Page 15 - Focus SPM KSSM 2021 Tingkatan 5 - Maths DLP

Mathematics Form 5 Chapter 2 Matrices

5. If ad – bc = 0, then the inverse matrix of A, A 24
–1
does not exist.
Find the inverse matrix for each of the following
matrices.
REMEMBER! 2 6
(a) E = 3 4
1 3 10
A ≠
–1
A
(b) F = 3 –8 –3 4
12 4
23 Solution Chapter 2
10 –6
1
3

Determine whether the following matrices are the (a) E = (2 × 10) – (6 × 3) –3 2 4
–1
Chapter 2
inverse matrices of 3 2 5 4 . = 1 10 –6 4
3
1 3
(a) 3 3 –5 4 (b) 3 –2 1 4 2 –3 2
–3
5 –3
5
–1 2

3
= 3 – 14
Solution 2
43
4
1
3
(a) 3 2 5 3 –5 4 (b) F = [(–8) × 4] – [(–3) × 12] –12 –8 4
3

–1
1 3 –1 2
= 3 6 + (–5) –10 + 10 4 = 1 4 3 4
3
3 + (–3)
–5 + 6
= 3 1 0 4 4 –12 –8
3
4
0 1

1
= 3 4
43
3 3 –5 2 5 4 –3 –2
1 3
–1 2
25
= 3 6 + (–5) 15 + (–15) 4 Find the value of m if P does not have an inverse
–2 + 2
–5 + 6
= 3 1 0 4 matrix.
0 1
(a) P = 3 m 12 4
Hence, 3 3 –5 4 is the inverse matrix of 3 2 5 4 3 4
–1 2
1 3
10 1 – 2m
because the product is an identity matrix. (b) P = 3 2 1 4
43
(b) 3 2 5 –2 1 4 Solution
(a) P does not have an inverse matrix, therefore
1 3 5 –3
ad – bc = 0
= 3 –4 + 25 2 + (–15) 4 |P| = 0.
1 + (–9)
–2 + 15
m(4) – 12(3) = 0
= 3 21 –13 4 4m – 36 = 0
13 –8
4m = 36
m = 9
≠ 3 1 0 4 (b) P does not have an inverse matrix, therefore
0 1
Hence, 3 –2 1 4 is not the inverse matrix of |P| = 0. ad – bc = 0
5 –3
3 2 5 4 because the product is not an identity 10(1) – (1 – 2m)(2) = 0
1 3
10 – 2 + 4m = 0

matrix. 4m = –8
m = –2
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02 Focus Math F5_E2021.indd 27 18/02/2021 2:32 PM