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:
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
Applying $C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-C_{1},$ we have:
$\Delta=(a+b+c)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c)\end{array}\right|$
$=(a+b+c)^{3}\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 b & -1 & 0 \\ 2 c & 0 & -1\end{array}\right|$
Expanding along $C_{3},$ we have:
$\Delta=(a+b+c)^{3}(-1)(-1)=(a+b+c)^{3}$
Hence, the given result is proved.
Similar Questions
If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.
If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.
$\left| {\,\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}\,} \right| = $
Prove that
$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$
Prove that
$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$
If $\left| {\begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
{{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\
{{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}}
\end{array}} \right|$ $ = \,k\lambda \,\,\left| {{\mkern 1mu} {\mkern 1mu} \begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
a&b&c \\
1&1&1
\end{array}} \right|,\lambda \, \ne \,0$ then $k$ is equal to
If $\left| {\begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
{{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\
{{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}}
\end{array}} \right|$ $ = \,k\lambda \,\,\left| {{\mkern 1mu} {\mkern 1mu} \begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
a&b&c \\
1&1&1
\end{array}} \right|,\lambda \, \ne \,0$ then $k$ is equal to
- [JEE MAIN 2014]
By using properties of determinants, show that:
$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$