Q94 / 200 · WBJEE 2013 · Mathematics
The value of determinant $\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$ is
- A0
- B$\left(1+a^{2}+b^{2}\right)$
- C$\left(1+a^{2}+b^{2}\right)^{2}$
- Correct answerD$\left(1+a^{2}+b^{2}\right)^{3}$
Solution
Let
$\Delta=\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$
Apply $\mathrm{C}_{\mathrm{1}} \rightarrow \mathrm{C}_{1}-\mathrm{bC}_{3}$ and $\mathrm{C}_{2} \rightarrow \mathrm{aC}_{3}+\mathrm{C}_{2}$
$=\left|\begin{array}{ccc}1+a^{2}-b^{2}+2 b^{2} & 2 a b-2 a b & -2 b \\ 2 a b-2 a b & 1-a^{2}+b^{2}+2 a^{2} & 2 a \\ 2 b-b+a^{2} b+b^{3} & -2 a+a-a^{3}-a b^{2} & 1-a^{2}-b^{2}\end{array}\right|$
$=\left|\begin{array}{ccc}\left(1+a^{2}+b^{2}\right) & 0 & -2 b \\ 0 & \left(1+a^{2}+b^{2}\right) & 2 a \\ b\left(1+a^{2}+b^{2}\right) & -a\left(1+a^{2}+b^{2}\right) & \left(1-a^{2}-b^{2}\right)\end{array}\right|$
$=\left(1+a^{2}+b^{2}\right)^{2}\left|\begin{array}{ccc}1 & 0 & -2 b \\ 0 & 1 & 2 a \\ b & -a & \left(1-a^{2}-b^{2}\right)\end{array}\right|$
$=\left(1+a^{2}+b^{2}\right)^{2}\left\{\left(1-a^{2}-b^{2}+2 a^{2}\right)+2 b^{2}\right\}$
$=\left(1+a^{2}+b^{2}\right)^{2}\left(1+a^{2}+b^{2}\right)=\left(1+a^{2}+b^{2}\right)^{3}$
Hence, the correct answer is option: 'D'.