નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરી  સાબિત કરો કે, $\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

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$\Delta=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-2 R_{1}$ and $R_{3} \rightarrow R_{3}-3 R_{1},$ we have:

$\Delta=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3 & 7+3 p\end{array}\right|$

Applying $R_{3} \rightarrow R_{3}-3 R_{2},$ we have:

$\Delta=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 0 & 1\end{array}\right|$

Expanding along $C_{1},$ we have:

$\Delta=1\left|\begin{array}{cc}1 & 2+p \\ 0 & 1\end{array}\right|=1(1-0)=1$

Similar Questions

શૂન્યતર $a$ માટે સમીકરણ $\left| {\begin{array}{*{20}{c}}
{x + a}&x&x\\
x&{x + a}&x\\
x&x&{x + a}
\end{array}} \right| = $ ઉકેલો.

સાબિત કરો કે $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$

જો $\left| {\,\begin{array}{*{20}{c}}{y + z}&x&y\\{z + x}&z&x\\{x + y}&y&z\end{array}\,} \right| = k(x + y + z){(x - z)^2}$, તો $k = $

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 $\mathrm{a, b, c}$ પૈકી પ્રત્યેક બે અસમાન અને પ્રત્યેક ધન હોય, તો સાબિત કરો કે નિશ્ચાયક $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ નું મૂલ્ય ઋણ છે.