By using properties of determinants, show that:
$\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$
$\Delta=\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+b R_{3}$ and $R_{2} \rightarrow R_{2}-a R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}1+a^{2}+b^{2} & 0 & -b\left(1+a^{2}+b^{2}\right) \\ 0 & 1+a^{2}+b^{2} & a\left(1+a^{2}+b^{2}\right) \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$
$=\left(1+a^{2}+b^{2}\right)\left|\begin{array}{ccc}1 & 0 & -b \\ 0 & 1 & a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$
Expanding along $R_{1},$ we have:
$\Delta=\left(1+a^{2}+b^{2}\right)^{2}\left[(1)\left|\begin{array}{cc}1 & a \\ -2 a & 1-a^{2}-b^{2}\end{array}\right|-b\left|\begin{array}{cc}0 & 1 \\ 2 b & -2 a\end{array}\right|\right]$
$=\left(1+a^{2}+b^{2}\right)^{2}\left[1-a^{2}-b^{2}+2 a^{2}-b(-2 b)\right]$
$=\left(1+a^{2}+b^{2}\right)^{2}\left(1+a^{2}+b^{2}\right)$
$=\left(1+a^{2}+b^{2}\right)^{3}$
Similar Questions
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:
$\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^{3} & b^{3} & c^{3}
\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ then $K = $
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ then $K = $
Evaluate $\Delta=\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$
Evaluate $\Delta=\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$
If $a, b, c,$ are non zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac}\\
{ab}&{{c^2} + {a^2}}&{bc}\\
{ac}&{bc}&{{a^2} + {b^2}}
\end{array}} \right| = k{a^2}{b^2}{c^2},$ then $k$ is equal to
If $a, b, c,$ are non zero complex numbers satisfying $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac}\\
{ab}&{{c^2} + {a^2}}&{bc}\\
{ac}&{bc}&{{a^2} + {b^2}}
\end{array}} \right| = k{a^2}{b^2}{c^2},$ then $k$ is equal to
- [AIEEE 2012]