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By using properties of determinants, show that $$\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac} \\ {ab}&{{b^2} + 1}&{bc} \\ {ca}&{cb}&{{c^2} + 1} \end{array}} \right| = 1 + {a^2} + {b^2} + {c^2}$$
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## QuestionMathsClass 12

By using properties of determinants, show that $$\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac} \\ {ab}&{{b^2} + 1}&{bc} \\ {ca}&{cb}&{{c^2} + 1} \end{array}} \right| = 1 + {a^2} + {b^2} + {c^2}$$

see analysis
4.6
4.6

## Solution

Given: $$\left\vert \begin{array}{ccc}{a}^{2}+1& ab& ac\\ ab& {b}^{2}+1& bc\\ ca& cb& {c}^{2}+1\end{array}\right\vert ,1+{a}^{2}+{b}^{2}+{c}^{2}$$
To prove: By using properties of determinant, show that $$\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac} \\ {ab}&{{b^2} + 1}&{bc} \\ {ca}&{cb}&{{c^2} + 1} \end{array}} \right| = 1 + {a^2} + {b^2} + {c^2}$$
Solution:
LHS$$=\left\vert \begin{array}{ccc}{a}^{2}+1& ab& ac\\ ab& {b}^{2}+1& bc\\ ca& cb& {c}^{2}+1\end{array}\right\vert \;$$
(Multiplying $${R}_{1},{R}_{2}\;and\;{R}_{3\;}by\;a,b\;and\;c$$  respectively)
$$=\frac{1}{abc}\left\vert \begin{array}{ccc}{a}^{3}+a& {a}^{2}b& {a}^{2}c\\ a{b}^{2}& {b}^{3}+b& {b}^{2}c\\ {c}^{2}a& {c}^{2}b& {c}^{3}+c\end{array}\right\vert$$
(Taking $$a, b, c$$ common from $$C{}_{1},C{}_{2},C{}_{3}$$  respectively)
$$=\frac{abc}{abc}\left\vert \begin{array}{ccc}{a}^{2}+1& {a}^{2}& {a}^{2}\\ {b}^{2}& {b}^{2}+1& {b}^{2}\\ {c}^{2}& {c}^{2}& {c}^{2}+1\end{array}\right\vert$$($${R_1} \to {R_1} + {R_2} + {R_3}$$)
$$= \left| {\begin{array}{*{20}{c}} {1 + {a^2} + {b^2} + {c^2}}&{1 + {a^2} + {b^2} + {c^2}}&{1 + {a^2} + {b^2} + {c^2}} \\ {{b^2}}&{{b^2} + 1}&{{b^2}} \\ {{c^2}}&{{c^2}}&{{c^2} + 1} \end{array}} \right|$$
$$= \left( {1 + {a^2} + {b^2} + {c^2}} \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\ {{b^2}}&{{b^2} + 1}&{{b^2}} \\ {{c^2}}&{{c^2}}&{{c^2} + 1} \end{array}} \right|$$($${C_1} \to {C_1} - {C_3},{C_2} \to {C_2} - {C_3}$$)

$$= \left( {1 + {a^2} + {b^2} + {c^2}} \right)\left| {\begin{array}{*{20}{c}} 0&0&1 \\ 0&1&{{b^2}} \\ { - 1}&{ - 1}&{{c^2} + 1} \end{array}} \right|$$(Expanding along $$R{}_{1}$$)
$$=\left(1+{a}^{2}+{b}^{2}+{c}^{2}\right)\left(\;1\times \left\vert \begin{array}{ll}0& 1\\ -1& -1\end{array}\right\vert \right)$$
$$=(1+a{}^{2}+b{}^{2}+c{}^{2})(0+1)$$
$$=(1+a{}^{2}+b{}^{2}+c{}^{2})\;=RHS$$