सारणिकों के गुणधर्मों का प्रयोग करके सिद्ध कीजिए :
$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3 x+k)$
$\Delta=\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|$
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.
समीकरण
$\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0,(0 < x < \pi)$ के हल है
यदि $f(x) = \left| {\begin{array}{*{20}{c}}{x - 3}&{2{x^2} - 18}&{3{x^3} - 81}\\{x - 5}&{2{x^2} - 50}&{4{x^3} - 500}\\1&2&3\end{array}} \right|$ then $f(1).f(3) + f(3).f(5) + f(5).f(1)$=
दर्शाइए कि $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ तो $K = $
यदि $a \neq 0$ हो तो समीकरण $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$ को हल कीजिए।