Show that if $A$ is invertible, then $\det(A^{−1} )=\det(A)^{−1}$ . Deduce a formula for the determinant of $4A^{−1}$ , when $A$ is an $n \times n$ -matrix.

I know that when a matrix is invertible it's determinant is equal to zero. Seems to me like the final answer would just be zero as well, but that seems too easy.

  • 1
    $\begingroup$ Actually, it's just the opposite. When a matrix is invertible, its determinant is NONZERO. Also, are you supposed to show $\det(A^{-1}) = \det(A)^{-1}$? $\endgroup$ – Alex Wertheim Nov 7 '14 at 22:04

Use the fact that determinant of product is product of determinants.

$\det(AA^{-1}) = \det(A)\det(A^{-1}) = 1 \implies \det(A^{-1}) = \det(A)^{-1}$.

For the second part...

$\det(cA) = c^n\det(A)$ for n x n matrix $A$.

So you can use that idea with the previous derivation to find a formula for $\det(4A^{-1})$.


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