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algebra

How many 3 by 3 determinants can be formed whose value is always zero , using two non zero distinct elements say a and b ??? Please help !!!

Note by Sachin Arora
4 years ago

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66 ways. · 4 years ago

how, u came to it...? · 4 years ago

Big solution, too lousy to write it here. Is there any other way to send you my solution.? · 4 years ago

if you don't feel disturbed to send it to my email...metronetizen@gmail.com · 4 years ago

Sure, I will work it on paper. Please wait. · 4 years ago

o.k thanx...waiting for your email... · 4 years ago

Sent. · 4 years ago

got it....i really misunderstood the problem ....i thought that there would be only once the number " a" and "b" would be used and as the questioner said that the value would be always zero i thought that the rest terms would be zero....but you showed me my mistake.....thanx for sending... · 4 years ago

There is no mistake in it right? · 4 years ago

so far i understand your logic you are right ....great thinking .... (probably) no mistake in technique .....awesome.... · 4 years ago

Please email the solution of this question at aryanarora02@gmail.com · 4 years ago

But the question is asked on brilliant. · 4 years ago

This question is already asked by me only .No body relplied . I posted again . · 4 years ago

are you sure it is not a question asked on brilliant? I am not able to understand what you say. · 4 years ago

I am vry sure , this question is not asked on brilliant · 4 years ago

Sent. · 4 years ago

You wanted to know how many determinants can be formed....see any two determinant can be said different whether its orientation of its terms are different....you are using only 2 non zero element ,you assumed "a" and "b"...now watch that in any place out of the 9 places of a 3 by 3 determinant you put them them and as a result you would get the result zero for every representation......so there are 9 places (3 rows 3 columns ) you are choosing 2 different places out of them .....you could do this in 9C2 ways....so there should (i'm not using "would" as i am not sure for my own logic and apprehension ) be only 36 ways to form different determinant....let's see if any one backs me up for this procedure to be right or wrong...... · 4 years ago