The determinant $\left| {\begin{array}{*{20}{c}}{^x{C_1}}&{^x{C_2}}&{^x{C_3}}\\ {^y{C_1}}&{^y{C_2}}&{^y{C_3}}\\{^z{C_1}}&{^z{C_2}}&{^z{C_3}}\end{array}} \right|$ $=$
The determinant $\left| {\begin{array}{*{20}{c}}{^x{C_1}}&{^x{C_2}}&{^x{C_3}}\\ {^y{C_1}}&{^y{C_2}}&{^y{C_3}}\\{^z{C_1}}&{^z{C_2}}&{^z{C_3}}\end{array}} \right|$ $=$
- A
$\frac{1}{3} \,xyz (x + y) (y + z) (z + x)$
- B
$\frac{1}{4} \,xyz (x + y - z) (y + z - x)$
- C
$\frac{1}{12} \, xyz (x - y) (y - z) (z - x)$
- D
none
Similar Questions
The value of $\left| {\,\begin{array}{*{20}{c}}a&{a + b}&{a + 2b}\\{a + 2b}&a&{a + b}\\{a + b}&{a + 2b}&a\end{array}\,} \right|$ is equal to
The value of $\left| {\,\begin{array}{*{20}{c}}a&{a + b}&{a + 2b}\\{a + 2b}&a&{a + b}\\{a + b}&{a + 2b}&a\end{array}\,} \right|$ is equal to
Value of $\left| {\begin{array}{*{20}{c}}
{{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\
{{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\
{{c^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right|$ is equal to
Value of $\left| {\begin{array}{*{20}{c}}
{{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\
{{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\
{{c^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right|$ is equal to
If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 3}\\{x + 2}&{x + 3}&{x + 4}\\{x + a}&{x + b}&{x + c}\end{array}\,} \right| = 0$, then $a,b,c$ are in
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Using properties of determinants, prove this:
$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$
Using properties of determinants, prove this:
$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$
The value of $\sum\limits_{n = 1}^N {{U_n},} $ if ${U_n} = \left| {\,\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}\,} \right|$ is
The value of $\sum\limits_{n = 1}^N {{U_n},} $ if ${U_n} = \left| {\,\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}\,} \right|$ is