Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In my textbook it says that if you multiply a row in a matrix $A$ by a nonzero constant $c$ to obtain $B$, then $\det{B}=c\det{A}$.

Later on it says that if you obtain $B = cA$ by adding $c$ times the $k^{\text{th}}$ row of $A$ to the $j^{\text{th}}$ row, $\det{B}=\det{A}$.

Isn't this a contradiction though? Is not adding $c$ times the $k^{\text{th}}$ row of $A$ to the $j^{\text{th}}$ row equivalent to multiplying the $k^{\text{th}}$ row by $c$, which increases the determinant by a factor of $c$, and then adding the row down?

In other words, is (I) the same as the (II) with the $2$ steps combined?
I. $cR_k + R_j \rightarrow R_j$. $1$ step in total.
II. First, do $cR_k \rightarrow R_k$. Second, do $cR_k + R_j \rightarrow R_j$. $2$ steps in total.

share|cite|improve this question
+1 for questioning maxims and thinking independently :) – gt6989b Mar 21 '13 at 18:02
@Eric No, of course not. The $k$-th row of matrix you get by doing the operations in your last paragraph is $c$ times what you started with – so you need to divide that row through by $c$ in order to get the matrix in your second paragraph. – Zhen Lin Mar 21 '13 at 18:08
up vote 6 down vote accepted

They are not equivalent operations, if I add $c$ times the $2^\text{nd}$ row of $$\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$ to the $1^\text{st}$ I get $$\begin{bmatrix}1 & c \\ 0 & 1\end{bmatrix}.$$ On the other hand if I just multiply the $2^\text{nd}$ row by $c$ I get $$\begin{bmatrix}1 & 0 \\ 0 & c\end{bmatrix}$$ and then adding it to the $1^\text{st}$ gives $$\begin{bmatrix}1 & c \\ 0 & c\end{bmatrix}.$$

share|cite|improve this answer

the difference is that in the latter case the $k^{th}$ row would also be altered.

what would give the same result, would be to multiply the $k^{th}$ row by $c$, then add it to the $j^{th}$ row, and then divide the $k^{th}$ row by $c$...

but that would lead to $\det A=\frac{1}{c}.c.\det A$...
no contradiction here...

share|cite|improve this answer

This would only be a contradiction if you manage (through adding linear combinations or rows to other rows as above) to scale one of the rows, without altering others.

The only cases I am aware of where you can do that (e.g. repeating rows upto a constance factor) feature a non-invertible matrix, hence have 0 determinant, so linear scale would not matter.

share|cite|improve this answer

The point is that the determinant of matrix $A$ is not altered if you add $c$ times row $k$ to row $j$ for $j$ different from $k$.

The reason is that if $j \ne k$, this operation is accomplished by multiplying $A$ on the left by $I + c e_{jk}$, so $B = (I + c e_{jk}) A$. Now $I + c e_{jk}$ has determinant $1$ if $j \ne k$, but determinant $c+1$ if $j = k$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.