InterviewSolution
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InterviewSolution
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The greatest value of n for which the determinant `Delta = |(1,1,1),(.^(n)C_(1),.^(n+3)C_(1),.^(n+6)C_(1)),(.^(n)C_(2),.^(n+3)C_(2),.^(n+6)C_(2))|` is divisible by `3^(n)`, isA. 7B. 5C. 3D. 1 |
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Answer» Correct Answer - C We have, `Delta = (1)/(2) |(1,1,1),(n,n +3,n +6),(n(n-1),(n+2) (n+3),(n+5) (n +6))|` `rArr Delta = (1)/(2) |(1,0,0),(n,3,6),(n(n-1),6n + 6,12n + 30)|` `rArr Delta = 9 |(1,0,0),(n,1,1),(n(n-1),2(n +1),2n +5)| = 27` Clearly, `Delta = 27` is divisible by 3, `3^(2) and 3^(3)`. Hence, n = 3 |
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