Let A be a square matrix such that AA ^ T = I . Then (1)/(2)A [ (A + A ^ T ) ^ 2 + (A - A ^ T ) ^ 2 ] is equal to

Correct answer: D. A ^ 3 + A ^ T Given: AA ^ T = 1 ⇒ A ^ T = A ^ - 1 Now, solving ⇒ (1)/(2)A[ (A + A ^ T ) ^ 2 + (A - A ^ T ) ^ 2 ] = (1)/(2)A × [2 A ^ 2 + 2 ( A ^ T ) ^ 2 + 2 AA ^ T - 2 AA ^ T ] ⇒ (1)/(2)A[ (A + A ^ T ) ^ 2 + (A - A ^ T ) ^ 2 ] = (1)/(2)A × 2[ A ^ 2 + ( A ^ T ) ^ 2 ] ⇒ (1)/(2)A[ (A + A ^ T ) ^ 2 + (A - A ^ T ) ^ 2 ] = A ^ 3 + A ( A ^ T ) ^ 2 Now using, AA ^ T = 1 , ⇒ (1)/(2)A[ (A + A ^ T ) ^ 2 + (A - A ^ T ) ^ 2 ] = A ^ 3 + A ^ T

JEE Main 202429 JanuaryMorning Shift

Exam: JEE Main
Year: 2024
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2024 · 29 January · Morning Shift · Mathematics · Matrices

Let A be a square matrix such that AA ^ T = I . Then (1)/(2)A [ (A...

Let

A

be a square matrix such that

{AA}^{T} = I

. Then

\dfrac{1}{2}A

\left[{

\left(A + {A}^{T}\right)}^{2} + {\left(A - {A}^{T}\right)}^{2}\right]

is equal to

Options

A ${A}^{2} + I$
B ${A}^{3} + I$
C ${A}^{2} + {A}^{T}$
D ${A}^{3} + {A}^{T}$

Correct answer

D. ${A}^{3} + {A}^{T}$

Step-by-step solution

Given: ${AA}^{T} = 1$

$⇒ {A}^{T} = {A}^{- 1}$

Now, solving

$⇒ \dfrac{1}{2}A\left[{\left(A + {A}^{T}\right)}^{2} + {\left(A - {A}^{T}\right)}^{2}\right] = \dfrac{1}{2}A \times \left[2{A}^{2} + 2{\left({A}^{T}\right)}^{2} + 2{AA}^{T} - 2{AA}^{T}\right]$ $⇒ \dfrac{1}{2}A\left[{\left(A + {A}^{T}\right)}^{2} + {\left(A - {A}^{T}\right)}^{2}\right] = \dfrac{1}{2}A \times 2\left[{A}^{2} + {\left({A}^{T}\right)}^{2}\right]$

$⇒ \dfrac{1}{2}A\left[{\left(A + {A}^{T}\right)}^{2} + {\left(A - {A}^{T}\right)}^{2}\right] = {A}^{3} + A{\left({A}^{T}\right)}^{2}$

Now using, ${AA}^{T} = 1$ ,

$⇒ \dfrac{1}{2}A\left[{\left(A + {A}^{T}\right)}^{2} + {\left(A - {A}^{T}\right)}^{2}\right] = {A}^{3} + {A}^{T}$

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