| , * 20 c a + b & a + 2b & a + 3b a + 2b & a + 3b & a + 4b a + 4… - Vidyadip

MCQ

$\left| {\,\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}\,} \right| = $

A
${a^2} + {b^2} + {c^2} - 3abc$
B
$3ab$
C
$3a + 5b$
✓
$0$

✓

Answer

Correct option: D.

$0$

d
(d) $\left| {\,\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}\,} \right|\, = \,\left| {\,\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\b&b&b\\{2b}&{2b}&{2b}\end{array}\,} \right|$ = 0

$\left\{ {{\rm{by }}\begin{array}{*{20}{c}}{{R_2} \to {R_2} - {R_1}}\\{{R_3} \to {R_3} - {R_2}}\end{array}} \right\}$

Trick: Putting $a = 1 = b$. The determinant will be $\left| {\,\begin{array}{*{20}{c}}2&3&4\\3&4&5\\5&6&7\end{array}\,} \right| = 0$. Obviously answer is $ (d)$

Note : Students remember while taking the values of $a,\,b,\,\,c,.......$ that for there values, the options $(a), (b), (c) $ and $(d) $ should not be identical.

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