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3 and 4 .Determinants and Matrices
hard
Let $A=\left[\begin{array}{lll}x & y & z \\ y & z & x \\ z & x & y\end{array}\right], \quad$ where $x, y$ and $z$ are real numbers such that $x + y + z >0$ and $xyz =2$ If $A ^{2}= I _{3},$ then the value of $x ^{3}+ y ^{3}+ z ^{3}$ is ............
A
$7$
B
$5$
C
$9$
D
$6$
(JEE MAIN-2021)
Solution
$A ^{2}= I$
$\left(\right.$ as $\left.A^{\prime}=A\right)$
$\Rightarrow AA ^{\prime}= I$
$\Rightarrow A$ is orthogonal
So, $x^{2}+y^{2}+z^{2}=1$ and $x y+y z+z x=0$
$\Rightarrow(x+y+z)^{2}=1+2 \times 0$
$\Rightarrow x+y+z=1$
Thus,
$x^{3}+y^{3}+z^{3}=3 \times 2+1 \times(1-0)$
$=7$
Standard 12
Mathematics
