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


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


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?


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Oct 15, 2019 at 11:38
  • 2
    $\begingroup$ @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. $\endgroup$
    – Arthur
    Oct 15, 2019 at 11:38
  • 2
    $\begingroup$ @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. $\endgroup$
    – J.G.
    Oct 15, 2019 at 11:39

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.


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