MATHEMATICSMatrices

Form5-Chapter 4Group Member

• Celia chia• Tan wei hao• Tan kim guan• Ema

MULTIPLICATION OF TWO MATIRCES

Determining whether the two matrices can be multiplied The multiplication of two matrices is possible if and only if the number of columns in the left matrix is the same as the number of rows in the right matrix.If two matrices can be multiplied, then the number of rows of the product (matrix R) will be the same as the number of rows of the left matrix (matrix P). The number of columns of the product (matrix R) will be the same as the number of columns on the right matrix (matrix Q). See below.

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

Solutiona)Order of matrix:2 × 2 and 1 × 2

Not the sameThus,Matrix multiplication is not possible

b) 4 7 3 8 4 5 9 6 6 7

Order of matrix :3 × 2 and 2 × 2 SameThus,matrix multiplication is possibleThe order of matrix formed is 3 × 2

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.

Example 2Find the product of each of the followinga) 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

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.

Example 3If 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

Exercise1)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 unknowsa) a 3 1 2 = -13 4 2 b -3 4 5 0b) 4 y 2 2 = 5 17 x -1 -1 3 5 1

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

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

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

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

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 1Hence: 8-y=5 2x+1=5 -y=-3 2x=4 y=3 x=2

Exercise 2GivenD= 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

Exercise 3A= 4 2 2 ,B= 8 2 4 2 10 6 2 0 12Find 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