Applied Grand Test 2

Question

What is the value of the given determinant ?

 

Comments

$\det(3(\frac{1}{4}A)^{-1}) = 3^3 \det((\frac{1}{4}A)^{-1}) = 3^3 \times \frac{1}{\det(\frac{1}{4}A)} = 3^3 \times \frac{1}{(\frac{1}{4})^3\det(A)} = 3^3 \times 4^3 \times \frac{1}{\det(A)} = 3^3 \times 4^3 \times \frac{1}{(-2)} = 27 \times 16\times -2 =-864$

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But det is always positive na? . Here it should be 864 na ?

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Determinants are always positive. Where have you read this?

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@ankitgupta.1729 Sir, I understood your approach towards the problem. I have one doubt that according to property : det(kA) = k^3 . |A| , where A is 3x3 matrix, which is the one you have used I believe. Sir, if det( 3 x det( 1/4 A )^(−1) ) and det(A) = -2 , then answer would be -96, right?

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determinant can not be always positive,we have to have some notation to show shrinking and negative represents that.

someone correct me if im wrong

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Yes correct @anshuman66.

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@ankit3009, determinant is a number associated with square matrices. Since, $3 \times \det( \frac{1}{4} A )^{−1}$ is a value, not a square matrix, so you can’t find determinant of it.

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Understood Sir. Thank you :)

Answer 1

det(3(1/4A)-1) = det(3(1/4)-1A-1) = det(3.4A-1) = det(12 A-1) = 123 * |A-1| = 123 * 1/|A| = 123 * 1/(-2) =-864