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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.

bryan.blackbee
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2 Answers2

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

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

copper.hat
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