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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 $det(D)$?

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2 Answers 2

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Since the matrix $C$ is non-singular, its row reduced echelon form is just $I$.

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  • $\begingroup$ Ohh. I overthought that way to much. Thanks. $\endgroup$ Dec 16, 2018 at 9:56
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$\det(C)=\det(\frac12A^{-1}B^3)=\frac1{2^3}\det(A^{-1}B^3)=\frac18\cdot\frac1{|A|}\cdot|B|^3=1/3$

$C$ is invertible. The row-reduced echelon form of $C$ is the identity matrix. So $D=I_3,\det(D)=1$.

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