1. Determinants Question

Hi I ran across this problem and could not figure it out. Could someone please give me an explanation? I feel like I have something close but no cigar.
Compute the following determinants by using row and/or column reduction

$\begin{bmatrix}3&-6&15\\-4&-5&-2\\2&7&3\end{bmatrix}$

Then, I applied the following row operations.
R2->R2+2R3
R3->3R3-2R1
I also factored out the 3 from the top row.

3 $\begin{bmatrix}1&-2&5\\0&9&4\\0&33&-21\end{bmatrix}$

Then, R3->R3-33/9R2

3 $\begin{bmatrix}1&-2&5\\0&9&4\\0&0&\frac{-107}{3}\end{bmatrix}$

Det=3*1*9*-107/3=-963

However, the answer should be -321.

2. Originally Posted by egshih
Hi I ran across this problem and could not figure it out. Could someone please give me an explanation? I feel like I have something close but no cigar.
Compute the following determinants by using row and/or column reduction

$\begin{vmatrix}3&-6&15\\-4&-5&-2\\2&7&3\end{vmatrix}$

Then, I applied the following row operations.
R2->R2+2R3
R3->3R3-2R1 Here's where it goes wrong. When you multiply R3 by 3 you multiply the determinant by 3. That's why you end up with 3 times the correct answer.
I also factored out the 3 from the top row.

3 $\begin{vmatrix}1&-2&5\\0&9&4\\0&33&-21\end{vmatrix}$

Then, R3->R3-33/9R2

3 $\begin{vmatrix}1&-2&5\\0&9&4\\0&0&\frac{-107}{3}\end{vmatrix}$

Det=3*1*9*-107/3=-963

However, the answer should be -321.
Also, in LaTeX you should use vmatrix rather than bmatrix to get a determinant.

3. $\left| \begin{matrix}
\phantom{-}3 & -6 & \phantom{-}15 \\
-4 & -5 & -2 \\
\phantom{-}2 & \phantom{-}7 & \phantom{-}3
\end{matrix} \right|=3\left| \begin{matrix}
\phantom{-}1 & -2 & \phantom{-}5 \\
-4 & -5 & -2 \\
\phantom{-}2 & \phantom{-}7 & \phantom{-}3
\end{matrix} \right|,$
now the determinant equals $3\left| \begin{matrix}
1 & -2 & \phantom{-}5 \\
0 & \phantom{-}9 & \phantom{-}4 \\
0 & \phantom{-}11 & -7
\end{matrix} \right|=3\left| \begin{matrix}
9 & \phantom{-}4 \\
11 & -7
\end{matrix} \right|=-321.$

Don't waste your time producing more zeros.