JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 22)
Let $$A = \left[ {\matrix{
{{a_1}} \cr
{{a_2}} \cr
} } \right]$$ and $$B = \left[ {\matrix{
{{b_1}} \cr
{{b_2}} \cr
} } \right]$$ be two 2 $$\times$$ 1 matrices with real entries such that A = XB, where
$$X = {1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & { - 1} \cr 1 & k \cr } } \right]$$, 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 __________.
$$X = {1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & { - 1} \cr 1 & k \cr } } \right]$$, 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 __________.
Answer
1
Explanation
$$XB = A$$
$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & { - 1} \cr 1 & k \cr } } \right]\left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$$
$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}\left[ {\matrix{ {{b_1} - {b_2}} \cr {{b_1} + k{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$$
$${b_1} - {b_2} = \sqrt 3 {a_1} \Rightarrow 3a_1^2 = b_1^2 + b_2^2 - 2{b_1}{b_2}$$
$${b_1} + k{b_2} = \sqrt 3 {a_2} \Rightarrow 3a_2^2 = b_1^2 + {k^2}b_2^2 + 2k{b_1}{b_2}$$
$$3\left( {a_1^2 + a_2^2} \right) = 2b_1^2 + \left( {{k^2} + 1} \right)b_2^2 + 2{b_1}{b_2}(k - 1)$$
$$ \Rightarrow $$ $${a_1^2 + a_2^2} $$ = $${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$$
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
$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & { - 1} \cr 1 & k \cr } } \right]\left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$$
$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}\left[ {\matrix{ {{b_1} - {b_2}} \cr {{b_1} + k{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$$
$${b_1} - {b_2} = \sqrt 3 {a_1} \Rightarrow 3a_1^2 = b_1^2 + b_2^2 - 2{b_1}{b_2}$$
$${b_1} + k{b_2} = \sqrt 3 {a_2} \Rightarrow 3a_2^2 = b_1^2 + {k^2}b_2^2 + 2k{b_1}{b_2}$$
$$3\left( {a_1^2 + a_2^2} \right) = 2b_1^2 + \left( {{k^2} + 1} \right)b_2^2 + 2{b_1}{b_2}(k - 1)$$
$$ \Rightarrow $$ $${a_1^2 + a_2^2} $$ = $${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$$
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
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