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Maths-

General

Easy

Question

If $a not equal to 6 comma b comma c$ satisfy $open vertical bar table row a cell 2 b end cell cell 2 c end cell row 3 b c row 4 a b end table close vertical bar equals 0 comma$ then $a b c equals$

$a + b + c$
0
$b^{3}$
$a b + b c$

hint Hint:

Determinants is a function which accepts nxn matrix as an input and output is a real or complex number .
$E x p a n s i o n space o f space a space d e t e r m i n a n t space open vertical bar table row a b c row d e f row g h i end table close vertical bar space w i l l space b e space e q u a l space t o space open square brackets a open parentheses e i minus h f close parentheses space minus space b open parentheses d i minus g f close parentheses space plus space c open parentheses d h minus e g close parentheses close square brackets$

The correct answer is: $b to the power of 3 end exponent$

$open vertical bar table row a cell 2 b end cell cell 2 c end cell row 3 b c row 4 a b end table close vertical bar space equals space 0 2 R subscript 2 space rightwards arrow space R subscript 1 open vertical bar table row cell a minus 6 end cell 0 0 row 3 b c row 4 a b end table close vertical bar space equals space 0 open parentheses a minus 6 close parentheses open parentheses b squared space minus space 4 a c close parentheses space equals space 0 H e r e comma a minus 6 equals 0 space space o r space b squared space minus space 4 a c space equals space 0 a space equals space 6 space space space space o r space space space space 4 a c space equals space b squared B u t space i t space i s space g i v e n space t h a t space a space not equal to space 6 comma space h e n c e 4 a c space equals space b squared M u l t i p l y i n g space b o t h space s i d e s space b y space b comma therefore 4 a b c space equals space b cubed$

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