3RD TERM

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SPECIFIC TOPIC: REVISION
REFERENCE BOOK : New General Mathematics For S.S.2
CONTENT: REVISION
Calculate the area of each of the following figures below

(a) 11cm 7cm (b) 1060

21cm 18cm
Solution
(a) Base = 21cm
Height = 7cm
Area = base x height
Area = 21 x 7
Area = 147cm2
(b) Base = 18cm
Other side 7cm
Included angle = 1060
Area = ABsinθ
Area = 18 x 7 x sin 1060
= 18 x 7 x 0.9613
= 121.1cm2
Inverse of a Matrix
Matrix Inverse
2. Augmented matrix method
Use Gauss-Jordan elimination to transform [ A | I ] into [ I | A-1 ].

Example: The following steps result in .

so we see that .
3. Adjoint method
A-1 = (adjoint of A) or A-1 = (cofactor matrix of A)T

Example: The following steps result in A-1 for .
The cofactor matrix for A is , so the adjoint is . Since det A = 22, we get
.

EVALUATION: The lesson is evaluated as the students are asked to solve the following problems. Find the inverse of the matrices using Gauss John method



2 3 1 -2 5 1
1. 3 1 0 2. 5 0 2
4 2 1 1 2 1
ASSIGNMENT:
Find the inverse of the matrices using Gauss John method

8 6 1 5 -9 3
1. 2 7 2 2. 1 2 7
9 9 1 4 3 -4
INDICIAL EQUATION
Example 1: Solve (1/2)X = 8
Solution
(1/2)X = 8
2-X = 23
-x = 3
X =-3
Example 2: Solve 3(3)x = 27
Solution
3(3)x = 27
3x+1 = 33
X + 1 = 3
X = 3 – 1
X = 2
EXPONENTIAL EQUATION
Example 1: Solve 52x – 26(5x) + 25 = 0
Solution
52x – 26(5x) + 25 =0
Therefore, (5x)2 – 26(5x) + 25 = 0
Let p = 5x then
P2 – 26P + 25 =0
(P – 1) (P – 25) =0
P = 1 or p = 25
5x = 1 = 50
X = 0
Or 5x = 25 = 52
X = 2
Hence x =0 or 2
Example 2: solve 52x+1 – 26(5X) + 5 =0
Solution
52x +1 -26(5x) + 5 = 0
(5x)2 x 51 – 26(5x) + 5 = 0
Let P = 5x
5p2 – 26p + 5 = 0
(5p – 1)(p – 5) = 0
P= 1/5 or p = 5
5x = 5-1
X = -1
Or 5x = 51
X = 1 Hence x = -1, or 1
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
(i) 25(5x) = 625
(ii) 3-x = 243
ASSIGNMENT:
(i) 22x - 6(2x) + 8 = 0
(ii) 32x – 4(3x + 1) + 27 = 0

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REVISION