Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question in my notes go like this:

Let $S$ be some square matrix with a length of $s$. Show that, if we multiply all members of some row of $S$ or some column of $S$ by a value, say $a$, the determinant of the new matrix is $a|S|$. What is the determinant of $aS$?

I know that if I multiply some row by $a$, I can apply the variation of the identity matrix by substituting the $1$ in that particular row to $a$. Multiply the two and that row will absorb the new multiplier. But I don't know how to show for columns.

Also, what is the last part trying to ask? I really have no clue.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

$\textbf{Hint:}$ $aS$ means multiplying each row of $S$ by $a$.

share|improve this answer
Would the answer, then be $a^2|S|$ ? –  bryansis2010 Feb 17 '13 at 16:43
@bryansis2010 No, unless $S$ is a $2\times 2$ matrix. Suppose $S$ is a $3\times 3$ matrix. You're multiplying the first, second and third rows by $a$. Can you find the determinant of $aS$ in this conditions? –  Git Gud Feb 17 '13 at 16:44
How about $a^s|S|$ –  bryansis2010 Feb 17 '13 at 16:46
I typed small $s$ as the exponent of $a$, with $s$ representing the number of rows or columns in the matrix $S$. –  bryansis2010 Feb 17 '13 at 16:48
@bryansis2010 Ok, couldn't figure it out. Yes, that answer is correct. –  Git Gud Feb 17 '13 at 16:48

It depends on which properties you are allowed to use.

One way is to notice that $a S = a I S = \text{diag}(a,a,...,a) S$ and use the property $\det (AB) = \det A \det B$. Then $\det (aS) = \det \text{diag}(a,a,...,a) \det S = a^s \det S$.

share|improve this answer
It took me one night to understand this explanation...but now that I do, yes, I understand. This to me is sufficient. –  bryansis2010 Feb 18 '13 at 6:29
@GitGud's approach is equivalent but far more succinct. –  copper.hat Feb 18 '13 at 6:33

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.