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#1

$Old$ December 4th, 2008, 01:08 PM

eiktmywib $eiktmywib is offline$

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$Default$ LinAlg - Determinants/Adjoints?

If A is 3x3 and detA=2, then det(A^-1-(adjA)) = ?

I know that the answer is -(1/2)
and my teacher did it on the board, saying that adjA=(detA)(A^-1)

but I don't understand that part of it!
i'd really appreciate it if someone would explain it to me

thanks in advance

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#2

$Old$ December 4th, 2008, 02:30 PM

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Use these facts:

$\det(AB) = \det (A) \cdot \det (B)$
$\det(A^{-1}) = \frac{1}{\det (A)}$
$\det (cA) = c^n \det (A)$ where $c$ is a constant, $n$ is the size of the matrix
$\det \left( \text{adj} (A)\right) = \left[\det (A)\right]^{n-1}$ ..... Can be shown here

______________

You should know that: $A^{-1} = \frac{1}{\det (A)} \cdot \text{adj} (A)$

Multiply both sides by $\det (A)$ to get: $\det (A) \cdot A^{-1} = \text{adj} (A)$

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