The determinant $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|$ is not equal to
The determinant $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|$ is not equal to
- A
$\left| {\,\begin{array}{*{20}{c}}2&1&1\\2&2&3\\2&3&6\end{array}\,} \right|$
- B
$\left| {\,\begin{array}{*{20}{c}}2&1&1\\3&2&3\\4&3&6\end{array}\,} \right|$
- C
$\left| {\begin{array}{*{20}{c}}1&2&1\\1&5&3\\1&9&6\end{array}} \right|$
- D
$\left| {\,\begin{array}{*{20}{c}}3&1&1\\6&2&3\\{10}&3&6\end{array}} \right|\,$
Similar Questions
If the system of equations $x+y+z=6 \,; \,2 x+5 y+\alpha z=\beta \,; \, x+2 y+3 z=14$ has infinitely many solutions, then $\alpha+\beta$ is equal to.
If the system of equations $x+y+z=6 \,; \,2 x+5 y+\alpha z=\beta \,; \, x+2 y+3 z=14$ has infinitely many solutions, then $\alpha+\beta$ is equal to.
- [JEE MAIN 2022]
Let $x, y, z > 0$ are respectively $2^{nd}, 3^{rd}, 4^{th}$ term of $G.P.$ and $\Delta = \left| {\begin{array}{*{20}{c}}
{{X^k}}&{{X^{k + 1}}}&{{X^{k + 2}}}\\
{{Y^k}}&{{Y^{k + 1}}}&{{Y^{k + 2}}}\\
{{Z^k}}&{{Z^{k + 1}}}&{{Z^{k + 2}}}
\end{array}} \right| = {\left( {r - 1} \right)^2}\left( {1 - \frac{1}{{{r^2}}}} \right)$ , (where $r$ is common ratio), then $k=$ .......
Let $x, y, z > 0$ are respectively $2^{nd}, 3^{rd}, 4^{th}$ term of $G.P.$ and $\Delta = \left| {\begin{array}{*{20}{c}}
{{X^k}}&{{X^{k + 1}}}&{{X^{k + 2}}}\\
{{Y^k}}&{{Y^{k + 1}}}&{{Y^{k + 2}}}\\
{{Z^k}}&{{Z^{k + 1}}}&{{Z^{k + 2}}}
\end{array}} \right| = {\left( {r - 1} \right)^2}\left( {1 - \frac{1}{{{r^2}}}} \right)$ , (where $r$ is common ratio), then $k=$ .......
If $\alpha , \beta \, and \, \gamma$ are real numbers , then $D = \left|{\begin{array}{*{20}{c}}1&{\cos \,(\beta \, - \,\alpha )}&{\cos \,(\gamma \, - \,\alpha )}\\{\cos \,(\alpha \, - \,\beta )}&1&{\cos \,(\gamma \, - \,\beta )}\\{\cos \,(\alpha \, - \,\gamma )}&{\cos \,(\beta \, - \,\gamma )}&1 \end{array}} \right|$ =
If $\alpha , \beta \, and \, \gamma$ are real numbers , then $D = \left|{\begin{array}{*{20}{c}}1&{\cos \,(\beta \, - \,\alpha )}&{\cos \,(\gamma \, - \,\alpha )}\\{\cos \,(\alpha \, - \,\beta )}&1&{\cos \,(\gamma \, - \,\beta )}\\{\cos \,(\alpha \, - \,\gamma )}&{\cos \,(\beta \, - \,\gamma )}&1 \end{array}} \right|$ =
If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :
If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :
- [JEE MAIN 2024]
Statement $-1 :$Determinant of a skew-symmetric matrix of order $3$ is zero
Statement $-2 :$ For any matrix $A,$ $\det \left( {{A^T}} \right) = {\rm{det}}\left( A \right)$ and $\det \left( { - A} \right) = - {\rm{det}}\left( A \right)$ Where $\det \left( A \right) = A$. Then :
Statement $-1 :$Determinant of a skew-symmetric matrix of order $3$ is zero
Statement $-2 :$ For any matrix $A,$ $\det \left( {{A^T}} \right) = {\rm{det}}\left( A \right)$ and $\det \left( { - A} \right) = - {\rm{det}}\left( A \right)$ Where $\det \left( A \right) = A$. Then :
- [AIEEE 2011]