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My homework question says to find the determinant given that $A$ is a $4\times 4$ matrix and its determinant is $-3$.

The question is: compute $\det(3A^{-1}).$

I got an answer of $-\frac 1{243}$ and I'm not sure its right.

My steps were this

$$\det(3A^{-1}) = 3^4 \cdot (-3)\det(A) = -\frac{3}{3^4}.$$

Can anyone confirm that I am doing this properly? My answer seems a little wonky to me.

Thanks in advance

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$\det(A^{-1}) \neq -\det(A)$. That's your mistake. – Arturo Magidin Oct 31 '11 at 21:22
up vote 6 down vote accepted

No. Your steps are wrong. Here are the right steps: ${\rm det} (3A^{-1})=3^4{\rm det} (A^{-1})=81 ({\rm det} A)^{-1}= 81 (-1/3)=-27$.

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thanks, just to confirm... If the question was det(2A^T) then the steps would be... 2^4(detA)^T 16(-3) = -48 ? – Cheesegraterr Oct 31 '11 at 21:23
What is T in ^T? Do you mean transpose of A? In this case yes the steps are true. – user17090 Oct 31 '11 at 21:27
sorry, ^T indicates Transpose. Thank you very much Ali – Cheesegraterr Oct 31 '11 at 21:40

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