Gaussian Elimination of Singular Matrix with partial pivoting

[ver_mathstats]

Homework Statement

Given the matrix A=[0.1 0.3 0.7; 0.3 0.6 0.9; 0.6 1.5 3] and b=[1.4; 1.8; 6] Part A: Show that A is singular.

Part B: Using exact arithmetic show where Gaussian elimination with partial pivoting fails to solve this system.

Part C: Although A is singular, cond(A) does not return Inf. Why?

Relevant Equations

A=[0.1 0.3 0.7; 0.3 0.6 0.9; 0.6 1.5 3] and b=[1.4; 1.8; 6]

Part (A): The matrix is a singular matrix because the determinant is 0 with my calculator.

Part (B): Once I perform Gauss Elimination with my pivot being 0.6 I arrive at the last row of matrix entries which are just 0's. So would this be why Gauss Elimination for partial pivoting fails for this matrix in particular? Because I now have a free variable with an infinite amount of solutions?

Part (C): This one I am confused because I know that if a matrix is singular it tends to inf. I also know that cond(A)=||A|| ||A^-1||, but the matrix does not have an inverse matrix, so I am really confused why inf is not showing up. On MATLAB I get the value that cond(A)=3.6114e+16 and the determinant is 2.4980e-18 on matlab, so I suppose that is enough for the cond(A) not to be inf. Is this enough of a reason?

Any help would be appreicated.

 

[lippyka]

Yes, the fact that the matrix does not have an inverse is enough to explain why the condition number of A is not infinite. The condition number of A measures how sensitively the output of a system of equations changes with small changes in the inputs. If the matrix had an inverse, then small changes in the input would lead to drastic changes in the output, resulting in an infinite condition number. Since the matrix does not have an inverse, small changes in the input do not necessarily lead to large changes in the output, so the condition number is finite.