QJEEM21Mac48f: JEE Main 2021 Mathematics - Class 12 JEEM21Mac48f Solution: Let A = [ a 1 a 2 ] and B = [ b 1 b 2 ] be two 2 1 matrices…

Question

Verified Solution • 2025–26 Session moderate

Let

A = [\begin{matrix} a_{1} \\ a_{2} \end{matrix}]

and

B = [\begin{matrix} b_{1} \\ b_{2} \end{matrix}]

be two 2

\times

1 matrices with real entries such that A = XB, where

X = \frac{1}{\sqrt{3}} [\begin{matrix} 1 & - 1 \\ 1 & k \end{matrix}]

, and k

\in

R.

If

a_1^2

+

a_2^2

=

{2 \over 3}

(b

_1^2

+ b

_2^2

) and (k² + 1) b

_2^2

\ne

-

2b₁b₂, then the value of k is __________.

Accepted Answer

XB = A

\Rightarrow

\frac{1}{\sqrt{3}} [\begin{matrix} 1 & - 1 \\ 1 & k \end{matrix}] [\begin{matrix} b_{1} \\ b_{2} \end{matrix}] = [\begin{matrix} a_{1} \\ a_{2} \end{matrix}]

\Rightarrow

\frac{1}{\sqrt{3}} [\begin{matrix} b_{1} - b_{2} \\ b_{1} + k b_{2} \end{matrix}] = [\begin{matrix} a_{1} \\ a_{2} \end{matrix}]

{b_1} - {b_2} = \sqrt 3 {a_1} \Rightarrow 3a_1^2 = b_1^2 + b_2^2 - 2{b_1}{b_2}

{b_1} + k{b_2} = \sqrt 3 {a_2} \Rightarrow 3a_2^2 = b_1^2 + {k^2}b_2^2 + 2k{b_1}{b_2}

3\left( {a_1^2 + a_2^2} \right) = 2b_1^2 + \left( {{k^2} + 1} \right)b_2^2 + 2{b_1}{b_2}(k - 1)

\Rightarrow

{a_1^2 + a_2^2}

=

{2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)

Given

a_1^2

+

a_2^2

=

{2 \over 3}

(b

_1^2

+ b

_2^2

)

\therefore

{2 \over 3}

(b

_1^2

+ b

_2^2

) =

{2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)

\Rightarrow

{2 \over 3}b_2^2 = {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)

Comparing both sides, We get

{{\left( {{k^2} + 1} \right)} \over 3} = {2 \over 3}

\Rightarrow

k² = 1

\Rightarrow

k =

\pm

1 ......(1)

and

{2 \over 3}\left( {k - 1} \right) = 0

\Rightarrow

k = 1 ....(2)

From (1) and (2),

k = 1

Step-by-Step Verified Solution