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?

Related Test: Test: Introduction To Matrices

16 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
Feb 16, 2020

⇒ 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

Gudiya Jain
answered
Mar 23, 2020

Comparing the elements of matrices A and B, we get two different sets of values of y, where nothing is common. So, we can say that no real values exist satisfying the given matrices. Note that it doesn't mean that values obtained are not real.

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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