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?

JEE Question

By Sanjay Kumar · Jun 05, 2018 ·JEE
14 Answers
Sanjay Kumar answered May 30, 2018
y^2-4y+3=0 (y-1)(y-3)=0 y=1,3

Anu answered Jun 03, 2018
Hint:-solve 5y=6y^2+1 & y^2-4y=-3

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

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
⇒ y2 - 4y = -3
⇒ y2 - 4y + 3 = 0
⇒ y2 -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 = 6y2 + 1
⇒ 6y2 -5y +1 = 0
⇒ 6y2 -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 .

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)

Amaan Ansari answered Jan 30, 2019
Yes no real value should be correct answer

Dev Garwa answered Oct 09, 2019
How 1 and 3 are not real numbers?

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