 1 )
Think first. What should P × P⁻¹ give?
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Solving simultaneous equations with an inverse
Two equations such as 3 x + y = 7 3x + y = 7 3 x + y = 7 2 x + y = 5 2x + y = 5 2 x + y = 5 ( 3 1 2 1 ) ( x y ) = ( 7 5 ) \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 5 \end{pmatrix} ( 3 2 1 1 ) ( x y ) = ( 7 5 ) ( x y ) \begin{pmatrix} x \\ y \end{pmatrix} ( x y )
3 × 3 determinants
For a 3 × 3 3 \times 3 3 × 3 2 × 2 2 \times 2 2 × 2 + , − , + +, -, + + , − , +
a b c d e f g h i + − + a(ei − fh) − b(di − fg) + c(dh − eg) 3 × 3 determinant Top row, signs + − +, each times the 2 × 2 left over
Your turn
(b) Given that X = ( 2 4 8 6 ) X = \begin{pmatrix} 2 & 4 \\ 8 & 6 \end{pmatrix} X = ( 2 8 4 6 ) Y = ( a b c d ) Y = \begin{pmatrix} a & b \\ c & d \end{pmatrix} Y = ( a c b d ) X Y = ( 12 16 8 10 ) XY = \begin{pmatrix} 12 & 16 \\ 8 & 10 \end{pmatrix} X Y = ( 12 8 16 10 ) Y Y Y Y Y Y
Worked solution (try it first) (b)(i) Multiply row by column:
X Y = ( 2 a + 4 c 2 b + 4 d 8 a + 6 c 8 b + 6 d ) XY = \begin{pmatrix} 2a + 4c & 2b + 4d \\ 8a + 6c & 8b + 6d \end{pmatrix} X Y = ( 2 a + 4 c 8 a + 6 c 2 b + 4 d 8 b + 6 d ) = ( 12 16 8 10 ) = \begin{pmatrix} 12 & 16 \\ 8 & 10 \end{pmatrix} = ( 12 8 16 10 ) .
The first column gives
2 a + 4 c = 12 2a + 4c = 12 2 a + 4 c = 12 and
8 a + 6 c = 8 8a + 6c = 8 8 a + 6 c = 8 .
Multiply the first by 4:
8 a + 16 c = 48 8a + 16c = 48 8 a + 16 c = 48 .
Take away the second:
10 c = 40 10c = 40 10 c = 40 , so
c = 4 c = 4 c = 4 , and
2 a = 12 − 16 = − 4 2a = 12 - 16 = -4 2 a = 12 − 16 = − 4 , so
a = − 2 a = -2 a = − 2 .
The second column gives
2 b + 4 d = 16 2b + 4d = 16 2 b + 4 d = 16 and
8 b + 6 d = 10 8b + 6d = 10 8 b + 6 d = 10 .
Multiply the first by 4:
8 b + 16 d = 64 8b + 16d = 64 8 b + 16 d = 64 .
Take away the second:
10 d = 54 10d = 54 10 d = 54 , so
d = 27 5 d = \frac{27}{5} d = 5 27 , and
2 b = 16 − 108 5 = − 28 5 2b = 16 - \frac{108}{5} = -\frac{28}{5} 2 b = 16 − 5 108 = − 5 28 , so
b = − 14 5 b = -\frac{14}{5} b = − 5 14 .
So
Y = ( − 2 − 2 4 5 4 5 2 5 ) Y = \begin{pmatrix} -2 & -2\frac45 \\ 4 & 5\frac25 \end{pmatrix} Y = ( − 2 4 − 2 5 4 5 5 2 ) .
Watch out
In (a), move every term to one side before factorising, and give both values of y y y In (b), the first column of Y Y Y a a a c c c b b b d d d Report a problem with this question
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