નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3 x+k)$
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3 x+k)$
$\Delta=\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2} R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}3 y+k & 3 y+k & 3 y+k \\ y & y+k & y \\ y & y & y+k\end{array}\right|$
$=(3 y+k)\left|\begin{array}{ccc}1 & 1 & 1 \\ y & y+k & y \\ y & y & y+k\end{array}\right|$
Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1},$ we have:
$\Delta=(3 y+k)\left|\begin{array}{lll}1 & 0 & 0 \\ y & k & 0 \\ y & 0 & k\end{array}\right|$
$=k^{2}(3 x+k)\left|\begin{array}{lll}1 & 0 & 0 \\ y & 1 & 0 \\ y & 0 & 1\end{array}\right|$
Expanding alone $C_{3},$ we have:
$\Delta=k^{2}(3 y+k)\left|\begin{array}{ll}1 & 0 \\ y & 1\end{array}\right|=k^{2}(3 y+k)$
Hence, the given result is proved.
Similar Questions
$\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$ નું મૂલ્ય શોધો.
$\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$ નું મૂલ્ય શોધો.
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$
$\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$ માટે ગુણધર્મ $1$ ચકાસો.
$\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$ માટે ગુણધર્મ $1$ ચકાસો.
જો $\left| {\begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
{{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\
{{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}}
\end{array}} \right|$ $ = \,k\lambda \,\,\left| {{\mkern 1mu} {\mkern 1mu} \begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
a&b&c \\
1&1&1
\end{array}} \right|,\lambda \, \ne \,0$ તો $k$ મેળવો.
જો $\left| {\begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
{{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\
{{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}}
\end{array}} \right|$ $ = \,k\lambda \,\,\left| {{\mkern 1mu} {\mkern 1mu} \begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
a&b&c \\
1&1&1
\end{array}} \right|,\lambda \, \ne \,0$ તો $k$ મેળવો.
- [JEE MAIN 2014]
જો $a, b, c $ એ દરેક એકબીજાથી ભિન્ન હોય અને $\left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4} - 1}\\b&{{b^3}}&{{b^4} - 1}\\c&{{c^3}}&{{c^4} - 1}\end{array}\,} \right|=0$ , તો $abc(ab + bc + ca)$ =
જો $a, b, c $ એ દરેક એકબીજાથી ભિન્ન હોય અને $\left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4} - 1}\\b&{{b^3}}&{{b^4} - 1}\\c&{{c^3}}&{{c^4} - 1}\end{array}\,} \right|=0$ , તો $abc(ab + bc + ca)$ =