If the matrix A = [ cases 0 & 2 K & - 1 cases ] satisfies A( A ^ 3 + 3I) = 2I, then the value of K is

Correct answer: A. (1)/(2) A( A ^ 3 + 3I) = 2I A ^ 2 = [ cases 0 & 2 K & - 1 cases ][ cases 0 & 2 K & - 1 cases ] = [ cases 2K & - 2 - K & 2K + 1 cases ] A ^ 3 = [ cases 2K & - 2 - K & 2K + 1 cases ][ cases 0 & - 1 K & 2 cases ] = [ cases - 2K & 4K + 2 2 K ^ 2 + K & - 2K - 2K - 1 cases ] A ^ 4 = [ cases 2K & - 2 - K & 2K + 1 cases ][ cases 2K & - 2 - K & 2K + 1 cases ] = [ cases 4 K ^ 2 + 2K & - 4K - 4K - 2 - 2 K ^ 2 - 2 K ^ 2 - K & 2K + 4 K ^ 2 + 1 + 4K cases ] A ^ 4 + 3AI = 2I 2I = [ cases 4 K ^ 2 + 2K & - 8K - 2 - 4 K ^ 2 - K & 4 K ^ 2 + 6K + 1 cases ] + 3[ cases 0 & 2 K & - 1 cases ][ cases 1 & 0 0 & 1 cases ] 2I = [ cases 4 K ^ 2 + 2K & - 8K - 2 - 4 K ^ 2 - K & 4 K ^ 2 + 6K + 1 cases ] + 3[ cases 0 & 2 K & - 1 cases ] 2I = [ cases 4 K ^ 2 + 2K & 6 - 8K - 2 3K - K - 4 K ^ 2 & 4 K ^ 2 + 6K + 1 - 3 cases ] 4 - 8K = 0 K = (1)/(2)

JEE Main 202127 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2021 · 27 August · Morning Shift · Mathematics · Matrices

If the matrix A = [ cases 0 & 2 K & - 1 cases ] satisfies A( A ^ 3...

If the matrix

A =

\left[

\begin{cases}0 & 2 \\ K & - 1\end{cases}

\right]

satisfies

A\left({A}^{3} + 3I\right) = 2I, then the value of

K

Options

A $\dfrac{1}{2}$
B $1$
C $- 1$
D $- \dfrac{1}{2}$

Correct answer

A. $\dfrac{1}{2}$

Step-by-step solution

$A\left({A}^{3} + 3I\right) = 2I$

${A}^{2} = \left[$ \begin{cases}0 & 2 \\ K & - 1\end{cases} $\right]\left[$ \begin{cases}0 & 2 \\ K & - 1\end{cases} $\right]$

$= \left[$ \begin{cases}2K & - 2 \\ - K & 2K + 1\end{cases} $\right]$

${A}^{3} = \left[$ \begin{cases}2K & - 2 \\ - K & 2K + 1\end{cases} $\right]\left[$ \begin{cases}0 & - 1 \\ K & 2\end{cases} $\right]$

$= \left[$ \begin{cases}- 2K & 4K + 2 \\ 2{K}^{2} + K & - 2K - 2K - 1\end{cases} $\right]$

${A}^{4} = \left[$ \begin{cases}2K & - 2 \\ - K & 2K + 1\end{cases} $\right]\left[$ \begin{cases}2K & - 2 \\ - K & 2K + 1\end{cases} $\right]$

$= \left[$ \begin{cases}4{K}^{2} + 2K & - 4K - 4K - 2 \\ - 2{K}^{2} - 2{K}^{2} - K & 2K + 4{K}^{2} + 1 + 4K\end{cases} $\right]$

${A}^{4} + 3AI = 2I$

$2I = \left[$ \begin{cases}4{K}^{2} + 2K & - 8K - 2 \\ - 4{K}^{2} - K & 4{K}^{2} + 6K + 1\end{cases} $\right] + 3\left[$ \begin{cases}0 & 2 \\ K & - 1\end{cases} $\right]\left[$ \begin{cases}1 & 0 \\ 0 & 1\end{cases} $\right]$

$2I = \left[$ \begin{cases}4{K}^{2} + 2K & - 8K - 2 \\ - 4{K}^{2} - K & 4{K}^{2} + 6K + 1\end{cases} $\right] + 3\left[$ \begin{cases}0 & 2 \\ K & - 1\end{cases} $\right]$

$2I = \left[$ \begin{cases}4{K}^{2} + 2K & 6 - 8K - 2 \\ 3K - K - 4{K}^{2} & 4{K}^{2} + 6K + 1 - 3\end{cases} $\right]$

$4 - 8K = 0$

$K = \dfrac{1}{2}$

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