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anonymous
 4 years ago
Can someone help me with determinants
anonymous
 4 years ago
Can someone help me with determinants

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What do you think they want me to do here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for 3x3 matrices, the formula for the determinant is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know the formula. What do u think they want me to show?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Just try it. What's the determinant of \[\left[\begin{matrix}1 & 1 & 4 \\2 &0 & 2 \\ 1 & 1 & 1\end{matrix}\right]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Very 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, exactly. Now, are you familiar with Cramer's rule?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya just watch a video of it on youtube so i like i know the steps

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Good. 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
 4 years ago
Best ResponseYou've already chosen the best response.0Well then it wld be undefined

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since you divide dx dy and dz by the determinant of A

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So what would that mean about the system? Does it have solutions?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that there is no solution or there is an infinite amount of solutions. I am not sure lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In 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.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2i think it really depends on what \(\vec{b}\) is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0U were so quiet i didnt think u were really here

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2that is if A is the matrix when k=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Zarkon that's right, my mistake Pippa. But the point is that the solution is not uniquely / welldefined, either way.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2you would need to check to see if \(\vec{b}\) is in the column space of A

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2Jemurray3 is definitely correct that it precludes the possibility of exactly one solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0Well what do u mean by patterns?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Look for similarities between the matrices. They will have something in common. Compare the rows and columns of each matrix to each other.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok will do. I will report back :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol i dont see a pattern

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0both were positive numbers but that means nothing

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What do you notice about the columns of the matrix A that I wrote down, once we replaced the 4 with the 1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Then, try this one: \[\left[\begin{matrix}1 & 2 & 3 \\7 & 1 & 8 \\ 1 & 1 & 2\end{matrix}\right]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And 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
 4 years ago
Best ResponseYou've already chosen the best response.0No, the determinant for that is not 16, check your work over again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but what does this have to do with the cramer rule?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The Cramer's rule illustrates what happens when the determinant is zero, but the question seems to be geared towards WHY determinants are zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhh umm the first one is zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mhmm, and the second?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay. 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
 4 years ago
Best ResponseYou've already chosen the best response.0so basically i shld say in order to get a unique system the determinant cant equal 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks jemurray that was very kind of you to go step by step with me :D I really appreciate your time

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No problem, glad to help
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