# Q : 14            By using properties of determinants, show that:                       $\dpi{100} \begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}=1+a^2+b^2+c^2$

Given determinant:

$\dpi{100} \begin{vmatrix} a^2+1 &ab &ac \\ ab &b^2+1 &bc \\ ca & cb &c^2+1 \end{vmatrix}$

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

Then we can clearly see that each column can be reduced by taking common factors like a,b, and c respectively from C1,C2,and C3.

We then get;

$=abc \begin{vmatrix} \left ( a+\frac{1}{a} \right ) &a &a \\ b &(b+\frac{1}{b}) &b \\ c & c &(c+\frac{1}{c}) \end{vmatrix}$

Now, applying column transformations: $C_{1} \rightarrow C_{1} -C_{2}$  and $C_{2} \rightarrow C_{2} -C_{3}$

then we have;

$=abc \begin{vmatrix} \left ( \frac{1}{a} \right ) &0 &a \\ -\frac{1}{b} &(\frac{1}{b}) &b \\ 0 & -\frac{1}{c} &(c+\frac{1}{c}) \end{vmatrix}$

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

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

Now, expanding the remaining determinant:

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

$= 1[(c^2+1)+b^2] + a^2(1)=a^2+b^2+c^2+1$.

Hence proved.

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