# By using the properties of determinants show that $\begin{array}{l} \begin{vmatrix} a^2+1&ab&ac \\ ab&b^2+1&bc \\ ca&cb&c^2+1 \end{vmatrix} = 1+a^2+b^2+c^2 \end{array}$

Toolbox:
• If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any one column of the given determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\bigtriangleup=\begin{vmatrix}a^2+1 & ab & ac\\ab & b^2+1 & bc\\ca & cb & c^2+1\end{vmatrix}$

Let us take the common factors a from $R_1$,b from $R_2$ and c from $R_3$.

$\bigtriangleup=abc\begin{vmatrix}a+1/a & b & c\\a & b+1/b & c\\a & b & c+1/c\end{vmatrix}$

Apply $R_2\rightarrow R_2-R_1$ and $R_3\rightarrow R_3-R_1$

$\bigtriangleup=abc\begin{vmatrix}a+1/a & b & c\\-1/a & 1/b & 0\\-1/a & 0 & 1/c\end{vmatrix}$

Multiply $C_1$ by a and apply $C_1\rightarrow aC_1$

Multiply $C_2$ by b and apply $C_2\rightarrow bC_2$

Multiply $C_3$ by c and apply $C_3\rightarrow cC_3$

We have $\bigtriangleup=abc\times \frac{1}{abc}\begin{vmatrix}a^2+1 & b ^2& c^2\\-1 & 1 & 0\\-1 & 0 & 1\end{vmatrix}$

$\quad=\begin{vmatrix}a^2+1 & b ^2& c^2\\-1 & 1 & 0\\-1 & 0 & 1\end{vmatrix}$

Now expand along $R_1$ we get

$\bigtriangleup=(a^2+1)(1-0)-b^2(-1-0)+c^2(0-(-9)]$

$\quad=a^2+1+b^2+c^2$

$\quad=1+a^2+b^2+c^2$

Hence $\begin{vmatrix}a^2+1& ab & ac\\ab & b^2+1 & bc\\ca &cb & c^2+1\end{vmatrix}=1+a^2+b^2+c^2$

Hence proved.

edited Feb 24, 2013