# Determinant of a $4\times4$ invertible matrix

Let $A$ be a $4$ by $4$ invertible matrix, such that $\det(3A)=3\det(A^4)$. Then $\det(A)=3$.

Would somebody please give me some clues on this? Thanks

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It's only true if the matrix is a real matrix, not if it might be complex. –  Thomas Andrews Nov 19 '12 at 4:00

Hints: for a matrix $\,n\times n\,$:

$$(1)\;\;\;\;\;\;\det(kA)=k^n\det A$$

$$(2)\;\;\;\;\;\;\;\det A^k=(\det A)^k$$

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Thank you for your hint –  mathwannabe Nov 20 '12 at 16:42

Recall the fact that $$\det(AB) = \det(A) \det(B)\tag{1}$$ and $$\det(k A) = k^n \det(A)\tag{2}$$ where $k \in \mathbb{C}$ and $A \in \mathbb{C}^{n \times n}$.

From $(1)$, we get that $\det(A^4) = \det(A)^4$ and using $(2)$, we get that $\det(3A) = 3^4 \det(A)$. Hence, we get that $$3^4 \det(A) = 3 \det(A)^4 \implies \det(A)^3 = 3^3 \implies \det(A) = 3,3 \omega \text{ or }3w^2.$$ Assuming $A$ has only real entries, the determinant also has to be real and hence $$\det(A) = 3$$

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thank you Michael and Marvis –  mathwannabe Nov 20 '12 at 16:42

$3det(A^4) = 3(detA)^4$

$= det(3A) = 3^4det(A)$

$(detA)^4 - 27 det(A) = 0$

Since $A$ is invertible we can divide by $det(A)$

$(detA)^3 =27$

$detA = 3$

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