Determinants of $3A, -A, A^2, A^{-1}$, where A is an $4\times 4$ matrix and $\det(A) = \frac{1}{3}$?

Question

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.

Answer 1

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 $\det(A) \neq 0$.

Answer 2

Hint

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: $$\det(\lambda I) = \lambda^n.$$

Answer 3

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

$\det(3A)=3^4\det(A)=3^3=27$

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

$\det(A^2)=\det(A)^2=\frac{1}{9}$

$\det(AA^{-1})=\det(I)=1$

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