Page 18 - Top Class F5 - Mathematics (Chapter 2)
P. 18
Mathematics Form 5 Chapter 2 Matrices
21. Determine whether each of the following pairs of matrices are the inverse matrices of each other. PL 2
Tentukan sama ada setiap pasangan matriks berikut ialah matriks songsang antara satu sama lain.
Example 2 6 –4 –6
,
(a)
–1 3 –3 –4 3 2
3 5
2 –5
,
1 2
If AB = BA = I, then A and B 2 6 –4 –6 10 0
=
are the inverse matrices of –3 –4 3 2 10 0
1 0
3 5 2 –5
0 10
=
each other.
6
1 2 –1 3
–4 –6 2
0 1
Jika AB = BA = I, maka A dan B
=
–1 3 1 2 = 1 0 ialah matriks songsang antara 3 2 –3 –4 0 10
2 –5 3 5
satu sama lain.
0 1
The product of the matrices is an identity matrix. The product of the matrices is not an identity
matrix. Hence, the matrices are not the inverse
Hence, the matrices are the inverse matrices of each matrices of each other.
other. Hasil darab matriks itu bukan matriks identiti. Maka, matriks
Hasil darab matriks itu ialah matriks identiti. Maka, matriks itu itu bukan matriks songsang antara satu sama lain.
ialah matriks songsang antara satu sama lain.
1 2
7 –2
2 1 –2 4
(b) , (c) ,
3 7 –3 1 4 1 1 –1
2 1 –2 4
1 0
1 2 7 –2
–3 7
3 7 –3 1 = 0 1 4 1 1 –1 = –7 15
7 –2 1 2 1 0 –2 4 2 1 12 2
–3 1 3 7 = 0 1 1 –1 4 1 = –2 0
The product of the matrices is an identity matrix. The product of the matrices is not an identity
Hence, the matrices are the inverse matrices of matrix. Hence, the matrices are not the inverse
each other. matrices of each other.
Hasil darab matriks itu ialah matriks identiti. Maka, matriks Hasil darab matriks itu bukan matriks identiti. Maka, matriks
itu ialah matriks songsang antara satu sama lain. itu bukan matriks songsang antara satu sama lain.
22. Find the inverse matrix of each of the following matrices. PL 3
Cari matriks songsang bagi setiap matriks yang berikut. 1 d –b
A = ad – bc –c a , ad – bc ≠ 0
–1
Example 3 7 7 6
(a) Q = –1 –3 (b) R =
4 3
4 1
P =
6 2
–1
–1
1 2 –1 Q = 1 –3 –7 R = 1 3 –6
–1
P = 3(–3) – 7(–1) 1 3 7(3) – 6(4) –4 7
4(2) – 1(6) –6 4 1 –3 –7 1 3 –6
1 2 –1 = – = –
= 2 1 3 3 –4 7
2 –6 4
3 7 –1 –
2
1 – 1 = 2 2 3 = 4 7
= 2 1 3 3
–3 2 – 2 – 2
2 1 8 –5
4 2
(c) S = –2 4 (d) T = 5 3 (e) U = 4 –3
1 4 –1 1 3 –2 1 –3 5
–1
–1
–1
S = T = U =
2(4) – 1(–2) 2 2 4(3) – 2(5) –5 4 8(–3) – (–5)(4) –4 8
1 4 –1 1 3 –2 1 –3 5
= = = –
10 2 2 2 –5 4 4 –4 8
3 –1 = 4 – 5
2
2
3
1
–
2
4
5
10
= 1 1 = 5 1 –2
5 5 – 2
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