Page 10 - Focus SPM KSSM 2021 Tingkatan 5 - Maths DLP

Mathematics Form 5 Chapter 2 Matrices

3
–2
6. Zero matrix is a matrix with all its elements (b) 5(P – Q) = 5 13 4 3 42
–
5
2
are zero, for example 3 0 0 4 and is represented –2 – 3
0
0
by O. = 5 3 5 – 2 4
7. The results of addition and subtraction of a = 5 3 4
–5
matrix with a zero matrix are the matrix itself. 3
Minion Pro 10pt = 1 A + O = A and A – O = A = 3 5 × (–5) 4
2 5 × 3
–25
= 1 0 0 2 Chapter 2 11 = 3 4 Chapter 2
0 –0
15
6 3
Arial 8.5 pt = 1 Given that T = 3 1 –9 4 , find Alternative Method
2 (a) 2T (b) 1 T
= 1 0 0 2 Solution 3 (a) Addition of matrices obeys the Distributive Law,
which is n(P + Q) = nP + nQ.
0
0
(a) 2T = 2 3 1 –9 4 Hence, 5(P + Q) = 5P + 5Q
3
–2
Kaedah Alternatif 6 3 = 5 3 4 3 4
+ 5
5
= 3 2 × 1 2 × (–9) 4 5 × (–2) 2 5 × 3
2 × 3
2 × 6
+
= 3 2 –18 4 = 3 5 × 5 4 3 5 × 2 4
12
6
= 3 –10 + 15 4
25 + 10
3
(b) 1 T = 1 1 –9 4
5
3 3 6 3 = 3 4
35
1
× 1 1 × (–9) (b) Subtraction of matrices obeys the Distributive
= 3 3 × 3 4
1
3 × 6 1 Law, which is n(P – Q) = nP – nQ.
3 3 Hence, 5(P – Q) = 5P – 5Q
3
–2
3 1 –3 4 = 5 3 4 3 4
– 5
3
= 2 1 5 2
–
= 3 5 × (–2) 4 3 5 × 3 4
5 × 5
5 × 2
12 –10 – 15
= 3 4
25 – 10
–2
3
Given that P = 3 4 and Q = 3 4 , calculate –25
5
2
15
(a) 5(P + Q) = 3 4
(b) 5(P – Q)
13
Solution Express each of the following as a single matrix.
–2
3
(a) 5(P + Q) = 5 13 4 3 42 (a) 3[–7 5] + 0.5[20 8]
+
2
5
= 5 3 –2 + 3 4 (b) 5 3 –1 2 4 3 4 –6 4
– 2
–2 3
8
–5
5 + 2
1
= 5 3 4 Solution
7
(a) 3[–7 5] + 0.5[20 8]
= 3 5 × 1 4 = [–21 15] + [10 4]
5 × 7
= [–21 + 10 15 + 4]
5
= 3 4 = [–11 19]
35
22
02 Focus Math F5_E2021.indd 22 18/02/2021 2:32 PM