6. Find the values of x, y and z from the following equations:

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1 Answer
V
8 months ago

(i) [43x5]=[yz15]

corresponding

By equating the elements of the matrices, we get,

x= 1

y= 4

z= 3.

(ii) [x+y25+zxy]=[6258]

By equating the corresponding elements of the matrices we get,

x+ y = 6 (I)

5 + Z = 5 z=55z=0

xy = 8

 x =8y →(2)

putting eqn(2) in (1) we get

8y + y = 6.

8 + y2 = 6y

y2 6y + 8 = 0.

y2 - 4y - 2y + 8 = 0

y (y-4) -2 (y-4) = 0

(y-4) (y-2) = 0

 y=

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f ( x ) = a x 2 + b x + c  

g (x) = px + q

  f ( g ( x ) ) = a ( p x + q ) 2 + b ( p ) ( + q ) + c

8 x 2 2 x = a ( p x + q ) 2 + b ( p x + q ) + c

Compare 8 = ap2 …………… (i)

-2 = a (2pq) + bp

0 = aq2 + bq + c

  9 ( f ( x ) ) = p ( a x 2 + b x + c ) + q

=>4x2 + 6x + 1 = apx2 + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x2 + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

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System of equation is

( 2 3 1 1 1 1 1 1 | λ | ) ( x y z ) = [ 2 4 4 λ 4 ]

R1 – 2 R2, R3 – R2

( 0 1 3 1 1 1 0 2 | λ | 1 ) ( x y z ) = ( 1 0 4 4 λ 8 )

System of equation will have no solution for = -7.

 

...Read more

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Maths Ncert Solutions class 12th 2026

Maths Ncert Solutions class 12th 2026

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