# Question on determinants

• Jul 9th 2010, 05:37 PM
Doktor_Faustus
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.
• Jul 9th 2010, 06:12 PM
Prove It
You should know \$\displaystyle \det{(AB)} = \det{(A)}\det{(B)}\$.

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

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

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

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

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

\$\displaystyle \det{(cA)} = c^n\det{(A)}\$, which completes the proof.
• Jul 9th 2010, 07:58 PM
Doktor_Faustus
Quote:

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

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

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

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

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

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

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

Thank you. Much appreciated.
• Jul 10th 2010, 07:30 AM
HallsofIvy
Quote:

Originally Posted by Doktor_Faustus
Hi guys,