Hard INTEGER +4 / -1 PYQ · JEE Mains 2021

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 ______.

Answer (integer) 17

Solution

As $PQ = kI \Rightarrow Q = k{P^{ - 1}}I$<br><br>now $Q = {k \over {|P|}}(adjP)I$ <br><br>$$\Rightarrow Q = {k \over {(20 + 12\alpha )}}\left[ {\matrix{ - &amp; - &amp; - \cr - &amp; - &amp; {( - 3\alpha - 4)} \cr - &amp; - &amp; - \cr } } \right]\left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 1 &amp; 0 \cr 0 &amp; 0 &amp; 1 \cr } } \right]$$<br><br>$\because$ ${q_{23}} = {{ - k} \over 8}$ <br><br>$\Rightarrow {k \over {(20 + 12\alpha )}}( - 3\alpha - 4) = {{ - k} \over 8}$ <br><br>$\Rightarrow 2(3\alpha + 4) = 5 + 3\alpha$<br><br>$3\alpha = - 3 \Rightarrow \alpha = - 1$<br><br>also $$|Q| = {{{k^3}|I|} \over {|P|}} \Rightarrow {{{k^2}} \over 2} = {{{k^3}} \over {(20 + 12\alpha )}}$$<br><br>$\Rightarrow$ $(20 + 12\alpha ) = 2k \Rightarrow 8 = 2k \Rightarrow k = 4$

About this question

Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations

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