If $a, b, c$ are all different and $\left| {\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4}\, - \,1}\\b&{{b^3}}&{{b^4}\, - \,1}\\c&{{c^3}}&{{c^4}\, - \,1}\end{array}} \right|$ $= 0$ , then :
- A$abc (ab + bc + ca) = a + b + c$
- B$(a + b + c) (ab + bc + ca) = abc$
- C$abc (a + b + c) = ab + bc + ca$
- Dnone of these
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Evaluate $\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$
Evaluate $\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$
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
$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$ is divisor of
$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$ is divisor of
By using properties of determinants, show that:
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By using properties of determinants, show that:
$\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$
$\left| {\,\begin{array}{*{20}{c}}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}\,} \right| = $