How do I find the determinants of the matrices $3A, -A, A^2$, and $A^{-1}$, where $A$ is a $4\times4$ matrix and $\det(A) = \frac{1}{3}$?

I know that to find the determinant we can use elimination to bring the matrix to an upper (or lower) triangular and then multiply the pivots.

Or we can use cofactors, which is like a recursive definition of determinant, where the base case is when we have a $2\times 2$ matrix.

This is very nice, but I am no seeing how can I use this information to calculate the determinants above.

  • $\begingroup$ The determinant is the product of the eigenvalues' moduluses (absolute values). Multiplication with a scalar multiplies every eigenvalue by that scalar. Power function also carry over to the eigenvalues so that each eigenvalue is raised to the same power. These facts together should allow you to solve the problem fast. $\endgroup$ – mathreadler Apr 25 '15 at 13:02

The calculation of the determinants does involve basic properties of the determinant function:

  • Multilinearity: Let $\alpha \in \mathbb{R}$ and $A$ an $n \times n$ matrix. Then

$$\det( \alpha A) = \alpha^n\det(A).$$

  • Multiplication of determinants: For two $n \times n$ matrices $A,B$

$$\det(A B ) = \det(A) \cdot \det(B).$$

Let now $A$ be an invertible matrix, i.e. $I = A \cdot A^{-1}$. This gives

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

Note that a matrix $A$ is invertible if and only if $\det(A) \neq 0$.



  1. For any $n \times n$ matrices $A, B$, we have $$\det(AB) = \det(A) \det(B).$$
  2. Any scalar matrix $\lambda I$ is diagonal, so its determinant is the product of its diagonal entries: $$\det(\lambda I) = \lambda^n.$$
  • 2
    $\begingroup$ And $cA\equiv (cI)A, c \in \mathbb{F}$ $\endgroup$ – Archaick Apr 25 '15 at 12:53
  • 1
    $\begingroup$ For the (2), you can also use the fact that it's a multilinear form. $\endgroup$ – Jean-Claude Arbaut Apr 25 '15 at 13:20
  • $\begingroup$ @Jean-ClaudeArbaut You're right, of course, though in my experience students don't even know the term "multilinear form" when they're first learning about properties of the determinant. But of course, they may well know the fact that the row operation of multiplying a row of a square matrix by $\lambda$ scales its determinant by a factor of $\lambda$, and this is essentially the property you mention. $\endgroup$ – Travis Apr 25 '15 at 13:26

$\det(cA)=c^nA$ where $A$ is an $n\times n$ matrix.





That is, $\det(A^{-1})=\frac{1}{\det(A)}=3$

  • $\begingroup$ $\det(A^2)=\det(AA)=\det(A)\det(A)=\det(A)^2$ By induction, this follows for all integers $n$, as in $\det(A^n)=\det(A)^n$ $\endgroup$ – Hasan Saad Apr 25 '15 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.