નિશ્વાયક — ગણિત ધોરણ 12 સાયન્સ — Question

$\xrightarrow[{{R_3}\left( {\frac{1}{c}} \right)}]{{{R_1}\left( {\frac{1}{a}} \right),\,\,{R_2}\left( {\frac{1}{b}} \right)}}\,\, = $ $D \ \begin{vmatrix}a^2+1 & ab &ac \\ab & b^2+1 & bc \\ac& bc & c^2+1\end{vmatrix}$

$\xrightarrow[{{c_3}\left( c \right)}]{{{c_1}\left( a \right),\,\,{c_2}\left( b \right)}}\,$$D=abc\begin{vmatrix}a+\frac{1}{a} & b &c \\a & b+\frac{1}{b} & c \\a& b & c+\frac{1}{c}\end{vmatrix}$

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

$ = \xrightarrow[{{c_1}\left( {\frac{1}{{{a^2} + {b^2} + {c^2} + 1}}} \right)}]{{{c_{21}}\left( 1 \right),\,\,{c_{31}}\left( 1\right)}}\begin{vmatrix}1 & b^2 &c^2 \\1 & b+1 & c^2 \\1& b^2 & (c^2+1)\end{vmatrix}(a^2+b^2+c^2+1)$

$ = \,\xrightarrow[{{R_{32}}\left( { - 1} \right)}]{{{R_{21}}\left( { - 1} \right)}} \ a^2+b^2+c^2+1\begin{vmatrix}0 & -1 &0 \\0 & 1 & -1 \\1& b^2 & c^2+1\end{vmatrix}$

=$(a^2+b^2+c^2+1)(1)$

=$a^2+b^2+c^2+1$