Easy matrix/determinants question

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SUMMARY

The discussion centers on the calculation of the determinant of the matrix C, defined as C = ((2A)^-1)B, where A and B are 3x3 matrices with known determinants det A = x and det B = y. The correct approach involves the property of determinants that states det(αA) = α^n det(A) for an n x n matrix, leading to det(2A) = 2^3 * x = 8x. Consequently, det C is calculated as det C = det(B) / det(2A) = y / (8x), confirming that the correct answer is option a) y/8x.

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Homework Statement



Suppose A and B are 3 x 3 matrices and det A = x ≠ 0 while det B = y. Let
C be the matrix ((2A)^-1 )B <-- (2A) inverse x B

then det C is:

Homework Equations

The Attempt at a Solution



det(2A) = 2x, so det 2A inverse = 1/(2x)
det C = y/(2x)...which is not one of the solutions

a) y/8x b) 2xy c) -2y/x d) 2y/x e) 8y/x

what am I missing here?
 
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You are very close; however,

\det(\alpha A)=\alpha^{n} \det(A)

where n is the order of the matrix A, in this case 3. To understand why this happens, think of the determinant of the identity and multiply it by a scalar.
 

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