If M = ( cases (5)/(2) & (3)/(2) - (3)/(2) & - (1)/(2) cases ) , then which of the following matrices is equal to M ^ 2022 ?

Correct answer: A. ( cases 3034 & 3033 - 3033 & - 3032 cases ) Given M = ( cases (5)/(2) & (3)/(2) - (3)/(2) & - (1)/(2) cases ) M ^ 2 = ( cases (5)/(2) & (3)/(2) - (3)/(2) & - (1)/(2) cases )( cases (5)/(2) & (3)/(2) - (3)/(2) & - (1)/(2) cases ) = ( cases 4 & 3 - 3 & - 2 cases ) M ^ 3 = ( cases (5)/(2) & (3)/(2) - (3)/(2) & - (1)/(2) cases )( cases 4 & 3 - 3 & - 2 cases ) = ( cases (11)/(2) & (9)/(2) - (9)/(2) & - (7)/(2) cases ) M ^ 4 = ( cases (11)/(2) & (9)/(2) - (9)/(2) & - (7)/(2) cases )( cases (11)/(2) & (9)/(2) - (9)/(2) & - (7)/(2) cases ) = ( cases 7 & 6 - 6 & - 5 cases ) and so on i.e. M ^ n = ( cases (3n)/(2) + 1 & (3n)/(2) - (3n)/(2) & - (3n)/(2) + 1 cases ) Hence, M ^ 2022 = ( cases (3 × 2022)/(2) + 1 & (3 × 2022)/(2) - (3 × 2022)/(2) & - (3 × 2022)/(2) + 1 cases ) = ( cases 3034 & 3033 - 3033 & - 3032 cases )

JEE Advanced 2022Paper 2

Exam: JEE Advanced
Year: 2022
Paper: Paper 2
Subject: Mathematics
Chapter: Matrices

JEE Advanced 2022 · Paper 2 · Mathematics · Matrices

If M = ( cases (5)/(2) & (3)/(2) - (3)/(2) & - (1)/(2) cases ) ,...

M = \left(

\begin{cases}\dfrac{5}{2} & \dfrac{3}{2} \\ - \dfrac{3}{2} & - \dfrac{1}{2}\end{cases}

\right)

, then which of the following matrices is equal to

{M}^{2022}

Options

A $\left($ \begin{cases}3034 & 3033 \\ - 3033 & - 3032\end{cases} $\right)$
B $\left($ \begin{cases}3034 & - 3033 \\ 3033 & - 3032\end{cases} $\right)$
C $\left($ \begin{cases}3033 & 3032 \\ - 3032 & - 3031\end{cases} $\right)$
D $\left($ \begin{cases}3032 & 3031 \\ - 3031 & - 3030\end{cases} $\right)$

Correct answer

A. $\left($ \begin{cases}3034 & 3033 \\ - 3033 & - 3032\end{cases} $\right)$

Step-by-step solution

Given $M = \left($ \begin{cases}\dfrac{5}{2} & \dfrac{3}{2} \\ - \dfrac{3}{2} & - \dfrac{1}{2}\end{cases} $\right)$
${M}^{2} = \left($ \begin{cases}\dfrac{5}{2} & \dfrac{3}{2} \\ - \dfrac{3}{2} & - \dfrac{1}{2}\end{cases} $\right)\left($ \begin{cases}\dfrac{5}{2} & \dfrac{3}{2} \\ - \dfrac{3}{2} & - \dfrac{1}{2}\end{cases} $\right) = \left($ \begin{cases}4 & 3 \\ - 3 & - 2\end{cases} $\right)$
${M}^{3} = \left($ \begin{cases}\dfrac{5}{2} & \dfrac{3}{2} \\ - \dfrac{3}{2} & - \dfrac{1}{2}\end{cases} $\right)\left($ \begin{cases}4 & 3 \\ - 3 & - 2\end{cases} $\right) = \left($ \begin{cases}\dfrac{11}{2} & \dfrac{9}{2} \\ - \dfrac{9}{2} & - \dfrac{7}{2}\end{cases} $\right)$
${M}^{4} = \left($ \begin{cases}\dfrac{11}{2} & \dfrac{9}{2} \\ - \dfrac{9}{2} & - \dfrac{7}{2}\end{cases} $\right)\left($ \begin{cases}\dfrac{11}{2} & \dfrac{9}{2} \\ - \dfrac{9}{2} & - \dfrac{7}{2}\end{cases} $\right) = \left($ \begin{cases}7 & 6 \\ - 6 & - 5\end{cases} $\right)$
and so on
i.e. ${M}^{n} = \left($ \begin{cases}\dfrac{3n}{2} + 1 & \dfrac{3n}{2} \\ - \dfrac{3n}{2} & - \dfrac{3n}{2} + 1\end{cases} $\right)$
Hence,
${M}^{2022} = \left($ \begin{cases}\dfrac{3 \times 2022}{2} + 1 & \dfrac{3 \times 2022}{2} \\ - \dfrac{3 \times 2022}{2} & - \dfrac{3 \times 2022}{2} + 1\end{cases} $\right)$
$= \left($ \begin{cases}3034 & 3033 \\ - 3033 & - 3032\end{cases} $\right)$

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