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September 27th, 2008, 08:43 AM  #1 
Newbie Joined: Sep 2008 Posts: 5 Thanks: 0  Finding values for non singular matrix
Dear Math Forum, regarding the 3x3 matrix shown below. I need to find out for which values of a the matrix is non singular. I have tried reducing it to figure out if the set is linearly independent but I can't seem to get it right. Can you help? 1 1 1 0 1 2 1 1 a Best regards, Aske PS. I am new in matrix calculus (1'st year at college) so sorry if the question is too simple. 
September 27th, 2008, 09:02 AM  #2 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0  Re: Finding values for non singular matrix
A matrix M is nonsingular if its determinant is nonzero. Do you know how to find the determinant of a 3x3 matrix?

September 27th, 2008, 10:04 AM  #3 
Newbie Joined: Sep 2008 Posts: 5 Thanks: 0  Re: Finding values for non singular matrix
I know how to find the determinant for a 3x3 matrix. According to the assignment we all ready know that the matrix is nonsingular but I need to find out for which values of "a" it is nonsingular. It is to be said that we haven't yet been through determinants of 3x3 matrices in our curriculum so there must be another way to solve it.

September 27th, 2008, 10:21 AM  #4 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0  Re: Finding values for non singular matrix
If you have covered elementary row operations, it is very simple to convert your matrix into an upper triangular matrix, whose determinant  the trace  will be the same as the original matrix.

September 27th, 2008, 03:14 PM  #5 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: Finding values for non singular matrix
Otherwise, think about linear systems: For any matrix A, Ax=0 has only the trivial solution of x if and only if it is nonsingular; so do basic row operations to get it into row echelon form, and you want the last row to be nozero, that way z must be zero, meaning only one solution. You can also set it up as an augmented matrix with the last column being all zeroes, but that column won't do anything: (I just subtracted the first row from the third). If a=1, the last row is the 0=0 identity, which means we need to pick a parameter for our third variable > infinite solutions > singular. So any a?1 will give a nonsingular matrix. 

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