Question

6. Find the values of x, y and z from the following equations:rtyz 9| X+2 |=152x+y 21「6 215+z xy| |5825(iii)x 51 5

Answer 1

(i)

The given equation is,

[ 4 3 x 5 ]=[ y z 1 5 ]

Since the given matrices are equal, their corresponding elements will be also equal.

x=1

y=4

z=3

Thus, the value of x=1 , y=4 and z=3 .

(ii)

The given equation is,

[ x+y 2 5+z xy ]=[ 6 2 5 8 ]

Since the given matrices are equal, their corresponding elements will be also equal.

x+y=6 (1)

xy=8 (2)

5+z=5 (3)

Solve the equation (3) for z , we get:

5+z=5 z=5−5 =0

Solve the equation (1) for x , we get:

x+y=6 x=6−y (4)

By substituting the value of x in equation (2), we get:

xy=8 ( 6−y )y=8 6y− y 2 =8 y 2 −6y+8=0

y 2 −6y+8=0 y 2 −4y−2y+8=0 y( y−4 )−2( y−4 )=0 ( y−2 )( y−4 )=0

If y – 2 = 0, y = 2. Then x is: x=6−y =6−2 =4

If y –4 = 0, y = 4. Then x is: x=6−y =6−4 =2

Hence, the values of x, y, z are 4, 2, 0 or 2, 4, 0 respectively.

(iii)

The given equation is,

[ x+y+z x+z y+z ]=[ 9 5 7 ]

Since the given matrices are equal, their corresponding elements will be also equal.

x+y+z=9 (5)

x+z=5 (6)

y+z=7 (7)

By substituting the value of equation (6) in equation (5), we get:

x+y+z=9 x+z+y=9 5+y=9 y=4

By substituting the value of equation (7) in equation (5), we get:

x+y+z=9 x+7=9 x=2

By substituting x=2 and y=4 in equation (5), we get:

x+y+z=9 2+4+z=9 z=3

Thus, the value of x=2 , y=4 and z=3 .