નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$

$\Delta=\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+b R_{3}$ and $R_{2} \rightarrow R_{2}-a R_{3},$ we have:

$\Delta=\left|\begin{array}{ccc}1+a^{2}+b^{2} & 0 & -b\left(1+a^{2}+b^{2}\right) \\ 0 & 1+a^{2}+b^{2} & a\left(1+a^{2}+b^{2}\right) \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$

$=\left(1+a^{2}+b^{2}\right)\left|\begin{array}{ccc}1 & 0 & -b \\ 0 & 1 & a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$

Expanding along $R_{1},$ we have:

$\Delta=\left(1+a^{2}+b^{2}\right)^{2}\left[(1)\left|\begin{array}{cc}1 & a \\ -2 a & 1-a^{2}-b^{2}\end{array}\right|-b\left|\begin{array}{cc}0 & 1 \\ 2 b & -2 a\end{array}\right|\right]$

$=\left(1+a^{2}+b^{2}\right)^{2}\left[1-a^{2}-b^{2}+2 a^{2}-b(-2 b)\right]$

$=\left(1+a^{2}+b^{2}\right)^{2}\left(1+a^{2}+b^{2}\right)$

$=\left(1+a^{2}+b^{2}\right)^{3}$