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Suppose A and B are $5\times 5$ matrices with $\det(A) = -1/3$ and $\det(B) = 6$, find the determinant of $ 2AB$.

Solution:

$$= \det(2AB) $$ $$= 2^5 \det(A)\det(B) $$ $$= (32)(-1/3)(6)$$ $$= -64$$


I understand how all of this works, except for where $2^5$ comes from, can anyone explain how this happens?

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    $\begingroup$ You might want to check out Jamie Banks' answer to this question. $\endgroup$ – jkn May 30 '13 at 21:55
  • $\begingroup$ Thanks for that link, very helpful! $\endgroup$ – Matthew Brzezinski May 30 '13 at 21:57
  • $\begingroup$ I thought it's worth a read! $\endgroup$ – jkn May 30 '13 at 22:01
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When you multiply a matrix by $2$, that multiplies all the rows of the matrix by $2$. The determinant is linear in each row, and the matrix $AB$ has five rows, so it multiplies the resulting determinant by $2^5$.

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Hint: $\, 2AB = (2I)AB,\,$ so $\det(2AB) = \det(2I) \det(A) \det(B).\,$ $\,2I\,$ is diagonal with fives $\,2$'s.

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The determinant is a multilinear form so $$\det(\lambda_1 C_1,\ldots,\lambda_n C_n)=\lambda_1\times\cdots\times\lambda_n \det(C_1,\ldots,C_n)$$

in your case take $\lambda_i=2,\quad i=1,\ldots,5$.

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