# Determinant of this 5x5 matrix.

• Dec 4th 2010, 05:41 PM
s3a
Determinant of this 5x5 matrix.
I tried this probably like 20 times (not exaggerating) and I keep making a mistake. I usually get -30 but the answer is 30. I even checked with software and it's 30. So, could someone please point out what I'm doing wrong?

(My work is attached).

Any help would be greatly appreciated!
• Dec 4th 2010, 06:02 PM
Ackbeet
Two errors:

Third line, second matrix: should be a -4 multiplying the 2 x 2 determinant, because the sign associated with the expansion corresponding to -1 has itself a -1, thus keeping whatever sign you had before.

Third line, first 2 x 2 determinant: (0)(-1) - (-3)(-3) = 0 - 9 = -9. You had a +9.

The two errors were sign errors corresponding to the two different pieces, thus totally explaining the opposite sign.
• Dec 4th 2010, 06:06 PM
harish21
when you are taking the det of the second 3x3 matrix in you second line, you should have:

$\displaystyle -2(2)(-(-1))\left(\begin{array}{cc}2&0\\1&3\end{array}\righ t)=-4(6-0)=-24$

and $\displaystyle \left|\begin{array}{cc}0&-3\\-3&1\end{array}\right| = 0(1)-(-3)(-3) = -9$
• Dec 5th 2010, 11:03 AM
s3a
I got it wrong again but I tried to see why I was wrong and I cannot see how the -1 gets an additional negative like: -(-1). I tried to explain myself on the new attached paper. If more explanation is needed, please tell me.
• Dec 5th 2010, 01:03 PM
tonio
Quote:

Originally Posted by s3a
I got it wrong again but I tried to see why I was wrong and I cannot see how the -1 gets an additional negative like: -(-1). I tried to explain myself on the new attached paper. If more explanation is needed, please tell me.

Develop the determinant $\displaystyle \left|\begin{array}{rrrrr}2&0&0&2&0\\2&0&3&0&0\\0& 2&0&0&-3\\0&0&0&-3&-1\\0&1&3&0&0\end{array}\right|$ as follows:

first substract row 1 from row 2 and develop then by the first column:

$\displaystyle \left|\begin{array}{rrrrr}2&0&0&2&0\\0&0&3&-2&0\\0&2&0&0&-3\\0&0&0&-3&-1\\0&1&3&0&0\end{array}\right|=$ $\displaystyle 2\cdot\left|\begin{array}{rrrr}0&3&-2&0\\2&0&0&-3\\0&0&-3&-1\\1&3&0&0\end{array}\right|$ . Now substract twice the 4th row from the 2nd one

and, again, develop by the first column:

$\displaystyle =2\cdot\left|\begin{array}{rrrr}0&3&-2&0\\0&-6&0&-3\\0&0&-3&-1\\1&3&0&0\end{array}\right|$ $\displaystyle =2\cdot(-1)\left|\begin{array}{rrrr}3&-2&0\\-6&0&-3\\0&-3&-1\end{array}\right|$ $\displaystyle =-2\cdot(-27+12)=30$

Tonio
• Dec 5th 2010, 05:42 PM
SammyS
Quote:

Originally Posted by s3a
I got it wrong again but I tried to see why I was wrong and I cannot see how the -1 gets an additional negative like: -(-1). I tried to explain myself on the new attached paper. If more explanation is needed, please tell me.

The signs associated with the determinant: $\displaystyle \left|\begin{array}{ccc}2&0&-3\\0&0&-1\\1&3&0\end{array}\right|$

are: $\displaystyle \left|\begin{array}{ccc}+&-&+\\-&+&-\\+&-&+\end{array}\right|$.

The second row is: - + -