If $a, b$ and $c$ are real numbers, and
$\Delta=\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|=0$
Show that either $a+b+c=0$ or $a=b=c$.
If $a, b$ and $c$ are real numbers, and
$\Delta=\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|=0$
Show that either $a+b+c=0$ or $a=b=c$.
$\Delta=\left|\begin{array}{lll}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|=0$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|$
$=2(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|$
Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1},$ we have:
$\Delta=2(a+b+c)\left|\begin{array}{ccc}1 & 0 & 0 \\ c+a & b-c & b-a \\ a+b & c-a & c-b\end{array}\right|$
Expanding along $R_{1},$ we have:
$\Delta=2(a+b+c)(1)[(b-c)(c-b)-(b-a)(c-a)]$
$=2(a+b+c)\left[-b^{2}-c^{2}+2 b c-b c+b a+a c-a^{2}\right]$
$=2(a+b+c)\left[a b+b c+c a-a^{2}-b^{2}-c^{2}\right]$
It is given that $\Delta=0$ $(a+b+c)\left[a b+b c+c a-a^{2}-b^{2}-c^{2}\right]=0$
$\Rightarrow$ Either $a+b+c=0,$ or $a b+b c+c a-a^{2}-b^{2}-c^{2}=0$
Now, $a b+b c+c a-a^{2}-b^{2}-c^{2}=0$
$\Rightarrow-2 a b-2 b c-2 c a+2 a^{3}+2 b^{3}+2 c^{3}=0$
$\Rightarrow(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$
$\Rightarrow(a-b)=(b-c)^{2}=(c-a)^{2}=0 \quad\left[(a-b)^{2},(b-c)^{2},(c-a)^{2} \text { are non-negative }\right]$
$\Rightarrow(a-b)=(b-c)=(c-a)=0$
$\Rightarrow a=b=c$
Hence, if $\Delta=0,$ then either $a+b+c=0$ or $a=b=c$
Similar Questions
$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to
$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ 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 ${a_1},{a_2},{a_3},........,{a_n},......$ 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
By using properties of determinants, show that:
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$
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
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
Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$
Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$