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

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 ( $$-2R_3+R_1\rightarrowR_3$$). 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?

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.
• But my textbook says adding a multiple of one row to another doesn't change the determinant, I'm just adding a multiple of $R_3$ to $R_1$. I apologize if it's an obvious question. Oct 15, 2019 at 11:38
• @MohamedTlili If you had done $-2R_3+R_1\to R_1$, then the determinant would've been unchanged. That would be "adding a multiple of $R_3$ to $R_1$". However, $-2R_3+R_1\to R_3$ doesn't just add a multiple of $R_1$ to $R_3$, it also multiplies $R_3$ by $-2$ first. Oct 15, 2019 at 11:38
• @MohamedTlili First you replaced $R_3$ with $-2R_3$, multiplying the determinant by $-2$. Then you added $R_1$ to the third row, which had no effect on the determinant. Writing down the intermediate matrix may help you follow this argument.
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.