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4.10 SPM Practice (Long Questions)

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Question 5:
(a) Given  1 14 ( 2 s −4 t )( t −1 4 2 )=( 1 0 0 1 ), find the value of s and of t.
(b) Write the following simultaneous linear equations as matrix form:
3x – 2y = 5
9x + y = 1
Hence, using matrix method, calculate the value of x and y.

Solution:
(a) 1 14 ( 2 s −4 t )( t −1 4 2 )=( 1 0 0 1 ) 1 14 ( 2t+4s −2+2s −4t+4t 4+2t )=( 1 0 0 1 ) −2+2s 14 =0   2s=2      s=1 4+2t 14 =1 4+2t=14 2t=10 t=5

(b) ( 3 −2 9 1 )( x y )=( 5 1 )   ( x y )= 1 21 ( 1 2 −9 3 )( 5 1 )   ( x y )= 1 21 ( ( 1 )( 5 )+( 2 )( 1 ) ( −9 )( 5 )+( 3 )( 1 ) )   ( x y )= 1 21 ( 7 −42 )   ( x y )=( 1 3 −2 ) ∴x= 1 3 ,  y=−2



Question 6:
It is given that matrix P=( 6 −3 −5 2 ) and matrix Q= 1 m ( 2 3 5 n )  such that PQ=( 1 0 0 1 ).
(a) Find the value of m and of n.
(b) Write the following simultaneous linear equations as matrix form:
6x – 3y = –24
–5x + 2y = 18
Hence, using matrix method, calculate the value of x and y.

Solution:
(a) m=6( 2 )−( −3 )( −5 )   =12−15 m=−3 n=6

(b) ( 6 −3 −5 2 )( x y )=( −24 18 )   ( x y )= 1 12−15 ( 2 3 5 6 )( −24 18 )   ( x y )= 1 −3 ( ( 2 )( −24 )+( 3 )( 18 ) ( 5 )( −24 )+( 6 )( 18 ) )   ( x y )= 1 −3 ( 6 −12 )   ( x y )=( −2 4 ) ∴x=−2,  y=4