Let A = [ cases x & 1 1 & 0 cases ], xεR and A ^ 4 = [ a _ ij ] . If a _ 11 = 109, then a _ 22 is equal to_____________.

Correct answer: 0 A ^ 2 = [ cases x & 1 1 & 0 cases ] × [ cases x & 1 1 & 0 cases ] = [ cases x ^ 2 + 1 & x x & 1 cases ] A ^ 4 = [ cases x ^ 2 + 1 & x x & 1 cases ] × [ cases x ^ 2 + 1 & x x & 1 cases ] = [ cases x ^ 4 + 3 x ^ 2 + 1 & x ^ 3 + 2x x ^ 3 + 2x & x ^ 2 + 1 cases ] Given a _ 11 = 109 ⇒ x ^ 4 + 3 x ^ 2 + 1 = 109 ( x ^ 2 + 12)( x ^ 2 - 9) = 0 ⇒ x ^ 2 = 9 ∴ a _ 22 = x ^ 2 + 1 = 10

JEE Main 20203 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 3 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2020 · 3 September · Morning Shift · Mathematics · Matrices

Let A = [ cases x & 1 1 & 0 cases ], xεR and A ^ 4 = [ a _ ij ] ....

Let

A = \left[

\begin{cases}x & 1 \\ 1 & 0\end{cases}

\right], xεR

and

{A}^{4} = \left[{a}_{ij}\right] .

{a}_{11} = 109,

then

{a}_{22}

is equal to_____________.

Correct answer

Step-by-step solution

{A}^{2} = \left[ $\begin{cases}x & 1 \\ 1 & 0\end{cases}$ \right] \times \left[ $\begin{cases}x & 1 \\ 1 & 0\end{cases}$ \right] = \left[ $\begin{cases}{x}^{2} + 1 & x \\ x & 1\end{cases}$ \right] $$ {A}^{4} = \left[ $\begin{cases}{x}^{2} + 1 & x \\ x & 1\end{cases}$ \right] \times \left[ $\begin{cases}{x}^{2} + 1 & x \\ x & 1\end{cases}$ \right] = \left[ $\begin{cases}{x}^{4} + 3{x}^{2} + 1 & {x}^{3} + 2x \\ {x}^{3} + 2x & {x}^{2} + 1\end{cases}$ \right] $Given$ {a}_{11} = 109 ⇒ {x}^{4} + 3{x}^{2} + 1 = 109 $\left({x}^{2} + 12\right)\left({x}^{2} - 9\right) = 0 ⇒ {x}^{2} = 9$

$∴ {a}_{22} = {x}^{2} + 1 = 10$

Practice Matrices on Quantrex Academy →

More from Matrices