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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Prove that:
$\begin{vmatrix}\text{a}^2+1&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2+1&\text{bc}\\\text{ca}&\text{cb}&\text{c}^2+1 \end{vmatrix}=1+\text{a}^2+\text{b}^2+\text{c}^2$
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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{a}^2+1&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2+1&\text{bc}\\\text{ca}&\text{cb}&\text{c}^2+1 \end{vmatrix}$
$=(\text{abc})\begin{vmatrix}\text{a}+\frac{1}{\text{a}}&\text{b}&\text{c}\\\text{a}&\text{b}+\frac{1}{\text{b}}&\text{c}\\\text{a}&\text{b}&\text{c}+\frac{1}{\text{c}} \end{vmatrix}$
[Taking out a, b and c common from $R_1, R_2$ and $R_3$]
$=(\text{abc})\begin{vmatrix}\text{a}+\frac{1}{\text{a}}&\text{b}&\text{c}\\-\frac{1}{\text{a}}&\frac{1}{\text{b}}&0\\-\frac{1}{\text{a}}&0&\frac{1}{\text{c}} \end{vmatrix}$
[Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$]
$=(\text{abc})\Big(\frac{1}{\text{abc}}\Big)\begin{vmatrix}\text{a}^2+1&\text{b}^2&\text{c}^2\\-1&1&0\\-1&0&1\end{vmatrix}$
[Applying $C_1 \rightarrow aC_1, C_2 \rightarrow bC_2$ and $C_3 \rightarrow cC_3$]
$=\begin{vmatrix}\text{a}^2+1&\text{b}^2&\text{c}^2\\-1&1&0\\-1&0&1\end{vmatrix}$
$=(-1)\begin{vmatrix}\text{b}^2&\text{c}^2\\1&0\end{vmatrix}+(1)\begin{vmatrix}\text{a}^2+1&\text{b}^2\\-1&1\end{vmatrix}$
$=(-1)(-\text{c}^2)+(\text{a}^2+1+\text{b}^2)$
$=(\text{a}^2+1+\text{b}^2+\text{c}^2)$
$=(\text{a}^2+\text{b}^2+\text{c}^2+1)$
$=\text{R.H.S}$

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