Question:

By using properties of determinants,show that:\(\begin{bmatrix}1+a^2-b^2& 2ab& -2b\\ 2ab& 1-a^2+b^2& 2a\\ 2b& -2a& 1-a^2-b^2\end{bmatrix}\)\(=(1+a^2+b^2)^3\)

Updated On: Sep 5, 2023
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Solution and Explanation

\(\triangle=\begin{bmatrix}1+a^2-b^2& 2ab& -2b\\ 2ab& 1-a^2+b^2& 2a\\ 2b& -2a& 1-a^2-b^2\end{bmatrix}\)
Applying \(R_1 → R_1 + bR_3\) and \(R_2 → R_2 − aR_3\), we have:
\(\triangle=\begin{bmatrix}1+a^2+b^2& 0& -b(1+a^2+b^2)\\ 0& 1+a^2+b^2& a(1+a^2+b^2)\\ 2b& -2a& 1-a^2-b^2\end{bmatrix}\)
\(=(1+a^2+b^2)^2\begin{bmatrix}1&0&-b\\ 0&1&a\\ 2b& -2a& 1-a^2-b^2\end{bmatrix}\)
Expanding along \(R_1\), we have:
\(△=(1+a^2+b^2)^2\bigg[(1)\begin{bmatrix}1& a \\-2a& 1-a^2-b^2\end{bmatrix}-b\begin{bmatrix}0& 1\\ 2b& -2a\end{bmatrix}\bigg]\)
\(=(1+a^2+b^2)^2[1-a^2-b^2+2a^2-b(-2b)]\)
\(=(1+a^2+b^2)^2(1+a^2+b^2)\)
\(=(1+a^2+b^2)^3\)
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