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The last time I looked at a matrix was like three months ago but I would start by finding AB and seeing what happens and trying to figure out how I would know that from just the two matrices. I have absolutely no idea either, though. All those 1s look like they have something to do with it.
(edited 5 years ago)
Original post by yolkie
The last time I looked at a matrix was like three months ago but I would start by finding AB and seeing what happens and trying to figure out how I would know that from just the two matrices. I have absolutely no idea either, though
But it mentioned without calculating AB, so why would you do that?
Original post by Yatayyat
But it mentioned without calculating AB, so why would you do that?
just to see if it would help to get an idea of what is going on. it might not, though
(edited 5 years ago)
Original post by yolkie
just to see if it would help to get an idea of what is going on. it might not, though
Ok so I got this for AB, but still not really certain where I can go from there...
Original post by yolkie
just to see if it would help to get an idea of what is going on. it might not, though
How would it help if we know that the matrix AB is a square matrix, does it tell us something about the determinant?
Original post by Helllooo1212
I havent touched matrices in a while now but ill try my best. First find the determinant of A and then the determinant of B.
Since det(A)×det(B)=det(AB), if both det(A) and det(B) are zero, then det(AB) is also zero.
If a matrix has a determinant of zero, it has no inverse so is singular.
Hope this helps
Since det(A)×det(B)=det(AB), if both det(A) and det(B) are zero, then det(AB) is also zero.
If a matrix has a determinant of zero, it has no inverse so is singular.
Hope this helps
I haven't done this in a while either, so don't take my word for it, but I don't think you can do inverse of non-square matrices.
Original post by Yatayyat
Ok so I got this for AB, but still not really certain where I can go from there...
Try to find the determinant of this matrix and then equate it to 0. This should give the values of k.
I don't know if this is the right way for this question as it tells you not so solve AB, but it's the only thing that comes to mind atm.
Original post by I'm God
Try to find the determinant of this matrix and then equate it to 0. This should give the values of k.
I don't know if this is the right way for this question as it tells you not so solve AB, but it's the only thing that comes to mind atm.
I don't know if this is the right way for this question as it tells you not so solve AB, but it's the only thing that comes to mind atm.
How would finding values of K help?
Plus there is a range of K values, I need to know that all K values would still give a AB matrix that is singular no matter the K value.
(edited 5 years ago)
this thread is a mess lol
I deleted it because it made no sense, I thought finding BA would help something but it doesn't
Original post by Yatayyat
I'm not really sure what you mean by that?
Matrix A and matrix B are not square matrices. Do you mean that the matrix AB is a square matrix.
Matrix A and matrix B are not square matrices. Do you mean that the matrix AB is a square matrix.
I deleted it because it made no sense, I thought finding BA would help something but it doesn't
Original post by Yatayyat
How would finding values of K help?
Plus there is a range of K values, I need to know that all K values would still give a AB matrix that is singular no matter the K value.
Plus there is a range of K values, I need to know that all K values would still give a AB matrix that is singular no matter the K value.
I think all the values of k would cancel out meaning that it is singular for all values of k?
Again, haven't done this in ages, so just trying my best.
Original post by Yatayyat
I’m so lost! How would I do it?
Think about what makes a matrix singular. If its determinant is 0, it has no inverse. Subbing in values of k in my calculator, I get a Det of 0 for all possible matrices AB. What properties of A or B might be causing this?
Original post by Yatayyat
I’m so lost! How would I do it?
We are in the middle of August? Are you getting a head start for next year or are you just bored?
Original post by Helllooo1212
I havent touched matrices in a while now but ill try my best. First find the determinant of A and then the determinant of B.
Since det(A)×det(B)=det(AB), if both det(A) and det(B) are zero, then det(AB) is also zero.
If a matrix has a determinant of zero, it has no inverse so is singular.
Hope this helps
Since det(A)×det(B)=det(AB), if both det(A) and det(B) are zero, then det(AB) is also zero.
If a matrix has a determinant of zero, it has no inverse so is singular.
Hope this helps
the determinant only exists for square matrices.
Original post by plklupu
Think about what makes a matrix singular. If its determinant is 0, it has no inverse. Subbing in values of k in my calculator, I get a Det of 0 for all possible matrices AB. What properties of A or B might be causing this?
Maybe for matrix AB to be singular, it has to contain a row or column of all zero elements in it; to a get a determinant of zero.
Does that mean either matrix A or matrix B has to have a row that all contains all zero elements...
I don't know what else it could be...
Original post by the bear
the determinant only exists for square matrices.
AB = a square bracket 3x3
Original post by username31459
We are in the middle of August? Are you getting a head start for next year or are you just bored?
In OP's defence, I've been pretty damn bored over the last month and a half, and I can't wait to dig into some uni pre-reading!
Original post by plklupu
In OP's defence, I've been pretty damn bored over the last month and a half, and I can't wait to dig into some uni pre-reading!
Sweet! What degree?
Wait, couldn't I change a row of matrix A, lets say the bottom row to become a row of zero's. I heard that you could do that by adding a constant times a row to another row. Lets say that other row is the top row and the constant is '-1'.
Then we could get a bottom row of zero for matrix a. Doesn't that make square matrix AB to have a bottom row of zeros. Hence the determinate of square matrix AB has to be zero therefore singular.
I'm not sure if this is right. Might be completely wrong :s
Then we could get a bottom row of zero for matrix a. Doesn't that make square matrix AB to have a bottom row of zeros. Hence the determinate of square matrix AB has to be zero therefore singular.
I'm not sure if this is right. Might be completely wrong :s
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