# matrix determinants

• Jul 10th 2006, 05:39 PM
luckyc1423
matrix determinants
I am real confused on determants. There is a couple examples below and if someone could show me how this is done that would be great.

Find the determinant for this matrix:
example 1)
2 9 -5 3 9
3 1 5 -3 1
0 -4 1 2 -4
8 5 6 9 5
3 -2 -1 -3 -2

example 2)
3 0 0 0
1 -2 0 0
4 2 -1 0
3 7 2 4
• Jul 10th 2006, 06:15 PM
malaygoel
Quote:

Originally Posted by luckyc1423
I am real confused on determants. There is a couple examples below and if someone could show me how this is done that would be great.

Find the determinant for this matrix:
example 1)
2 9 -5 3 9
3 1 5 -3 1
0 -4 1 2 -4
8 5 6 9 5
3 -2 -1 -3 -2

example 2)
3 0 0 0
1 -2 0 0
4 2 -1 0
3 7 2 4

Second one is simple.
Expand it along the first row
You will get
3 multipied by
-2 0 0
2 -1 0
7 2 4
Again expand it along the first row
You wil get
6 multiplied by
-1 0
2 4
Hence determinant of the matrix comes out to be -24.

Keep Smiling
Malay
• Jul 11th 2006, 07:46 AM
Soroban
Hello, luckyc1423!

Quote:

I am real confused on determants. There are a couple examples below
and if someone could show me how this is done that would be great.

If you're really confused and know nothing about determinants,
. . you shoud not be playing by $4\times4$ and $5\times5$ determinants.

If you're waiting for someone to work out the 5x5 for you,
. . you may have a long wait.

Having said that, I must assume that you know the general procedure.
[If you don't, this is a total waste of time!]

Quote:

$2)\;\;\begin{vmatrix}3&0&0&0\\1&\text{-}2&0&0\\4&2&\text{-}1&0\\3&7&2&4\end{vmatrix}$

As Malay pointed out, using the top row is best choice.

. . . . . . . . + . - . .+ .-
We have: / $\begin{vmatrix}3&0&0&0\\1&\text{-}2&0&0\\4&2&\text{-}1&0\\3&7&2&4\end{vmatrix}$

. . $= \;3\cdot\begin{vmatrix}\text{-}2&0&0\\2&\text{-}1&0\\7&2&4\end{vmatrix} - 0\cdot\begin{vmatrix}1&0&0\\4&\text{-}1&0\\3&2&4\end{vmatrix} +$ $0\cdot\begin{vmatrix}1&\text{-}2&0\\4&2&0\\3&7&4\end{vmatrix} - 0\cdot\begin{vmatrix}1&\text{-}2&0\\4&2&\text{-}1\\3&7&2\end{vmatrix}$

. . . . . . . . . . . . . . . . . + . - . +
All we have left is: . $3\cdot\begin{vmatrix}\text{-}2&0&0\\2&\text{-}1&0\\7&2&4\end{vmatrix}$

. . $= \;3\left[-2\cdot\begin{vmatrix}\text{-}1&0\\2&4\end{vmatrix} - 0\cdot\begin{vmatrix}2&0\\7&4\end{vmatrix} + 0\cdot\begin{vmatrix}2&\text{-}1\\7&2\end{vmatrix}\right]$

. . $= \;3(-2)\cdot\begin{vmatrix}\text{-}1&0\\2&4\end{vmatrix}$

. . $= \;-6\left[(-1)(4) - (0)(2)\right] \;= \;-6(-4 - 0) \;= \;-6(-4) \;= \;\boxed{24}$

• Jul 11th 2006, 08:59 AM
luckyc1423
so you could use this form of find determinents for every matrix?
I did an example and could you tell me if im on the right track and if this is right......

+ - +
2 5 -2
3 1 2
-1 -2 4

= 2 * 1 2 - 5 * 3 2 + -2 * 3 1
-2 4 -1 4 -1 -2

= 2(4 + 4) - 5*(12 + 2) + (-2) *(-6 + 1)

= 16 - 70 + 10

= -44
• Jul 11th 2006, 10:02 AM
Soroban
Hello, luckyc1423!

Quote:

so you could use this form of find determinents for every matrix? . . . Yes!

I did an example and could you tell me if im on the right track and if this is right?

.+ . .- . +
$\begin{vmatrix}2&5&\text{-}2 \\ 3&1&2\\\text{-}1&\text{-}2&4\end{vmatrix}$

$= \;2\cdot\begin{vmatrix}1&2\\\text{-}2&4\end{vmatrix} - 5\cdot\begin{vmatrix}3&2\\ \text{-}1&4\end{vmatrix} - 2\cdot\begin{vmatrix}3&1\\ \text{-}1&\text{-}2\end{vmatrix}$

$= \;2(4 + 4) - 5*12 + 2) + (-2)(-6 + 1)$

$= \;16 - 70 + 10$

$= \;-44$

Absolutely correct!
. . . Nice work!

• Jul 11th 2006, 10:26 AM
luckyc1423
Thanks that helped out a bunch! I just wish my teacher could have explained it as well as you did. Once again, thanks!
• Jul 11th 2006, 02:37 PM
JakeD
Quote:

Originally Posted by luckyc1423
Find the determinant for this matrix:
example 1)
2 9 -5 3 9
3 1 5 -3 1
0 -4 1 2 -4
8 5 6 9 5
3 -2 -1 -3 -2

For this one, you're supposed to use a property of determinants instead of expanding it. Notice that the second and fifth columns are the same. That implies the determinant is zero.
• Jul 11th 2006, 04:35 PM
Soroban
Hello, JakeD!

Quote:

Notice that the second and fifth columns are the same.
That implies the determinant is zero.

Good eye, Jake!
I was trying row operations, but missed that completely . . . *blush*