Left| {,begin{array}{*{20}{c}}{a - 1}&a&{bc}{b - 1}&b&{ca}{c

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3 and 4 .Determinants and Matrices

easy

$\left| {\,\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}\,} \right| = $

$0$

$(a - b)(b - c)(c - a)$

${a^3} + {b^3} + {c^3} - 3abc$

None of these

Solution

(d) $\left| {\,\begin{array}{*{20}{c}}{a – 1}&a&{bc}\\{b – 1}&b&{ca}\\{c – 1}&c&{ab}\end{array}\,} \right| = \left| {\,\begin{array}{*{20}{c}}a&a&{bc}\\b&b&{ca}\\c&c&{ab}\end{array}\,} \right| – \left| {\,\begin{array}{*{20}{c}}1&a&{bc}\\1&b&{ca}\\1&c&{ab}\end{array}\,} \right|$

= $ – \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&1\\b&{{b^2}}&1\\c&{{c^2}}&1\end{array}\,} \right| = – \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&1\\{b – a}&{{b^2} – {a^2}}&0\\{c – a}&{{c^2} – {a^2}}&0\end{array}\,} \right|$

[By ${R_2} \to {R_2} – {R_1};\,{R_3} \to {R_3} – {R_1}$]

= $ – (a – b)\,(b – c)\,(c – a)$.

Standard 12

Mathematics

Find area of the triangle with vertices at the point given in each of the following: $(-2,-3),(3,2),(-1,-8)$

medium

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{{\omega ^2}}&\omega \\1&\omega &{{\omega ^2}}\end{array}\,} \right| = $

easy

If $a, b, c$ are sides of a scalene triangle, then the value of $\left| \begin{array}{*{20}{c}}
a&b&c\\
b&c&a\\
c&a&b
\end{array} \right|$ is

hard

(JEE MAIN-2013)

The value of a for which the system of equations ; $a^3x + (a +1)^3 y + (a + 2)^3 \, z = 0$ ,$ax + (a + 1) y + (a + 2)\, z = 0$ & $x + y + z = 0$ has a non-zero solution is :

normal

The values of $\theta, \lambda$ for which the following equations $\sin \theta x – cos\theta y + (\lambda +1)z = 0$; $\cos\theta x + \sin\theta\, y – \lambda z = 0$;$ \lambda x +(\lambda + 1)y + \cos\theta z = 0$ have non trivial solution, is