1. ## Question on determinants

Hi guys,

Let A be an nxn matrix and c a non zero scalar. If matrix B is obtained from A by multiplying the elements of a row by c then det (B) = cdet (A).

Show that det (cA) = c^n det (A).

Would appreciate any help whatsoever. I've looked through my book on linear algebra ( by Howard Anton...pretty good book too ), but I can't find anything besides it being stated without proof.

Cheers.

2. You should know $\det{(AB)} = \det{(A)}\det{(B)}$.

In your case, you are wanting to find $\det{(cA)}$.

Note that $cA = cI_nA$, which is now the product of two matrices, $cI_n$ and $A$.

So $\det{(cA)} = \det{(cI_nA)}$

$= \det{(cI_n)}\det{(A)}$.

Since $cI_n$ is a diagonal matrix, its determinant is the product of the diagonal entries. In this case they are all $c$, so their product is $c^n$. Thus $\det{(cI_n)} = c^n$, and finally we have...

$\det{(cA)} = c^n\det{(A)}$, which completes the proof.

3. Originally Posted by Prove It
You should know $\det{(AB)} = \det{(A)}\det{(B)}$.

In your case, you are wanting to find $\det{(cA)}$.

Note that $cA = cI_nA$, which is now the product of two matrices, $cI_n$ and $A$.

So $\det{(cA)} = \det{(cI_nA)}$

$= \det{(cI_n)}\det{(A)}$.

Since $cI_n$ is a diagonal matrix, its determinant is the product of the diagonal entries. In this case they are all $c$, so their product is $c^n$. Thus $\det{(cI_n)} = c^n$, and finally we have...

$\det{(cA)} = c^n\det{(A)}$, which completes the proof.
Thank you. Much appreciated.

4. Originally Posted by Doktor_Faustus
Hi guys,

If you are given this then you can prove $det(cA)= c^n(A)$ by repeating it n times. Let B1 be the matrix obtained by multiplying the elements of row 1 by c: det(B1)= cdet(A). Let B2 be the matrix obtained by multiplying every element of row 2 of B1 by c: det(B2)= cdet(B1)= c^2 det(A). Repeat as needed!