Evaluate $\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$
Evaluate $\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$
- A
$2\left(x^{3}+y^{3}\right)$
- B
$2\left(x^{3}-y^{3}\right)$
- C
$-2\left(x^{3}-y^{3}\right)$
- D
$-2\left(x^{3}+y^{3}\right)$
Similar Questions
Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}( A )=4$. Let $R _{ i }$ denote the $i ^{\text {th }}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R _{2} \rightarrow 2 R _{2}+5 R _{3}$ on $2 A ,$ then $\operatorname{det}( B )$ is equal to ...... .
Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}( A )=4$. Let $R _{ i }$ denote the $i ^{\text {th }}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R _{2} \rightarrow 2 R _{2}+5 R _{3}$ on $2 A ,$ then $\operatorname{det}( B )$ is equal to ...... .
- [JEE MAIN 2021]
$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = $
The determinant $\left| {\begin{array}{*{20}{c}}{{b_1}\, + \,\,{c_1}}&{{c_1}\, + \,\,{a_1}}&{{a_1}\, + \,\,{b_1}}\\{{b_2}\, + \,\,{c_2}}&{{c_2}\, + \,\,{a_2}}&{{a_2}\, + \,\,{b_2}}\\{{b_3}\, + \,\,{c_3}}&{{c_3}\, + \,\,{a_3}}&{{a_3}\, + \,\,{b_3}}
\end{array}} \right|$ $=$
Which of the following values of $\alpha$ satisfy the equation
$\left|\begin{array}{lll}(1+\alpha)^2 & (1+2 \alpha)^2 & (1+3 \alpha)^2 \\ (2+\alpha)^2 & (2+2 \alpha)^2 & (2+3 \alpha)^2 \\ (3+\alpha)^2 & (3+2 \alpha)^2 & (3+3 \alpha)^2\end{array}\right|=-648 \alpha$ ?
$(A)$ $-4$ $(B)$ $9$ $(C)$ $-9$ $(D)$ $4$
Which of the following values of $\alpha$ satisfy the equation
$\left|\begin{array}{lll}(1+\alpha)^2 & (1+2 \alpha)^2 & (1+3 \alpha)^2 \\ (2+\alpha)^2 & (2+2 \alpha)^2 & (2+3 \alpha)^2 \\ (3+\alpha)^2 & (3+2 \alpha)^2 & (3+3 \alpha)^2\end{array}\right|=-648 \alpha$ ?
$(A)$ $-4$ $(B)$ $9$ $(C)$ $-9$ $(D)$ $4$
- [IIT 2015]
Show that
$\Delta=\left|\begin{array}{ccc}
(y+z)^{2} & x y & z x \\
x y & (x+z)^{2} & y z \\
x z & y z & (x+y)^{2}
\end{array}\right|=2 x y z(x+y+z)^{3}$
Show that
$\Delta=\left|\begin{array}{ccc}
(y+z)^{2} & x y & z x \\
x y & (x+z)^{2} & y z \\
x z & y z & (x+y)^{2}
\end{array}\right|=2 x y z(x+y+z)^{3}$