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By using properties of determinants, show that
$\left|\begin{array}{ccc}1+a^2-b^2 & 2 a b & -2 b \\ 2 a b & 1-a^2+b^2 & 2 a \\ 2 b & -2 a & 1-a^2+b^2\end{array}\right|=\left(1+a^2+b^2\right)^3$
$\left|\begin{array}{ccc}1+a^2-b^2 & 2 a b & -2 b \\ 2 a b & 1-a^2+b^2 & 2 a \\ 2 b & -2 a & 1-a^2+b^2\end{array}\right|=\left(1+a^2+b^2\right)^3$
Solution:
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Verified Answer
L.H.S.
$\mathrm{C}_1-b \mathrm{C}_3 \rightarrow \mathrm{C}_1 \mathrm{C}_2+a \mathrm{C}_3 \rightarrow \mathrm{C}_2$
$=\left(1+a^2+b^2\right)^2\left|\begin{array}{ccc}1 & 0 & -2 b \\ 0 & 1 & 2 a \\ 0 & -a & 1-a^2+b^2\end{array}\right|$
$\mathrm{R}_3-b \mathrm{R}_1 \Rightarrow \mathrm{R}_3$
$=\left(1+a^2+b^2\right)^3=$ R.H.S.
$\mathrm{C}_1-b \mathrm{C}_3 \rightarrow \mathrm{C}_1 \mathrm{C}_2+a \mathrm{C}_3 \rightarrow \mathrm{C}_2$
$=\left(1+a^2+b^2\right)^2\left|\begin{array}{ccc}1 & 0 & -2 b \\ 0 & 1 & 2 a \\ 0 & -a & 1-a^2+b^2\end{array}\right|$
$\mathrm{R}_3-b \mathrm{R}_1 \Rightarrow \mathrm{R}_3$
$=\left(1+a^2+b^2\right)^3=$ R.H.S.
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