The Gateway to Computer Science Excellence
+15 votes
1.7k views

Consider the following determinant $\Delta = \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$

Which of the following is a factor of $\Delta$?

  1. $a+b$
  2. $a-b$
  3. $a+b+c$
  4. $abc$
in Linear Algebra by Veteran (52.1k points)
edited by | 1.7k views
+3
Take a=1,b=2,c=3

Det will come as 2

Only b will be factor(1-2=-1)
0
#rahul didnt get ur approach ??
+1
take some random example with a b and c values.

Solve determinant

use options to see what can be factor
+5

@rahul sharma 5 

just small addition to your approach , whenever you take numbers to solve question vy hit and trial , first check if all options yield different results so that you don't need to do backtrack with some other set of numbers. here for 1,2,3, both c and d gives 6 , better to select 2,3,4 for this question

4 Answers

+20 votes
Best answer
$\begin{vmatrix} 1 &a &bc \\ 1& b& ca\\ 1& c& ab \end{vmatrix}$

$\begin{vmatrix} 1 &a &bc \\ 0& b-a& ca-bc\\ 0& c-a& ab-bc \end{vmatrix} \ \ \ \ \ \ \ \ R_{2}\rightarrow R_{2}- R_{1} \ \ , \ R_{3}\rightarrow R_{3}- R_{1}$

$(b-a)( ab-bc) - (c-a)(ca-bc)$

$-(a-b) \ \ b \ \ ( a-c) + (a-c)\ \ c \ \ (a-b)$

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

Option$(B)a-b$ is the correct choice.
by Boss (41k points)
edited by
0
@leen how (c-b) in last line

it should be (b+c) na?
+3

 A_i_$_h my name is leen not leena. i am a boy not a girl.

0
$-(a-b) \ \ b \ \ ( a-c) + (a-c)\ \ c \ \ (a-b)$

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

$(a-b)( a-c) (c-b)$
0
oops sorry:(  dint notice that properly
0

@leen what is wrng in  Bhagirathi's approach

+1
We can't multiply two rows or column because after multiplying the value of determinant will change.we can multiply some constant to rows or column to solve the determinant and also add and subtract two column.
+13 votes
Answer is B

R2->R2 - R1

R3 -> R3 - R2

you will gt det = (a-b)*(a-c)*(b+c)

in matrix operations, you cannot multiply rows or columns. That will not yield the same matrix. So abc is not correct
by Active (1.2k points)
0
cant we multiply R1 by a

R2 by b and R3 by c. This will give abc as a factor.
+4 votes

Answer : (b)

by Active (2.4k points)
–5 votes

C3<-C2*C3

now take common abc from C3

now we can see C1 and C2 is equal so abc is a factor

by Boss (14.4k points)
Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,650 questions
56,208 answers
194,071 comments
95,107 users