Why is the determinant $1$ not $-2$

Question

I am trying to find the determinant of this matrix. Let $A$ be the matrix :

\begin{bmatrix}2&2&1\\1&0&5\\1&1&0\end{bmatrix}

Using row operations, I can change $R_3$ to ( $-2R_3$$+$$R_1$$\rightarrow$$R_3$). So, $B$ is the matrix :

\begin{bmatrix}2&2&1\\1&0&5\\0&0&1\end{bmatrix}

Then, using Laplace Expansion and expanding along the 2nd column,

$det(A) = (2)(-1)^3 det(\begin{bmatrix}1&5\\0&1\end{bmatrix})= (2)(-1)(1) = -2$

However, $det(A)=det(B)$ if a multiple of one row is added to another, but the determinant of $A$ is $1$. What am I doing wrong here?

Answer 1

Because of that $-2R_3$ that you wrote. When you do $-2R_3+R_1\to R_3$, part of you are doing is to multipliply the third row by $-2$, which changes the determinant of the matrix.

Answer 2

You're not adding a multiple of $R_3$ to $R_1$. You're replacing $R_3$ by a multiple of $R_3$ plus $R_1$.

I warn my students at the start that there's a place coming up later where it will be important to realize that (if $j\ne k$) $$R_j+cR_k\to R_j$$is a row operation, while $$R_j+cR_k\to R_k$$is not an official elementary row operation. This it that place. As long as $c\ne0$ both versions work fine for finding an echelon form, but the second changes the determinant, while the first does not.