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# Mathematics

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### Mathematics

1. 1. MATHEMATICS<br />Matrices<br />Form5-Chapter 4<br />Group Member<br /><ul><li>Celia chia
2. 2. Tan weihao
3. 3. Tan kim guan
4. 4. Ema</li></li></ul><li>MULTIPLICATION OF TWO MATIRCES <br />Determining whether the two matrices can be multiplied <br />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.<br />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. <br />
5. 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<br />a) 5 3<br /> 1 9 4 3 <br />Solution<br />a)Order of matrix:2 × 2 and 1 × 2<br /> Not the same<br />Thus,Matrix multiplication is not possible<br />
6. 6. b) <br /> 4 7 <br /> 3 8 4 5<br /> 9 6 6 7 <br />Order of matrix :3 × 2 and 2 × 2<br /> Same<br />Thus,matrix multiplication is possible<br />The order of matrix formed is 3 × 2 <br />
7. 7. Finding the product of two matrices <br />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.<br />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. <br />Repeat the process for all other rows in the first matrix.<br />
8. 8. Example 2<br />Find the product of each of the following<br />a) 3 <br /> 4 1 2<br />Solution<br /> 3<br /> 4 1 2 = 3 x 1 3 x 2<br /> 4 x 1 4 x 2<br /> = 3 6<br /> 4 8 <br />
9. 9. Solving matrix equations involving the multiplication of two matrices <br />To find the unknowns element in a matrix can be achieved by solving matrix equation involving the multiplication of two matrices as follows:<br />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. <br />
10. 10. Example 3<br />If 2 4 6 ,find the value of x+y<br /> x ( y 3)= 8 9 <br /> Solution:<br /> 2 4 6 <br /> x y 3 = 8 9 <br /> 2y 6 = 4 6 <br />xy 3x 8 9 <br />Compare the corresponding elements:<br />Hence,2y=4 3x=9 <br /> y=2 x=3 <br />Thus,x+y=2+3<br /> =5<br />
11. 11. Exercise<br />1)Give A= 4 2 ,B= -8 4 1 and C= 4 3<br /> 1 0 6 3 -2 7 -5 <br /> -3 5 <br />a)Find AB and BA.Is AB=BA?<br />b)Find C².<br />2)Find the unknows<br />a) a 3 1 2 = -13 4<br /> 2 b -3 4 5 0<br />b) 4 y 2 2 = 5 17 <br /> x -1 -1 3 5 1 <br />
12. 12. Solution:<br />1)AB= 4 2 -8 4 1<br /> 1 0 6 3 -2 <br /> -3 5<br /> 4×(-8)+2×6 4×4+2×3 4×1+2×(-2)<br /> = 1×(-8)+0×6 1×4+0×3 1×1+0×(-2)<br /> -3×(-8)+5×6 -3×4+5×3 -3×1+5×(-2)<br /> = -32+12 16+6 4+(-4)<br /> -8+0 4+0 1+0<br /> 24+30 -12+15 -3+(-10) <br />= -20 22 0<br /> -8 4 1 <br /> 54 3 -13 <br />
13. 13. BA= -8 4 1 4 2<br /> 6 3 -2 1 0<br /> -3 5<br />= (-8)×4+4×1+1×(-3) (-8)×2+4×0+1×5<br /> 6×4+3×1+(-2)×(-3) 6×2+3×0+(-2)×5<br />= -32+4+(-3) -16+0+5<br /> 24+3+6 12+0+(-10)<br />= -31 -11 <br /> 33 2 Therefore,AB≠BA<br />
14. 14. b) C²=CC<br /> = 4 3 4 3 <br /> 7 -5 7 -5<br /> = 4×4+3×7 4×3+3×(-5)<br /> 7×4+(-5)×7 7×3+5(-5)×(-5)<br /> = 16+21 12+(-15)<br /> 28+(-35) 21+25 <br /> = 37 -3<br /> -7 46 <br />
15. 15. 2a) a 3 1 2 = -13 4<br /> 2 b -3 4 5 0<br /> a(1)+3(-3) a(2)+3(4) = -13 4<br /> 2(1)+b(-3) 2(2)+b(4) 5 0<br /> a-9 2a+12 = -13 4<br /> 2-3b 4+4b 5 0 <br />Hence:<br /> a-9=-13 2-3b=5<br /> a=-13+9 -3b=5-2 <br /> a=-4 -3b=3<br /> b=-1 <br />
16. 16. 2b) 4 y 2 2 = 5 17<br /> x -1 -1 3 5 1<br /> 4(2)+y(-1) 4(2)+y(3) 5 17<br /> x(2)+-1(-1) x(2)-1(3) = 5 1<br /> 8-y 8+3y = 5 17<br /> 2x+1 2x-3 5 1<br />Hence:<br /> 8-y=5 2x+1=5<br /> -y=-3 2x=4<br /> y=3 x=2 <br />
17. 17. Exercise 2<br />Given<br />D= 4 3 ,E= 1 0 <br /> 2 -1 0 1<br />Find the product of DE<br />Solution <br />= 1×4+0×2 1×3+0×(-1)<br /> 0×4+1×2 0×3+1×(-1)<br /> = 4 3<br /> 2 -1<br />
18. 18. Exercise 3<br />A= 4 22 ,B= 8<br />24 2 10<br /> 6 2 0 12<br />Find the product of AB<br />= 4×8+2×10+2×12<br /> 2×8+4×10+2×12<br /> 6×8+2×10+0×12<br />= 16+10+12 = 38<br /> 8+20+12 40 <br />24+10+0 34<br />