By using properties of determinants, show that:
$\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$
$\Delta=\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|$
Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}2(x+y+z) & x & y \\ 2(x+y+z) & y+z+2 x & y \\ 2(x+y+z) & x & z+x+2 y\end{array}\right|$
$=2(x+y+z)\left|\begin{array}{ccc}1 & x & y \\ 1 & y+z+2 x & y \\ 1 & x & z+x+2 y\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1},$ we have:
$\Delta=2(x+y+z)\left|\begin{array}{ccc}1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z\end{array}\right|$
$=2(x+y+z)\left|\begin{array}{lll}1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$
Expanding along $R_{3},$ we have:
$\Delta=2(x+y+z)^{3}(1)(1-0)=2(x+y+z)^{3}$
Hence, the given result is proved.
Similar Questions
$\left| {\,\begin{array}{*{20}{c}}1&{1 + ac}&{1 + bc}\\1&{1 + ad}&{1 + bd}\\1&{1 + ae}&{1 + be}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}1&{1 + ac}&{1 + bc}\\1&{1 + ad}&{1 + bd}\\1&{1 + ae}&{1 + be}\end{array}\,} \right| = $
If $A, B, C$ are the angles of a triangle and $\left| {\begin{array}{*{20}{c}}1&1&1\\{1 + \sin A}&{1 + \sin B}&{1 + \sin C}\\{\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right|$ $= 0$, then the triangle is
If $A, B, C$ are the angles of a triangle and $\left| {\begin{array}{*{20}{c}}1&1&1\\{1 + \sin A}&{1 + \sin B}&{1 + \sin C}\\{\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right|$ $= 0$, then the triangle is
Verify Property $1$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$
Verify Property $1$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$
$\left| {\,\begin{array}{*{20}{c}}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}\,} \right| = $
If $a, b, c > 0 \, and \, x, y, z \in R$ , then the determinant $\left|{\begin{array}{*{20}{c}}{{{\left( {{a^x}\, + \,\,{a^{ - x}}} \right)}^2}}&{{{\left( {{a^x}\, - \,\,{a^{ - x}}} \right)}^2}}&1\\{{{\left( {{b^y}\, + \,\,{b^{ - y}}} \right)}^2}}&{{{\left( {{b^y}\, - \,\,{b^{ - y}}} \right)}^2}}&1\\{{{\left( {{c^z}\, + \,\,{c^{ - z}}} \right)}^2}}&{{{\left( {{c^z}\, - \,\,{c^{ - z}}} \right)}^2}}&1\end{array}} \right|$ $=$
If $a, b, c > 0 \, and \, x, y, z \in R$ , then the determinant $\left|{\begin{array}{*{20}{c}}{{{\left( {{a^x}\, + \,\,{a^{ - x}}} \right)}^2}}&{{{\left( {{a^x}\, - \,\,{a^{ - x}}} \right)}^2}}&1\\{{{\left( {{b^y}\, + \,\,{b^{ - y}}} \right)}^2}}&{{{\left( {{b^y}\, - \,\,{b^{ - y}}} \right)}^2}}&1\\{{{\left( {{c^z}\, + \,\,{c^{ - z}}} \right)}^2}}&{{{\left( {{c^z}\, - \,\,{c^{ - z}}} \right)}^2}}&1\end{array}} \right|$ $=$