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If $P$ is a $3 \times 3$ real matrix such that $P^T=a P+(a-1) I$, where $a>1$, then :

JEE · Math · previous-year question

  1. A.$|A d j P|=1$correct
  2. B.$|A d j P|>1$
  3. C.$|A d j P|=\frac{1}{2}$
  4. D.$P$ is a singular matrix

Answer

A. $|A d j P|=1$

Explanation

$$P = \left[ {\begin{matrix} {{a_1}} & {{b_1}} & {{c_1}} \\ {{a_2}} & {{b_2}} & {{c_2}} \\ {{a_3}} & {{b_3}} & {{c_3}} \\ \end{matrix} } \right]$$ Given : $${P^T} = aP + (a - 1)I$$ $$\left[ {\begin{matrix} {{a_1}} & {{a_2}} & {{a_3}} \\ {{b_1}} & {{b_2}} & {{b_3}} \\ {{c_1}} & {{c_2}} & {{c_3}} \\ \end{matrix} } \right] = \left[ {\begin{matrix} {a{a_1} + a - 1} & {a{b_1}} & {a{c_1}} \\ {a{a_2}} & {a{b_2} + a - 1} & {a{c_2}} \\ {a{a_3}} & {a{b_3}} & {a{c_3} + a - 1} \\ \end{matrix} } \right]$$ $$ \Rightarrow {a_1} = a{a_1} + a - 1 \Rightarrow {a_1}(1 - a) = a - 1 \Rightarrow {a_1} = - 1$$ Similarly, $${a_1} = {b_2} = {c_3} = - 1$$ Now, $$\left. \begin{matrix} {a_2} = a{b_1} \hfill \\ {b_1} = a{a_2} \hfill \\\end{matrix} \right] \to {a_2} = {a^2}{a_2} \Rightarrow {a_2} = 0 \Rightarrow {b_1} = 0$$ $${c_1} = a{a_3}$$ Similarly, all other elements will also be 0 $${a_2} = {a_3} = {b_1} = {b_3} = {c_1} = {c_2} = 0$$ $$\therefore$$ $$P = \left[ {\begin{matrix} { - 1} & 0 & 0 \\ 0 & { - 1} & 0 \\ 0 & 0 & { - 1} \\ \end{matrix} } \right]$$ $$|P| = - 1$$ $$|Adj(P){|_{n \times n}} = |A{|^{(n - 1)}}$$ $$ \Rightarrow |Adj(P)| = {( - 1)^2} = 1$$

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