2. Tan weihao

3. Tan kim guan

5. Example 1 :Determine whether the matrix multiplication is possible for each of the matrix equations shown below.State the order of the matrix formed if the multiplication is possible a) 5 3 1 9 4 3 Solution a)Order of matrix:2 × 2 and 1 × 2 Not the same Thus,Matrix multiplication is not possible

6. b) 4 7 3 8 4 5 9 6 6 7 Order of matrix :3 × 2 and 2 × 2 Same Thus,matrix multiplication is possible The order of matrix formed is 3 × 2

7. Finding the product of two matrices If two matrices of order m x n and n x p is multiplied then the matrix formed is of the order m x p. The multiplication process involves multiplying the elements of the 1st row of the first matrix with the elements of each column on the second matrix. Repeat the process for all other rows in the first matrix.

8. Example 2 Find the product of each of the following a) 3 4 1 2 Solution 3 4 1 2 = 3 x 1 3 x 2 4 x 1 4 x 2 = 3 6 4 8

9. Solving matrix equations involving the multiplication of two matrices To find the unknowns element in a matrix can be achieved by solving matrix equation involving the multiplication of two matrices as follows: I. Simplify the matrix equations so that the multiplication form two equal matrices. II. Compare their corresponding elements in the two equal matrices formed. The comparison allows to write down linear equation where the values of unknown elements can be determined.

10. Example 3 If 2 4 6 ,find the value of x+y x ( y 3)= 8 9 Solution: 2 4 6 x y 3 = 8 9 2y 6 = 4 6 xy 3x 8 9 Compare the corresponding elements: Hence,2y=4 3x=9 y=2 x=3 Thus,x+y=2+3 =5

11. Exercise 1)Give A= 4 2 ,B= -8 4 1 and C= 4 3 1 0 6 3 -2 7 -5 -3 5 a)Find AB and BA.Is AB=BA? b)Find C². 2)Find the unknows a) a 3 1 2 = -13 4 2 b -3 4 5 0 b) 4 y 2 2 = 5 17 x -1 -1 3 5 1

12. Solution: 1)AB= 4 2 -8 4 1 1 0 6 3 -2 -3 5 4×(-8)+2×6 4×4+2×3 4×1+2×(-2) = 1×(-8)+0×6 1×4+0×3 1×1+0×(-2) -3×(-8)+5×6 -3×4+5×3 -3×1+5×(-2) = -32+12 16+6 4+(-4) -8+0 4+0 1+0 24+30 -12+15 -3+(-10) = -20 22 0 -8 4 1 54 3 -13

13. BA= -8 4 1 4 2 6 3 -2 1 0 -3 5 = (-8)×4+4×1+1×(-3) (-8)×2+4×0+1×5 6×4+3×1+(-2)×(-3) 6×2+3×0+(-2)×5 = -32+4+(-3) -16+0+5 24+3+6 12+0+(-10) = -31 -11 33 2 Therefore,AB≠BA

14. b) C²=CC = 4 3 4 3 7 -5 7 -5 = 4×4+3×7 4×3+3×(-5) 7×4+(-5)×7 7×3+5(-5)×(-5) = 16+21 12+(-15) 28+(-35) 21+25 = 37 -3 -7 46

15. 2a) a 3 1 2 = -13 4 2 b -3 4 5 0 a(1)+3(-3) a(2)+3(4) = -13 4 2(1)+b(-3) 2(2)+b(4) 5 0 a-9 2a+12 = -13 4 2-3b 4+4b 5 0 Hence: a-9=-13 2-3b=5 a=-13+9 -3b=5-2 a=-4 -3b=3 b=-1

16. 2b) 4 y 2 2 = 5 17 x -1 -1 3 5 1 4(2)+y(-1) 4(2)+y(3) 5 17 x(2)+-1(-1) x(2)-1(3) = 5 1 8-y 8+3y = 5 17 2x+1 2x-3 5 1 Hence: 8-y=5 2x+1=5 -y=-3 2x=4 y=3 x=2

17. Exercise 2 Given D= 4 3 ,E= 1 0 2 -1 0 1 Find the product of DE Solution = 1×4+0×2 1×3+0×(-1) 0×4+1×2 0×3+1×(-1) = 4 3 2 -1

18. Exercise 3 A= 4 22 ,B= 8 24 2 10 6 2 0 12 Find the product of AB = 4×8+2×10+2×12 2×8+4×10+2×12 6×8+2×10+0×12 = 16+10+12 = 38 8+20+12 40 24+10+0 34