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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 non-singular if its determinant is non-zero. 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 non-singular but I need to find out for which values of "a" it is non-singular. 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 non-singular; so do basic row operations to get it into row echelon form, and you want the last row to be no-zero, 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: $$\begin{array} 1 &1 &-1\\ 0 &1 &2\\ 1 &1 &a \end{array}$ \rightarrow $\begin{array} 1 &1 &-1\\ 0 &1 &2\\ 0 &0 &a+1 \end{array}$$ (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 non-singular matrix.

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# for whatvalue ofx following matrix is nonsingular matrix

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