Suppose $A$ is a 4x4 matrix such that $\det(A)=\frac{1}{64}$

Question

Suppose A is a 4x4 matrix such that $\det(A)=\frac{1}{64}$ then $\det(4A^{-1})^T$

1. I created a 2x2 matrix $B$ and transposed it both had the same determinant
2. I then found $\det(B)$ and $\det(B^{-1})$ the results were inverses of each other
3. I then tried found the $\det(2B)$

but I cannot see what is the relationship between the determinant of a matrix and the determinant of that matrix multiplied by a constant?

Answer 1

Hint: Scaling a single row (or column) by a constant $k$ results in the determinant being scaled by $k$. So what happens if the entire matrix is scaled by $k$?

Answer 2

Use those properties:

For $A\in\mathcal M_n(\Bbb C)$ we have

• $\det(kA)=k^n\det(A)$
• If $A$ is invertible then $\det(A^{-1})=\frac{1}{\det (A)}$
• $\det(A^T)=\det(A)$

Answer 3

As a real-valued function on $n \times n$ matrices, the determinant is a polynomial of degree $n$ in the entries of a matrix, and the polynomial is homogeneous (all of its terms have the same degree), so $\det(\lambda A) = \lambda^n \det A$.

For example, for $n = 2$,

$\det\left(\begin{array}{cc}a & b \\ c & d \end{array} \right) = ad - bc$,

and

$\det \left[ \lambda \left(\begin{array}{cc}a & b \\ c & d \end{array} \right) \right] = \det\left(\begin{array}{cc}\lambda a & \lambda b \\ \lambda c & \lambda d \end{array} \right) = (\lambda a)(\lambda d) - (\lambda b)(\lambda c) = \lambda^2 (ad - bc)$.