# Cramer's rule row reducing?

An example would be:

Let this bas the $Ax = 0$ matrix

$$A= \begin{bmatrix} 3 & 2 & -1 \\ 1 & -4 & 1 \\ 5 & 2 & 0 \end{bmatrix}$$

If the b values ares

$$b=\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$$

Find the unknowns $Ax = b$.

The determinant of A i got from reducing is:

$$A= \begin{bmatrix} 3 & 2 & -1 \\ 1 & -4 & 1 \\ 5 & 2 & 0 \end{bmatrix} \leftrightarrow \begin{bmatrix} 4 & -2 & 0 \\ 1 & -4 & 1 \\ 5 & 2 & 0 \end{bmatrix} = -18 \text{ Let the reduced matrix be H}$$

My question is If I were to use H with cramers rule will I get the right answer?

Like the determinant of this matrix is $-4$. The answer in my textbook is zero.

$$H= \begin{bmatrix} 1 & -2 & 0 \\ -2 & -4 & 1 \\ 1 & 2 & 0 \end{bmatrix} = -4$$

For several examples this worked so I'm wondering if you shouldn't ever take the reduced matrix when using cramers rule or did I make a mistake?

Just row reducing $A$ in isolation will lead to incorrect solutions. For example, if we keep carrying out row reductions on $A$ we eventually get to $$A \leadsto I.$$ Clearly, solutions to $A \bf{x} = \bf{b}$ will be different than solutions solutions to $I\bf{x} = \bf{b}$, no matter what sort of Cramer's Rule trickery we employ.
If you want to make computation easier, you can do row reduction on the augmented matrix $[A ~|~\bf{b}] \leadsto [H ~|~ \bf{\tilde{b}} ]$. The key is that we have to do the same row operations to $\bf{b}$ as well! This is effectively the point of row operations. If we get from $[A ~|~\bf{b}] \leadsto [H ~|~ \bf{\tilde{b}} ]$ by doing row operations, then $A \bf{x} = \bf{b}$ has the some solution vector $\bf{x}$ as $H \bf{x} = \bf{\tilde b}$. We can solve this new system however we like, whether it be Cramer's Rule, row reduction, or even guess and check.