1. ## 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

2. 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

3. Hello, luckyc1423!

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!]

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

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

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

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

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

4. 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

5. Hello, luckyc1423!

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!

6. Thanks that helped out a bunch! I just wish my teacher could have explained it as well as you did. Once again, thanks!

7. 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.

8. Hello, JakeD!

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*