Question

Find $n$ if: $\left|\begin{array}{ccc}1+a^2-b^2& 2ab& -2b\\ 2ab& 1-a^2+b^2& 2a\\ 2b& -2a& 1-a^2-b^2\end{array}\right|=(1+a^2+b^2{)}^n$

Answer 1

find n ?

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

Solve determinant

$C_2\to C_2+a{C}_3$

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

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

$R_3\to R_3+a{R}_2$

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

Expand the determinant with respect to $C_{22}$

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

$=(1+a^2+b^2)\left[\right(1+a^2)+(b^2{)}^2-2(1+a^2)b^2+4b^2(1+a^2)]$

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

$(1+a^2+b^2)(1+a^2+b^2{)}^2$

$(1+a^2+b^2{)}^3$

Compare both sides

$n=3$