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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.

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2 Answers 2

up vote 3 down vote accepted

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

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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 $\det (aS) = \det \text{diag}(a,a,...,a) \det S = a^s \det S$.

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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

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