Question

Let $A=\begin{bmatrix}1+x^2-y^2-z^2 & 2(x y+z) & 2(z x-y) \\ 2(x y-z) & 1+y^2-z^2-x^2 & 2(y z+x) \\ 2(z x+y) & 2(y z-x) & 1+z^2-x^2-y^2\end{bmatrix}$ then det. $A$ is equal to

• A. $(1+x y+y z+z x)^3$
• B. $\left(1+x^2+y^2+z^2\right)^3$
• C. $(x y+y z+z x)^3$
• D. $\left(1+x^3+y^3+z^3\right)^2$

Answer 1

Answer: (B)

multiply $R_2$ by $z$ and $R_3$ by y and use $R_1 \rightarrow R_1-R_2+R_3$
Objective approach : put $z = y =0$ then choices are $A =1 ; B =\left(1+ x ^2\right)^3 ; C =0 ; D =\left(1+ x ^3\right)^2$ and determinant comes out to be $\left(1+x^2\right)^3 \Rightarrow(B)$