Let A = ( cases 2 & - 2 1 & - 1 cases ) and B = ( cases - 1 & 2 - 1 & 2 cases ) . Then the number of elements in the set (n, m): n, m ∈ \ 1, 2, . 10\ and n A ^ n + m B ^ m = I is _____.

Correct answer: 0 Given, A = ( cases 2 & - 2 1 & - 1 cases ) So, A ^ 2 = ( cases 2 & - 2 1 & - 1 cases )( cases 2 & - 2 1 & - 1 cases ) = ( cases 2 & - 2 1 & - 1 cases ) = A Also, B = ( cases - 1 & 2 - 1 & 2 cases ) B ^ 2 = ( cases - 1 & 2 - 1 & 2 cases )( cases - 1 & 2 - 1 & 2 cases ) = ( cases - 1 & 2 - 1 & 2 cases ) = B So, A + B = ( cases 2 & - 2 1 & - 1 cases ) + ( cases - 1 & 2 - 1 & 2 cases ) = ( cases 1 & 0 0 & 1 cases ) = I Therefore the equation n A ^ n + m B ^ m = I is true for n = 1 and m = 1 so, only one set is possible.

JEE Main 202225 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 25 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2022 · 25 June · Evening Shift · Mathematics · Matrices

Let A = ( cases 2 & - 2 1 & - 1 cases ) and B = ( cases - 1 & 2 - 1...

Let

A = \left(

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

\right)

and B = \left(

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

\right)

. Then the number of elements in the set {\left(n, m\right): n, m \in \left\{1, 2, \ldots \ldots . 10\right\}\text{ and } n{A}^{n} + m{B}^{m} = I}

is _____.

Correct answer

Step-by-step solution

Given,

$A = \left($ \begin{cases}2 & - 2 \\ 1 & - 1\end{cases} $\right)$ So, ${A}^{2} = \left($ \begin{cases}2 & - 2 \\ 1 & - 1\end{cases} $\right)\left($ \begin{cases}2 & - 2 \\ 1 & - 1\end{cases} $\right) = \left($ \begin{cases}2 & - 2 \\ 1 & - 1\end{cases} $\right) = A$

Also, $B = \left($ \begin{cases}- 1 & 2 \\ - 1 & 2\end{cases} $\right)$

${B}^{2} = \left($ \begin{cases}- 1 & 2 \\ - 1 & 2\end{cases} $\right)\left($ \begin{cases}- 1 & 2 \\ - 1 & 2\end{cases} $\right) = \left($ \begin{cases}- 1 & 2 \\ - 1 & 2\end{cases} $\right) = B$

So, $A + B = \left($ \begin{cases}2 & - 2 \\ 1 & - 1\end{cases} $\right) + \left($ \begin{cases}- 1 & 2 \\ - 1 & 2\end{cases} $\right) = \left($ \begin{cases}1 & 0 \\ 0 & 1\end{cases} $\right) = I$

Therefore the equation $n{A}^{n} + m{B}^{m} = I$ is true for $n = 1 \text{ and } m = 1$ so, only one set is possible.

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