Without expanding, prove that $\Delta=\left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|=0$
Without expanding, prove that $\Delta=\left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|=0$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}$ to $\Delta,$ we get
$\Delta=\left|\begin{array}{ccc}
x+y+z & x+y+z & x+y+z \\
z & x & y \\
1 & 1 & 1
\end{array}\right|$
since the elements of $R_{1}$ and $R_{3}$ are proportional, $\Delta=0$
Similar Questions
The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right|$ $= 11$ is
The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right|$ $= 11$ is
If $ab + bc + ca = 0$ and $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$, then one of the value of $x$ is
If $ab + bc + ca = 0$ and $\left| {\,\begin{array}{*{20}{c}}{a - x}&c&b\\c&{b - x}&a\\b&a&{c - x}\end{array}\,} \right| = 0$, then one of the value of $x$ is
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}\,} \right| = $
- [IIT 1986]
By using properties of determinants, show that:
$\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^{3} & b^{3} & c^{3}
\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^{3} & b^{3} & c^{3}
\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$
If the system of equations $ax + y + z = 0$, $x + by + z = 0$ and $x + y + cz = 0 $, where $a,b,c \ne 1,$ has a non trivial solution, then the value of $\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}}$is
If the system of equations $ax + y + z = 0$, $x + by + z = 0$ and $x + y + cz = 0 $, where $a,b,c \ne 1,$ has a non trivial solution, then the value of $\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}}$is