Using properties of determinants, prove that |
\[\left| \begin{matrix} {{a}^{2}}+1 & ab & ac \\ ab & {{b}^{2}}+1 & bc \\ ca & ca & {{c}^{2}}+1 \\ \end{matrix} \right|=1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}.\]? |
OR |
Prove that |
\[\left| \begin{matrix} {{(b\,+c)}^{2}} & {{a}^{2}} & {{a}^{2}} \\ {{b}^{2}} & {{(c\,+a)}^{2}} & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{(a\,+b)}^{2}} \\ \end{matrix} \right|=2abc{{(a+b+c)}^{3}}.\] |
Answer:
Let \[\Delta =\left| \begin{matrix} {{a}^{2}}+1 & ab & ac \\ ab & {{b}^{2}}+1 & bc \\ ca & ca & {{c}^{2}}+1 \\ \end{matrix} \right|\] On applying \[{{R}_{1}}\to a{{R}_{1}},\] \[{{R}_{2}}\to b{{R}_{2}}\,\,and\,\,{{R}_{3}}\to c{{R}_{3}},\] We get \[\Delta =\frac{1}{abc}\left| \begin{matrix} a({{a}^{2}}+1) & {{a}^{2}}b & {{a}^{2}}c \\ a{{b}^{2}} & b({{b}^{2}}+1) & {{b}^{2}}c \\ {{c}^{2}}a & {{c}^{2}}b & c({{c}^{2}}+1) \\ \end{matrix} \right|\] On applying \[{{C}_{1}}\to \frac{1}{a}{{C}_{1}},\] \[{{C}_{2}}\to \frac{1}{b}{{C}_{2}}\] and \[{{C}_{3}}\to \frac{1}{c}{{C}_{3}},\] we get \[\Delta =\frac{abc}{abc}\,\left| \begin{matrix} {{a}^{2}}+1 & {{a}^{2}} & {{a}^{2}} \\ {{b}^{2}} & {{b}^{2}}+1 & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{c}^{2}}+1 \\ \end{matrix} \right|\] On applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}},\] we get \[\Delta =\left| \begin{matrix} 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\ {{b}^{2}} & {{b}^{2}}+1 & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{c}^{2}}+1 \\ \end{matrix} \right|\]On taking common \[(1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}})\] from \[{{R}_{1}},\]we get \[\Delta =(1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}})\left| \begin{matrix} 1 & 1 & 1 \\ {{b}^{2}} & {{b}^{2}}+1 & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{c}^{2}}+1 \\ \end{matrix} \right|\] On applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}},\] we get \[\Delta =(1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}})\left| \begin{matrix} 1 & 0 & 0 \\ {{b}^{2}} & 1 & 0 \\ {{c}^{2}} & 0 & 1 \\ \end{matrix} \right|\] Now expanding along \[{{R}_{1}},\] we get \[\Delta =(1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}})\,1\,[1-0]\] \[=1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\] Hence proved. OR Let the given determinant be \[\Delta .\] then, \[\Delta =\left| \begin{matrix} {{(b\,+c)}^{2}} & {{a}^{2}} & {{a}^{2}} \\ {{b}^{2}} & {{(c\,+a)}^{2}} & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{(a\,+b)}^{2}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} {{(b\,+c)}^{2}}-{{a}^{2}} & 0 & {{a}^{2}} \\ 0 & {{(c\,+a)}^{2}}-{{b}^{2}} & {{b}^{2}} \\ {{c}^{2}}-{{(a\,+b)}^{2}} & {{c}^{2}}-{{(a\,+b)}^{2}} & {{(a\,+b)}^{2}} \\ \end{matrix} \right|\] \[[{{C}_{1}}\to {{C}_{1}}-{{C}_{3}}\,\,\,and\,\,\,{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}]\] \[=\left| \begin{matrix} (a+b+c)(b+c-a) & 0 & {{a}^{2}} \\ 0 & (a+b+c)(c+a-b) & {{b}^{2}} \\ (a+b+c)(c-a-b) & (a+b+c)(c-a-b) & {{(a+b)}^{2}} \\ \end{matrix} \right|\]\[={{(a\,+b\,+c)}^{2}}\cdot \left| \begin{matrix} (b\,+c\,-a) & 0 & {{a}^{2}} \\ 0 & c+a-b & {{b}^{2}} \\ -\,c-a-b & -\,c-a-b & 2ab \\ \end{matrix} \right|\] [taking \[(a\,+b\,+c)\]common from \[{{C}_{1}}\]and \[{{C}_{2}}\]both] \[={{(a\,+b\,+c)}^{2}}\cdot \left| \begin{matrix} (b\,+c\,-a) & 0 & {{a}^{2}} \\ 0 & c+a-b & {{b}^{2}} \\ -\,2b & -\,2a & 2ab \\ \end{matrix} \right|\] \[[{{R}_{3}}\to {{R}_{3}}-({{R}_{1}}+{{R}_{2}})]\] \[={{(a\,+b+c)}^{2}}\,[(b\,+c\,-a)\{(c\,+a-b)\cdot 2ab\,+2a{{b}^{2}}\}\] \[+{{a}^{2}}\{0+2b(c\,+a\,-b)\}]\] \[={{(a\,+b\,+c)}^{2}}[b\,+c\,-a]\cdot 2ab\{(c\,+a\,-b\,+b)\}\] \[+\,2{{a}^{2}}b(c\,+a\,-b)]\] \[=2ab{{(a\,+b\,+c)}^{2}}\{(b\,+c\,-a)\,(c\,+a)+a(c+a-b)\}\] \[=2ab{{(a\,+b\,+c)}^{2}}\cdot \{bc\,+ab+{{c}^{2}}+ac-ac-{{a}^{2}}\] \[+ac+{{a}^{2}}-ab\}\] \[=2ab\,{{(a\,+b\,+c)}^{2}}\{bc\,+{{c}^{2}}+ac\}\] \[=2abc\,{{(a\,+b\,+c)}^{3}}.\] Hence, \[\Delta =2abc\,{{(a\,+b\,+c)}^{3}}.\] Hence proved.
You need to login to perform this action.
You will be redirected in
3 sec