Solutions 11(b) 1
Complete solutions to Exercise 11(b)
1. Use (11.1) to find the determinants.
⎛1 3⎞ ⎛1 5⎞
(a) det ⎜ ⎟ = (1× 7 ) − ( 5 × 3) = −8 det ⎜ ⎟ = (1× 7 ) − ( 3 × 5 ) = −8
⎝5 7⎠ ⎝3 7⎠
⎛ −1 2 ⎞ ⎛ −1 5 ⎞
(b) det ⎜ ⎟ = ( −1× 3) − ( 5 × 2 ) = −13 det ⎜ ⎟ = ( −1× 3) − ( 2 × 5 ) = −13
⎝ 5 3⎠ ⎝ 2 3⎠
⎛ −1 −1⎞ ⎛ −1 0 ⎞
(c) A = ⎜ ⎟, B = ⎜ ⎟ , det A = detB = 0
⎝0 0⎠ ⎝ −1 0 ⎠
The matrix A is transposed (rows → columns) to give matrix B. The same numbers
on each of the diagonals, so the determinant is the same, det A = detB .
2. By (11.1)
⎛a b⎞ ⎛a c ⎞ ⎛a b⎞
det ⎜ ⎟ = ad − cb det ⎜ ⎟ = ad − bc = ad − cb = det ⎜ ⎟
⎝c d⎠ ⎝b d ⎠ ⎝c d⎠
3. Use (11.1) to evaluate the determinants.
(a) (i) det A = (1× 6 ) − (5 × 3) = −9 (ii) det B = (3 × 5) − (−1× 7) = 22
(iii) det A × detB = −9 × 22 = −198
⎛ 1 3 ⎞ ⎛ 3 7 ⎞ ⎛ 0 22 ⎞
(iv) AB = ⎜ ⎟⎜ ⎟=⎜ ⎟
⎝ 5 6 ⎠ ⎝ −1 5 ⎠ ⎝ 9 65 ⎠
det ( AB ) = ( 0 × 65 ) − ( 9 × 22 ) = −198
(b) (i) det A = ( −1× 1.5 ) − (10 × 170 ) = −1701.5
(ii) det B = ⎡⎣ −30 × ( −1.9 ) ⎤⎦ − [ −9.3 × 61] = 624.3
(iii) det A × det B = −1701.5 × 624.3 = −1062246.45
⎛ −1 10 ⎞ ⎛ −30 −9.3 ⎞ ⎛ 640 −9.7 ⎞
(iv) AB = ⎜ ⎟⎜ ⎟=⎜ ⎟
⎝170 1.5 ⎠ ⎝ 61 −1.9 ⎠ ⎝ −5008.5 −1583.85 ⎠
det ( AB ) = ⎡⎣ 640 × ( −1583.85 ) ⎤⎦ − ⎡⎣( −9.7 ) × ( −5008.5 ) ⎤⎦ = −1062246.45
(c) (i) det A = ⎡⎣5 × ( −5.6 ) ⎤⎦ − ⎡⎣ 2.2 × ( −3) ⎤⎦ = −21.4
(ii) det B = ⎡⎣ −7.1× ( −12.2 ) ⎤⎦ − ⎡⎣ −3.5 × ( −2.1) ⎤⎦ = 79.27
(iii) det A × det B = −21.4 × 79.27 = −1696.378
⎛ 5 −3 ⎞⎛ −7.1 −2.1 ⎞ ⎛ −25 26.1 ⎞
(iv) AB = ⎜ ⎟⎜ ⎟=⎜ ⎟
⎝ 2.2 −5.6 ⎠⎝ −3.5 −12.2 ⎠ ⎝ 3.98 63.7 ⎠
det ( AB ) = ( −25 × 63.7 ) − ( 3.98 × 26.1) = −1696.378
All the results satisfy det A det B = det (AB) .
⎛a b⎞
(11.1) det ⎜ ⎟ = ad − cb
⎝c d⎠
Solutions 11(b) 2
4. In each case det A = 1 so we use (11.3).
(a) Exchanging numbers 3 and 9 and placing a negative sign in front of the other
numbers gives:
⎛ 3 −2 ⎞
A −1 = ⎜ ⎟
⎝ −13 9 ⎠
−1 ⎛ 5 −7 ⎞
(b) A = ⎜ ⎟
⎝ −12 17 ⎠
⎛1 0⎞ −1 ⎛1 0⎞
(c) In this case A is the identity matrix, ⎜ ⎟ and so I = I = ⎜ ⎟.
⎝0 1⎠ ⎝0 1⎠
5. We first evaluate the determinant.
⎛5 4⎞
(a) By (11.1), det ⎜ ⎟ = ( 5 ×1) − ( 3 × 4 ) = −7 . So using (11.4)
⎝3 1⎠
1 ⎛ 1 −4 ⎞ ⎛ −1 7 4 7 ⎞
A −1 = − ⎜ ⎟=⎜ ⎟
7 ⎝ −3 5 ⎠ ⎝ 3 7 − 5 7 ⎠
⎛ 3 6⎞
(b) We have det ⎜ ⎟ = ( 3 × 8 ) − ( 7 × 6 ) = 24 − 42 = −18 . Thus applying (11.4)
⎝7 8⎠
−1
−1 ⎛ 3 6⎞ 1 ⎛ 8 −6 ⎞ ⎛ −8 18 6 18 ⎞ ⎛ −4 9 1 3 ⎞
A =⎜ ⎟ =− ⎜ ⎟=⎜ ⎟=⎜ ⎟
⎝7 8⎠ 18 ⎝ −7 3 ⎠ ⎝ 7 18 −3 18 ⎠ ⎝ 7 18 −1 6 ⎠
(c) det A = (7 × 2 ) − (14 × 1) = 0 . Since det A = 0 , so A cannot have an inverse.
6. Putting the equations into matrix form gives
⎛ 2 3 ⎞ ⎛ i1 ⎞ ⎛ 5 ⎞
⎜ ⎟⎜ ⎟ = ⎜ ⎟
⎝ 6 7 ⎠ ⎝ i2 ⎠ ⎝10 ⎠
⎛ 2 3⎞
We need to find the inverse of A = ⎜ ⎟ . By (11.4)
⎝6 7⎠
−1
⎛ 2 3⎞ 1 ⎛ 7 −3 ⎞ 1 ⎛ 7 −3 ⎞
⎜ ⎟ = =
⎝6 7⎠ ( 2 × 7 ) − ( 6 × 3) ⎜⎝ −6 2 ⎟⎠ −4 ⎜⎝ −6 2 ⎟⎠
Using (11.5) we have
⎛ i1 ⎞ 1 ⎛ 7 −3 ⎞⎛ 5 ⎞ 1 ⎛ 5 ⎞ ⎛ −5 4⎞
⎜ ⎟=− ⎜ ⎟⎜ ⎟ =− ⎜ ⎟=⎜ ⎟
⎝ i2 ⎠ 4 ⎝ −6 2 ⎠⎝10 ⎠ 4 ⎝ −10 ⎠ ⎝ 10 4 ⎠
i1 = −1.25 A and i2 = 2.5 A
(11.1) det A = (a × d ) − (b × c)
−1
⎛a b⎞ 1 ⎛ d −b ⎞
(11.4) ⎜ ⎟ = ⎜ ⎟
⎝c d⎠ ad − bc ⎝ −c a ⎠
(11.5) u = A−1b
Solutions 11(b) 3
7. These equations can be written as
30i1 − 10i2 = 12
−10i1 + 35i2 = 5
In matrix form we have
⎛ 30 −10 ⎞ ⎛ i1 ⎞ ⎛ 12 ⎞
⎜ ⎟⎜ ⎟ = ⎜ ⎟
⎝ −10 35 ⎠ ⎝ i2 ⎠ ⎝ 5 ⎠
⎛ 30 −10⎞
Let A = ⎜ . Then by (11.4)
⎝ −10 35 ⎠
−1
⎛ 30 −10 ⎞ 1 ⎛ 35 10 ⎞ 1 ⎛ 35 10 ⎞
⎜ ⎟ = ⎟=
⎝ −10 35 ⎠ ( 30 × 35) − (10 ×10 ) ⎝ 10 30 ⎠ 950 ⎜⎝ 10 30 ⎟⎠
⎜
Using (11.5) gives
⎛ i1 ⎞ 1 ⎛ 35 10 ⎞⎛ 12 ⎞ 1 ⎛ 470 ⎞
⎜ ⎟= ⎜ ⎟⎜ ⎟ = ⎜ ⎟
⎝ i2 ⎠ 950 ⎝ 10 30 ⎠⎝ 5 ⎠ 950 ⎝ 270 ⎠
470 47 270 27
i1 = = A and i2 = = A
950 95 950 95
8. Similar to solutions 6 and 7.
(a) x = 2 and y = 4 (b) x = −1 and y = 1
1 1
(c) x = and y = −
4 3
(11.1) det A = (a × d ) − (b × c)
−1
⎛a b⎞ 1 ⎛ d −b ⎞
(11.4) ⎜ ⎟ = ⎜ ⎟
⎝c d⎠ ad − bc ⎝ −c a ⎠
(11.5) u = A−1b
