# Determinant properties with row reduction

I have the following question here.

Let $$A$$ and $$B$$ be $$3 × 3$$ matrices with $$det(A) = 3$$ and $$det(B) = 2$$. Let $$C = \frac{1}{2}A^{-1}B^3$$ and let $$D$$ be the reduced row echelon form of $$C$$. Then:

$$(a)$$ $$det(C)=\frac{4}{3}$$, $$det(D)=1$$

$$(b)$$ $$det(C)=\frac{1}{3}$$, $$det(D)=1$$

$$(c)$$ $$det(C)=\frac{4}{3}$$, $$det(D)=\frac{4}{3}$$

$$(d)$$ $$det(C)=\frac{1}{3}$$, $$det(D)=3$$

$$(e)$$ $$det(C)=\frac{1}{3}$$, $$det(D)=\frac{1}{3}$$

The answer is supposed to be $$b$$. I know $$det(C)=\frac{1}{3}$$ just because of determinant properties. That was easy. I'm not 100% sure how the $$RREF$$ of $$D$$ comes into play here. I know that elementary row operations affect the determinant but HOW does that affects the determinant here.

Can someone provide any guidance as to how I would calculate $$det(D)$$?

Since the matrix $$C$$ is non-singular, its row reduced echelon form is just $$I$$.
$$\det(C)=\det(\frac12A^{-1}B^3)=\frac1{2^3}\det(A^{-1}B^3)=\frac18\cdot\frac1{|A|}\cdot|B|^3=1/3$$
$$C$$ is invertible. The row-reduced echelon form of $$C$$ is the identity matrix. So $$D=I_3,\det(D)=1$$.