Page 10 - Top Class F5 - Mathematics (Chapter 2)
P. 10
Mathematics Form 5 Chapter 2 Matrices
10. The product of matrix A and identity matrix, I, is the 14. ad – bc is known as the determinant of matrix A and is
matrix A itself. written as |A|, that is |A| = ad – bc.
Hasil darab suatu matriks A dengan matriks identiti, I, ialah matriks ad – bc dikenali sebagai penentu bagi matriks A dan ditulis sebagai
A sendiri. |A|, iaitu |A| = ad – bc.
AI = IA = A
15. If ad – bc = 0, then the inverse matrix, A does not exist.
–1
–1
11. When multiply two matrices, A and B, if Jika ad – bc = 0, maka matriks songsang, A tidak wujud.
Apabila mendarabkan dua matriks, A dan B, jika 16. The simultaneous linear equations can be solved by
AB = BA = I,
then B is the inverse matrix of A and A is the inverse matrix method. Given
Persamaan linear serentak boleh diselesaikan dengan kaedah
matrix of B. matriks. Diberi
maka B ialah matriks songsang bagi A dan A ialah matriks songsang ax + by = p
bagi B.
cx + dy = q
12. The inverse matrix of A can be written as A .
–1
Therefore, (a) Write the simultaneous linear equations in the matrix
Matriks songsang bagi A boleh ditulis sebagai A . Maka, form.
–1
–1
–1
AA = A A = I Tuliskan persamaan linear serentak dalam bentuk matriks.
p
a b x
13. If matrix A = a b , then inverse matrix of A, A , can c d y = q
–1
c d
x
be found by the formula: (b) Find the matrix .
a b
–1 y
Jika matriks A = c d , maka matriks songsang bagi A, A , boleh x
dicari dengan rumus: Cari matriks .
y
1 d –b x d –b p
–1
A = , ad – bc ≠ 0 = 1
ad – bc –c a y ad – bc –c a q
5. Determine whether addition and subtraction can be performed on the following pairs of matrices. PL 2
Tentukan sama ada penambahan dan penolakan boleh dilaksanakan pada pasangan matriks berikut.
Example 14 16 5 7 (b) –3 0 –6 5 9
,
,
–8 9
1
2 8
7
5 11 15 3 (a) 10 5 2 1 20
,
4 –6
3 –4
6 –2
Yes Yes No
Ya Ya Tidak
(c) [3 15 9], [7 –1 ] (d) [23], [12] 35 –9
(e)
,
21
16
No Yes Yes
Tidak Ya
Ya
6. Solve each of the following. PL 3
Selesaikan setiap yang berikut.
Example –2 11 6 –3 15 1 3 –6
(a)
+
2 –5
4
–4 5 9
16 7 8 –6 17 3 = 7 –2 + 6 11 + (–3) (b) –2 7 – –5 4
–
8 11
12 14
3 –2 1
–4 – (–6) 5 – 17 9 – 3 7 + 2 4 + (–5) 15 – 3 1 – (–6)
= 16 – 3 7 – (–2) 8 – 1 = 4 8 = –2 – (–5) 7 – 4
8 – 12
2 –12 6
= 13 9 7 9 –1 12 7 11 – 14
= 3 3
–4 –3
a b a ± p b ± q
p q
±
=
c ± r d ± s
s
r
c d
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