3 and 4 .Determinants and Matrices
medium

ધારો કે $A$ એક એવો ચોરસ શ્રેણિક છે કે જેથી $A A^T=I$. તો $\frac{1}{2} A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right]=$_____________.

A

 $A^2+I$

B

$A^3+I$

C

 $A^2+A^T$

D

 $A^3+A^T$

(JEE MAIN-2024)

Solution

$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$

On solving given expression, we get

$ \frac{1}{2} A\left[A^2+\left(A^T\right)^2+2 A A^T+A^2+\left(A^T\right)^2-2 A A^T\right] $

$ =A\left[A^2+\left(A^T\right)^2\right]=A^3+A^T$

Standard 12
Mathematics

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