Q : 13        By using properties of determinants, show that:

                \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}=(1+a^2+b^2)^3

Answers (1)

We have determinant:

                            \triangle = \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}

Applying row transformations, R_{1} \rightarrow R_{1} +bR_{3}    and   R_{2} \rightarrow R_{2} -aR_{3} then we have;

= \begin{vmatrix} 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{vmatrix}

taking common factor out of the determinant;

= (1+a^2+b^2)^2\begin{vmatrix} 1 &0 &-b \\ 0 &1 &a \\ 2b &-2a & 1-a^2-b^2 \end{vmatrix}

Now expanding the remaining determinant we get;

= (1+a^2+b^2)^2\left [ (1)\begin{vmatrix} 1& a\\ -2a&1-a^2-b^2 \end{vmatrix} - b\begin{vmatrix} 0&1 \\ 2b&-2a \end{vmatrix}\right ]

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

= (1+a^2+b^2)^2\left [ 1+a^2+b^2\right ] = (1+a^2+b^2)^3

Hence proved.

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