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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By using properties of determinants, show that \(\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac} \\ {ab}&{{b^2} + 1}&{bc} \\ {ca}&{cb}&{{c^2} + 1} \end{array}} \right| = 1 + {a^2} + {b^2} + {c^2}\)
Answer
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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 \(R{}_{1}\))
\(=\left(1+{a}^{2}+{b}^{2}+{c}^{2}\right)\left(\;1\times \left\vert \begin{array}{ll}0& 1\\ -1& -1\end{array}\right\vert \right)\)
\(=(1+a{}^{2}+b{}^{2}+c{}^{2})(0+1)\)
\(=(1+a{}^{2}+b{}^{2}+c{}^{2})\;=RHS\)
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 \(R{}_{1}\))
\(=\left(1+{a}^{2}+{b}^{2}+{c}^{2}\right)\left(\;1\times \left\vert \begin{array}{ll}0& 1\\ -1& -1\end{array}\right\vert \right)\)
\(=(1+a{}^{2}+b{}^{2}+c{}^{2})(0+1)\)
\(=(1+a{}^{2}+b{}^{2}+c{}^{2})\;=RHS\)
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