સાબિત કરો કે $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4 a b c$
સાબિત કરો કે $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4 a b c$
Let $\Delta=\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|$
Applying $\quad \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}-\mathrm{R}_{3}$ to $\Delta,$ we get
$\Delta=\left|\begin{array}{ccc}
0 & -2 c & -2 b \\
b & c+a & b \\
c & c & a+b
\end{array}\right|$
Expanding along $\mathrm{R}_{1},$ we obtain
$\Delta = 0\left| {\begin{array}{*{20}{c}}
{c + a}&b \\
c&{a + b}
\end{array}} \right| - ( - 2c)\left| {\begin{array}{*{20}{c}}
b&b \\
c&{a + b}
\end{array}} \right| + ( - 2b)\left| {\begin{array}{*{20}{c}}
b&{c + a} \\
c&c
\end{array}} \right|$
$ = 2c\left( {ab + {b^2} - bc} \right) - 2b\left( {bc - {c^2} - ac} \right)$
$ = 2abc + 2c{b^2} - 2b{c^2} - 2{b^2}c + 2b{c^2} + 2abc$
$ = 4abc$
Similar Questions
જો $A$, $B$ અને $C$ ત્રિકોણના ખૂણા હોય તો નિશ્ચાયક
$\left| {\begin{array}{*{20}{c}}
{ - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\
{ - 1 + \cos B}&{ - 1 + \cos A}&{ - 1}
\end{array}} \right|$ ની કિમંત મેળવો.
જો $A$, $B$ અને $C$ ત્રિકોણના ખૂણા હોય તો નિશ્ચાયક
$\left| {\begin{array}{*{20}{c}}
{ - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\
{ - 1 + \cos B}&{ - 1 + \cos A}&{ - 1}
\end{array}} \right|$ ની કિમંત મેળવો.
જો $ab + bc + ca = 0$ અને $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$, તો $x$ ની કોઈ એક કિમત મેળવો.
જો $ab + bc + ca = 0$ અને $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$, તો $x$ ની કોઈ એક કિમત મેળવો.
$\left| {\,\begin{array}{*{20}{c}}{441}&{442}&{443}\\{445}&{446}&{447}\\{449}&{450}&{451}\end{array}\,} \right|$
$\left| {\,\begin{array}{*{20}{c}}{441}&{442}&{443}\\{445}&{446}&{447}\\{449}&{450}&{451}\end{array}\,} \right|$
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ તો $K = $
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ તો $K = $
જો $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z - x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}\,} \right| = k\,xyz$, તો $k$ મેળવો.
જો $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z - x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}\,} \right| = k\,xyz$, તો $k$ મેળવો.