For what real value of y will matrix A be equal to

matrix B, where

a)1, 3

b)No real value

c)1/3, 1/2

d)2 and 3

Correct answer is option 'c'. Can you explain this answer?

a)1, 3

b)No real value

c)1/3, 1/2

d)2 and 3

Correct answer is option 'c'. Can you explain this answer?

14 Answers

Shikha Sony
answered
Dec 20, 2018

As both matrix are eaual given so each element are equal y^2-4y=-3

y^2 -3y-y+3=0

y(y-3)-1(y-3)=0

y-1=0, y-3=0

y=1,3 similarly on solving other quadratic equation on y we get

y=1/3,1/2

no common value so no real value

y^2 -3y-y+3=0

y(y-3)-1(y-3)=0

y-1=0, y-3=0

y=1,3 similarly on solving other quadratic equation on y we get

y=1/3,1/2

no common value so no real value

Naincy Tripathi
answered
Jun 06, 2018

(5y = 6y^2+1) (6y^2 - 5y +1 = 0) apply shri dharacharya method nd solve u will get 1/2 nd1/3 which is not real HENCE OPTION b IS CORRECT

Balister
answered
2 weeks ago

⇒ y^{2} - 4y = -3

⇒ y^{2} - 4y + 3 = 0

⇒ y^{2} -3y - y +3 = 0

⇒ y (y - 3) -1 (y-3) = 0

⇒(y - 1) (y - 3) = 0

⇒ y = 1,3

But these are not real numbers.

We have another equation:

⇒ 5y = 6y^{2} + 1

⇒ 6y^{2} -5y +1 = 0

⇒ 6y^{2} -3y - 2y + 1 = 0

⇒ 3y (2y - 1) - 1 (2y - 1) = 0

⇒ (3y - 1) (2y - 1) = 0

⇒ y = 1/2, 1/3

Hence value of y is 1/2, 1/3

Krishna Khadakumarge
answered
Oct 24, 2018

Y must satisfy both conditions 5y=6y(square)+1y(square)-4y=-3Which is not possible to have real number that's why answer is option B

Rishu Kumar
answered
Feb 20, 2019

Y2-4y=-3 on solving we get y=1,3

6y2-5y+1=0 on solving we get y=1/3,1/2

there is no common value so y is not real .

6y2-5y+1=0 on solving we get y=1/3,1/2

there is no common value so y is not real .

Zain Khan
answered
Feb 11, 2019

Consider y*y-4y =-3

y*y-4y+3=0

factors are -1,-3

=> y-3=0 ; y=3

y-1=0; y=1

a)

y*y-4y+3=0

factors are -1,-3

=> y-3=0 ; y=3

y-1=0; y=1

a)

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