# Math Help - Need help with determinant of 5X5 matrix

1. ## Need help with determinant of 5X5 matrix

I'm given his matrix and asked to find the determinant.

$\left[\begin{array}{ccccc}2&-2&0&0&-3\\3&0&3&2&-1\\0&1&-2&0&2\\-1&2&0&3&0\\0&4&1&0&0\end{array}\right]$

Now I do $-4C_{3} + C_{2} \longrightarrow C_{2}$which gives me this

$\left[\begin{array}{ccccc}2&-2&0&0&-3\\3&-12&3&2&-1\\0&9&-2&0&2\\-1&2&0&3&0\\0&0&1&0&0\end{array}\right]$

This eliminates the 5th row and 3rd column.

I'm now left with this:

$\left[\begin{array}{cccc}2&-2&0&-3\\3&-12&2&-1\\0&9&0&2\\-1&2&3&0\end{array}\right]$

Now I'm trying to get rid of row and column 1:

$R_{1} + R{2} \longrightarrow R_{2}$

$3R_{1} + 2R{4} \longrightarrow R_{4}$

$\left[\begin{array}{cccc}2&0&0&0\\3&-9&2&7\\0&9&0&4\\-1&1&3&-3\end{array}\right]$

And now I'm left with this:

$\left[\begin{array}{ccc}-9&2&7\\9&0&4\\1&3&-3\end{array}\right]$

Now I can solve this and this is what I get:
$-9 \left[\begin{array}{cc}0&4\\3&-3\end{array}\right]$ $-2 \left[\begin{array}{cc}9&4\\1&-3\end{array}\right]$ $+7 \left[\begin{array}{cc}9&0\\1&3\end{array}\right]$

That gives me:

$2 [- 9(-12) - 2(-31) + 7(27)]= 2(359)$

This is where I don't understand because the answer in the book is 359, but what about the cofactor of 2 when I eliminated the row and column 1?

Don't I have to multiply everything by 2 which would not be the answer in the book?

Thank you.

2. Originally Posted by Mrs. White
I'm given his matrix and asked to find the determinant.

$\left[\begin{array}{ccccc}2&-2&0&0&-3\\3&0&3&2&-1\\0&1&-2&0&2\\-1&2&0&3&0\\0&4&1&0&0\end{array}\right]$

Now I do $-4C_{3} + C_{2} \longrightarrow C_{2}$which gives me this

$\left[\begin{array}{ccccc}2&-2&0&0&-3\\3&-12&3&2&-1\\0&9&-2&0&2\\-1&2&0&3&0\\0&0&1&0&0\end{array}\right]$

This eliminates the 5th row and 3rd column.

I'm now left with this:

$\left[\begin{array}{cccc}2&-2&0&-3\\3&-12&2&-1\\0&9&0&2\\-1&2&3&0\end{array}\right]$

Now I'm trying to get rid of row and column 1:

$R_{1} + R{2} \longrightarrow R_{2}$

$3R_{1} + 2R{4} \longrightarrow R_{4}$ Awkward says: here you doubled the determinant.

$\left[\begin{array}{cccc}2&0&0&0\\3&-9&2&7\\0&9&0&4\\-1&1&3&-3\end{array}\right]$

And now I'm left with this:

$\left[\begin{array}{ccc}-9&2&7\\9&0&4\\1&3&-3\end{array}\right]$

Now I can solve this and this is what I get:
$-9 \left[\begin{array}{cc}0&4\\3&-3\end{array}\right]$ $-2 \left[\begin{array}{cc}9&4\\1&-3\end{array}\right]$ $+7 \left[\begin{array}{cc}9&0\\1&3\end{array}\right]$

That gives me:

$2 [- 9(-12) - 2(-31) + 7(27)]= 2(359)$

This is where I don't understand because the answer in the book is 359, but what about the cofactor of 2 when I eliminated the row and column 1?

Don't I have to multiply everything by 2 which would not be the answer in the book?

Thank you.
See above.