anonymous
  • anonymous
Can someone help me with determinants
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
What do you think they want me to do here?
anonymous
  • anonymous
for 3x3 matrices, the formula for the determinant is

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anonymous
  • anonymous
I know the formula. What do u think they want me to show?
anonymous
  • 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
  • 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
  • anonymous
Wait let me solve it
anonymous
  • anonymous
it is 6
anonymous
  • 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
  • anonymous
ok give me a sec :D
anonymous
  • anonymous
2k-2
anonymous
  • 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
  • anonymous
1
anonymous
  • anonymous
Yes, exactly. Now, are you familiar with Cramer's rule?
anonymous
  • anonymous
ya just watch a video of it on youtube so i like i know the steps
anonymous
  • 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
  • anonymous
Well then it wld be undefined
anonymous
  • anonymous
since you divide dx dy and dz by the determinant of A
anonymous
  • anonymous
So what would that mean about the system? Does it have solutions?
anonymous
  • anonymous
that there is no solution or there is an infinite amount of solutions. I am not sure lol
anonymous
  • 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
  • anonymous
ok
Zarkon
  • Zarkon
i think it really depends on what \(\vec{b}\) is
anonymous
  • anonymous
Zarkon perked up lol
anonymous
  • anonymous
U were so quiet i didnt think u were really here
Zarkon
  • Zarkon
that is if A is the matrix when k=1
anonymous
  • anonymous
@Zarkon that's right, my mistake Pippa. But the point is that the solution is not uniquely / well-defined, either way.
Zarkon
  • Zarkon
you would need to check to see if \(\vec{b}\) is in the column space of A
anonymous
  • anonymous
ok
Zarkon
  • Zarkon
Jemurray3 is definitely correct that it precludes the possibility of exactly one solution
anonymous
  • 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
  • anonymous
ya ok i get that
anonymous
  • 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
  • anonymous
Well what do u mean by patterns?
anonymous
  • 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
  • anonymous
ok will do. I will report back :D
anonymous
  • anonymous
lol i dont see a pattern
anonymous
  • anonymous
both were positive numbers but that means nothing
anonymous
  • 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
  • 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
  • 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
  • anonymous
Then, try this one: \[\left[\begin{matrix}1 & 2 & 3 \\7 & 1 & 8 \\ 1 & 1 & 2\end{matrix}\right]\]
anonymous
  • anonymous
so d=-16
anonymous
  • 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
  • anonymous
lol
anonymous
  • anonymous
No, the determinant for that is not -16, check your work over again.
anonymous
  • anonymous
but what does this have to do with the cramer rule?
anonymous
  • 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
  • anonymous
ohhh umm the first one is zero
anonymous
  • anonymous
Mhmm, and the second?
anonymous
  • anonymous
same zero
anonymous
  • 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
  • anonymous
ok get that
anonymous
  • anonymous
so basically i shld say in order to get a unique system the determinant cant equal 0
anonymous
  • anonymous
Thanks jemurray that was very kind of you to go step by step with me :D I really appreciate your time
anonymous
  • anonymous
*your help
anonymous
  • anonymous
No problem, glad to help

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