Can someone help me with determinants

- anonymous

Can someone help me with determinants

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- anonymous

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- anonymous

What do you think they want me to do here?

- anonymous

for 3x3 matrices, the formula for the determinant is

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- anonymous

I know the formula. What do u think they want me to show?

- anonymous

oh, well, not sure what arbitrary means, but when the det. is 0, when using matrices to solve systems of equations im pretty sure it means there is no solution. I havent used matrices in sytems for a year so I need refreshed on it. Cramers rule shows how to use matrices to solve systems of equations

- anonymous

Just try it. What's the determinant of
\[\left[\begin{matrix}1 & 1 & 4 \\2 &0 & 2 \\ 1 & 1 & 1\end{matrix}\right]\]

- anonymous

Wait let me solve it

- anonymous

it is 6

- anonymous

Alright, so let's choose an element to replace. How about that 4 in the upper right hand corner? Let's try the determinant of
\[\left[\begin{matrix}1 &1 & k \\ 2& 0 & 2 \\ 1& 1 & 1\end{matrix}\right]\]

- anonymous

ok give me a sec :D

- anonymous

2k-2

- anonymous

Very good. So if we let k = 4 like it used to be, then we get that the determinant is six. But what would k have to be if we wanted the determinant to be zero?

- anonymous

1

- anonymous

Yes, exactly. Now, are you familiar with Cramer's rule?

- anonymous

ya just watch a video of it on youtube so i like i know the steps

- anonymous

Good. Cramer's rule provides us with a solution to the equation
\[A \vec{x} = \vec{b}\]
where A is an nxn matrix and x and b are n dimensional vectors. Let's say we're solving it by using Cramer's rule. Why would it be bad if the determinant of A was zero?

- anonymous

Well then it wld be undefined

- anonymous

since you divide dx dy and dz by the determinant of A

- anonymous

So what would that mean about the system? Does it have solutions?

- anonymous

that there is no solution or there is an infinite amount of solutions. I am not sure lol

- anonymous

In this case it has an infinite number of solutions, but the point is it doesn't have one particular solution, which is typically what we look for.

- anonymous

ok

- Zarkon

i think it really depends on what \(\vec{b}\) is

- anonymous

Zarkon perked up lol

- anonymous

U were so quiet i didnt think u were really here

- Zarkon

that is if A is the matrix when k=1

- anonymous

@Zarkon that's right, my mistake Pippa. But the point is that the solution is not uniquely / well-defined, either way.

- Zarkon

you would need to check to see if \(\vec{b}\) is in the column space of A

- anonymous

ok

- Zarkon

Jemurray3 is definitely correct that it precludes the possibility of exactly one solution

- anonymous

You will find later (or maybe you already have) that a nonzero determinant means that the matrix A is invertible, i.e. that there exists some other matrix A' such that if you multiply A by A' you get the identity matrix. Such a system must then have exactly one solution, corresponding to x = A' * b. If, on the other hand, the matrix has determinant zero, then the matrix A' does not exist, and this corresponds to a system whose solutions are not unique, if they exist at all.

- anonymous

ya ok i get that

- anonymous

The next thing you should do with this is write down a few other matrices that you just make up off the top of your head and take the determinants. If they are not zero, replace one element with k and figure out how to make them zero. Look for patterns in the matrices with determinant zero and try to figure out how you might be able to see without solving for k.

- anonymous

Well what do u mean by patterns?

- anonymous

Look for similarities between the matrices. They will have something in common. Compare the rows and columns of each matrix to each other.

- anonymous

ok will do. I will report back :D

- anonymous

lol i dont see a pattern

- anonymous

both were positive numbers but that means nothing

- anonymous

What do you notice about the columns of the matrix A that I wrote down, once we replaced the 4 with the 1?

- anonymous

its the same as column 1 which makes the determinant equal zero. If you did a cloumn operation and multiplied the first row by negative one and added it to the third you wld get a column of zeroes

- anonymous

Yes, you're right. So if one of the columns is zero, then the determinant is zero. Check to see if the same is true if two of the rows are the same.

- anonymous

Then, try this one:
\[\left[\begin{matrix}1 & 2 & 3 \\7 & 1 & 8 \\ 1 & 1 & 2\end{matrix}\right]\]

- anonymous

so d=-16

- anonymous

And finally, this one
\[\left[\begin{matrix}3 & 1 & 2 \\ 2 & 1 & 1 \\ 5 &2 & 3\end{matrix}\right]\]
once you've tried them all, I'll tell you when you can expect the determinant to be zero.

- anonymous

lol

- anonymous

No, the determinant for that is not -16, check your work over again.

- anonymous

but what does this have to do with the cramer rule?

- anonymous

The Cramer's rule illustrates what happens when the determinant is zero, but the question seems to be geared towards WHY determinants are zero.

- anonymous

ohhh umm the first one is zero

- anonymous

Mhmm, and the second?

- anonymous

same zero

- anonymous

Okay. The general principle, which you should already have some idea of, is that the determinant of the matrix is equal to zero if any of its rows can be written as some linear combination of the others (like row 3 = 3* row 1 + 2 * row 2), or if any of the columns can be written as some linear combination of the other columns.

- anonymous

ok get that

- anonymous

so basically i shld say in order to get a unique system the determinant cant equal 0

- anonymous

Thanks jemurray that was very kind of you to go step by step with me :D I really appreciate your time

- anonymous

*your help

- anonymous

No problem, glad to help

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