Let \(A=\left[\begin{array}{l}a_{1} \\ a_{2}\end{array}\right]\) and \(B=\left[\begin{array}{l}b_{1} \\ b_{2}\end{array}\right]\) be two \(2 \times 1\) matrices with real entries such that \(A=X B\), where \(\mathrm{X}=\frac{1}{\sqrt{3}}\left[\begin{array}{c}1-1 \\ 1 \mathrm{k}\end{array}\right]\), and \(\mathrm{k} \in \mathrm{R}\). If \(\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}=\frac{2}{3}\left(\mathrm{~b}_{1}^{2}+\mathrm{b}_{2}^{2}\right)\) and \(\left(\mathrm{k}^{2}+1\right) \mathrm{b}_{2}^{2} \neq-2 \mathrm{~b}_{1} \mathrm{~b}_{2}\), then the value of k is. \_\_\_\_

Matrices — JEE Main · 2021 · 16 Mar 2021 · Shift 2 Question with Solution

Let

A = [a_{1} a_{2}]

and

B = [b_{1} b_{2}]

be two

2 \times 1

matrices with real entries such that

A = X B

, where

X = \frac{1}{3} [1 - 1 1 k]

, and

k \in R

. If

a_{1}^{2} + a_{2}^{2} = \frac{2}{3} (b_{1}^{2} + b_{2}^{2})

and

(k^{2} + 1) b_{2}^{2} \neq = - 2 b_{1} b_{2}

, then the value of k is. ____

Answer: 1

Given

A = X B

, we have:

a_{1} = \frac{1}{3} (b_{1} - b_{2})

a_{2} = \frac{1}{3} (b_{1} + k b_{2})

Then:

a_{1}^{2} + a_{2}^{2} = \frac{1}{3} [(b_{1} - b_{2})^{2} + (b_{1} + k b_{2})^{2}]

= \frac{1}{3} [b_{1}^{2} - 2 b_{1} b_{2} + b_{2}^{2} + b_{1}^{2} + 2 k b_{1} b_{2} + k^{2} b_{2}^{2}]

= \frac{1}{3} [2 b_{1}^{2} + (1 + k^{2}) b_{2}^{2} + 2 (k - 1) b_{1} b_{2}]

Given

a_{1}^{2} + a_{2}^{2} = \frac{2}{3} (b_{1}^{2} + b_{2}^{2})

\frac{1}{3} [2 b_{1}^{2} + (1 + k^{2}) b_{2}^{2} + 2 (k - 1) b_{1} b_{2}] = \frac{2}{3} (b_{1}^{2} + b_{2}^{2})

Simplifying:

2 b_{1}^{2} + (1 + k^{2}) b_{2}^{2} + 2 (k - 1) b_{1} b_{2} = 2 b_{1}^{2} + 2 b_{2}^{2}

(k^{2} - 1) b_{2}^{2} + 2 (k - 1) b_{1} b_{2} = 0

(k - 1) [(k + 1) b_{2}^{2} + 2 b_{1} b_{2}] = 0

Since

(k^{2} + 1) b_{2}^{2} \neq = - 2 b_{1} b_{2}

, we must have

k = 1

.

Answer: 1

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer