JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 14)
Let P = $$\left[ {\matrix{
3 & { - 1} & { - 2} \cr
2 & 0 & \alpha \cr
3 & { - 5} & 0 \cr
} } \right]$$, where $$\alpha $$ $$ \in $$ R. Suppose Q = [ qij] is a matrix satisfying PQ = kl3 for some non-zero k $$ \in $$ R.
If q23 = $$ - {k \over 8}$$ and |Q| = $${{{k^2}} \over 2}$$, then a2 + k2 is equal to ______.
If q23 = $$ - {k \over 8}$$ and |Q| = $${{{k^2}} \over 2}$$, then a2 + k2 is equal to ______.
Answer
17
Explanation
As $$PQ = kI \Rightarrow Q = k{P^{ - 1}}I$$
now $$Q = {k \over {|P|}}(adjP)I $$
$$\Rightarrow Q = {k \over {(20 + 12\alpha )}}\left[ {\matrix{ - & - & - \cr - & - & {( - 3\alpha - 4)} \cr - & - & - \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$ \because $$ $${q_{23}} = {{ - k} \over 8} $$
$$\Rightarrow {k \over {(20 + 12\alpha )}}( - 3\alpha - 4) = {{ - k} \over 8} $$
$$\Rightarrow 2(3\alpha + 4) = 5 + 3\alpha $$
$$3\alpha = - 3 \Rightarrow \alpha = - 1$$
also $$|Q| = {{{k^3}|I|} \over {|P|}} \Rightarrow {{{k^2}} \over 2} = {{{k^3}} \over {(20 + 12\alpha )}}$$
$$ \Rightarrow $$ $$(20 + 12\alpha ) = 2k \Rightarrow 8 = 2k \Rightarrow k = 4$$
now $$Q = {k \over {|P|}}(adjP)I $$
$$\Rightarrow Q = {k \over {(20 + 12\alpha )}}\left[ {\matrix{ - & - & - \cr - & - & {( - 3\alpha - 4)} \cr - & - & - \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$ \because $$ $${q_{23}} = {{ - k} \over 8} $$
$$\Rightarrow {k \over {(20 + 12\alpha )}}( - 3\alpha - 4) = {{ - k} \over 8} $$
$$\Rightarrow 2(3\alpha + 4) = 5 + 3\alpha $$
$$3\alpha = - 3 \Rightarrow \alpha = - 1$$
also $$|Q| = {{{k^3}|I|} \over {|P|}} \Rightarrow {{{k^2}} \over 2} = {{{k^3}} \over {(20 + 12\alpha )}}$$
$$ \Rightarrow $$ $$(20 + 12\alpha ) = 2k \Rightarrow 8 = 2k \Rightarrow k = 4$$
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