# Showing a matrix is singular HELP please! Watch

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#2

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.

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(Original post by

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

**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

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#4

(Original post by

But it mentioned without calculating AB, so why would you do that?

**Yatayyat**)But it mentioned without calculating AB, so why would you do that?

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(Original post by

just to see if it would help to get an idea of what is going on. it might not, though

**yolkie**)just to see if it would help to get an idea of what is going on. it might not, though

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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.

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**yolkie**)

just to see if it would help to get an idea of what is going on. it might not, though

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#8

(Original post by

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)Ã—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

**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)Ã—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

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#9

(Original post by

Ok so I got this for AB, but still not really certain where I can go from there...

**Yatayyat**)Ok so I got this for AB, but still not really certain where I can go from there...

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.

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(Original post by

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'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.

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.

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#11

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

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.

**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.

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#12

(Original post by

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.

**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.

Again, haven't done this in ages, so just trying my best.

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#13

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#14

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#15

**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)Ã—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

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(Original post by

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?

**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?

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...

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#17

(Original post by

the determinant only exists for square matrices.

**the bear**)the determinant only exists for square matrices.

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#18

(Original post by

We are in the middle of August? Are you getting a head start for next year or are you just bored?

**username31459**)We are in the middle of August? Are you getting a head start for next year or are you just bored?

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#19

(Original post by

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!

**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!

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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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