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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