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2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
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Comments
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......
@Aditya Parson
–
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...
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
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......
66 ways.
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how, u came to it...?
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Big solution, too lousy to write it here. Is there any other way to send you my solution.?
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if you don't feel disturbed to send it to myLog in to reply
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Please email the solution of this question atLog in to reply
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this is a question which is asked on brilliant