If for the matrix, A = [ cases 1 & - α α & β cases ], A A ^ T = I _ 2 , then the value of α ^ 4 + β ^ 4 is :

Correct answer: B. 1 A = [ cases 1 & - α α & β cases ]A A ^ T = I _ 2 ⇒ [ cases 1 & - α α & β cases ][ cases 1 & α - α & β cases ] = [ cases 1 & 0 0 & 1 cases ] ⇒ [ cases 1 + α ^ 2 & α - αβ α - αβ & α ^ 2 + β ^ 2 cases ] = [ cases 1 & 0 0 & 1 cases ] ⇒ 1 + α ^ 2 = 1 and α ^ 2 + β ^ 2 = 1 ⇒ α ^ 2 = 0 and β ^ 2 = 1 ∴ α ^ 4 = 0 and β ^ 4 = 1 ∴ α ^ 4 + β ^ 4 = 1

JEE Main 202125 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2021 · 25 February · Evening Shift · Mathematics · Matrices

If for the matrix, A = [ cases 1 & - α α & β cases ], A A ^ T = I _...

If for the matrix,

A = \left[

\begin{cases}1 & - \alpha \\ \alpha & \beta\end{cases}

\right], A{A}^{T} = {I}_{2}

, then the value of

{\alpha}^{4} + {\beta}^{4}

is :

Options

A $3$
B $1$
C $2$
D $4$

Correct answer

B. $1$

Step-by-step solution

$A = \left[$ \begin{cases}1 & - \alpha \\ \alpha & \beta\end{cases} $\right]A{A}^{T} = {I}_{2}$

$⇒ \left[$ \begin{cases}1 & - \alpha \\ \alpha & \beta\end{cases} $\right]\left[$ \begin{cases}1 & \alpha \\ - \alpha & \beta\end{cases} $\right] = \left[$ \begin{cases}1 & 0 \\ 0 & 1\end{cases} $\right]$

$⇒ \left[$ \begin{cases}1 + {\alpha}^{2} & \alpha - \alpha\beta \\ \alpha - \alpha\beta & {\alpha}^{2} + {\beta}^{2}\end{cases} $\right] = \left[$ \begin{cases}1 & 0 \\ 0 & 1\end{cases} $\right]$

$⇒ 1 + {\alpha}^{2} = 1 \text{ and } {\alpha}^{2} + {\beta}^{2} = 1$

$⇒ {\alpha}^{2} = 0 \text{ and } {\beta}^{2} = 1$

$∴ {\alpha}^{4} = 0 \text{ and } {\beta}^{4} = 1$

$∴ {\alpha}^{4} + {\beta}^{4} = 1$

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