Question

The value of the determinant
$\begin{vmatrix}1+a^{2}-b^{2}&2ab&-2b\\ 2ab&1-a^{2}+b^{2}&2a\\ 2b&-2a&1-a^{2}-b^{2}\end{vmatrix}$
is equal to

• A. 0
• B. $(1+a^2+b^2)$
• C. $(1+a^2+b^2)^2$
• D. $(1+a^2+b^2)^3$

Answer 1

Answer: (D)

. Let $\Delta=\begin{vmatrix}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{vmatrix}$
Apply $C_{1} \rightarrow C_{1}-b C_{3}$ and $C_{2}$
$ \rightarrow a C_{3}+C_{2}$
$\Delta=\begin{vmatrix}1+a^{2}-b^{2}+2 b^{2} & 2 a b-2 a b & -2 b \\ 2 a b-2 a b & 1-a^{2}+b^{2}+2 a^{2} & 2 a \\ 2 b-b+a^{2} b+b^{3} & -2 a+a-a^{3}-a b^{2} & 1-a^{2}-b^{2}\end{vmatrix}$
$=\begin{vmatrix}\left(1+a^{2}+b^{2}\right) & 0 & -2 b \\ 0 & \left(1+a^{2}+b^{2}\right) & 2 a \\ b\left(1+a^{2}+b^{2}\right) & -a\left(1+a^{2}+b^{2}\right) & \left(1-a^{2}-b^{2}\right)\end{vmatrix}$
$=\left(1+a^{2}+b^{2}\right)^{2}\begin{vmatrix}1 & 0 & -2 b \\ 0 & 1 & 2 a \\ b & -a & \left(1-a^{2}-b^{2}\right)\end{vmatrix}$
$=\left(1+a^{2}+b^{2}\right)^{2}\left\{\left(1-a^{2}-b^{2}+2 a^{2}\right)+2 b^{2}\right\}$
$=\left(1+a^{2}+b^{2}\right)^{2}\left(1+a^{2}+b^{2}\right)=\left(1+a^{2}+b^{2}\right)^{3}$