If $a + b + c = 0$, then the solution of the equation $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$ is
If $a + b + c = 0$, then the solution of the equation $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$ is
- A
$0$
- B
$ \pm \frac{3}{2}({a^2} + {b^2} + {c^2})$
- C
$0,\, \pm \sqrt {\frac{3}{2}({a^2} + {b^2} + {c^2})} $
- D
$0,\,\, \pm \sqrt {{a^2} + {b^2} + {c^2}} $
Similar Questions
Show that $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$
Show that $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$
If ${a_1},{a_2},{a_3},........,{a_n},......$ are in G.P. and ${a_i} > 0$ for each $i$, then the value of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 2}}}&{\log {a_{n + 4}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 8}}}&{\log {a_{n + 10}}}\\{\log {a_{n + 12}}}&{\log {a_{n + 14}}}&{\log {a_{n + 16}}}\end{array}} \right|$ is equal to
If ${a_1},{a_2},{a_3},........,{a_n},......$ are in G.P. and ${a_i} > 0$ for each $i$, then the value of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 2}}}&{\log {a_{n + 4}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 8}}}&{\log {a_{n + 10}}}\\{\log {a_{n + 12}}}&{\log {a_{n + 14}}}&{\log {a_{n + 16}}}\end{array}} \right|$ is equal to
If $p, q, r, s$ are in $A.P.$ and $f (x) =$ $\left| {\,\begin{array}{*{20}{c}}
{p\,\, + \,\,\sin \,x}&{q\,\, + \,\,\sin \,x}&{p\,\, - \,\,r\,\, + \,\,\sin \,x}\\
{q\,\, + \,\,\sin \,x}&{r\,\, + \,\,\sin \,x}&{ - \,1\,\, + \,\,\sin \,x}\\
{r\,\, + \,\,\sin \,x}&{s\,\, + \,\,\sin \,x}&{s\,\, - \,\,q\,\, + \,\,\sin \,x}
\end{array}\,} \right|$ such that $f (x)dx = - 4$ then the common difference of the $A.P.$ can be :
If $a+x=b+y=c+z+1,$ where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to
If $a+x=b+y=c+z+1,$ where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to
- [JEE MAIN 2020]
Using the property of determinants and without expanding, prove that:
$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$
Using the property of determinants and without expanding, prove that:
$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$