



 
Anna, Have a look at my response to Karlena's question a while ago. Her system has exactly one solution so the rows of the augmented matrix are linearly independent. Notice that after performing some row operations I wrote the resulting matrix back in equation form. If the final matrix had been
then the corresponding equations woud be
which is clearly impossible, zero is not equal to eight. If row operations on the augmented matrix result in a row of the form
where k is not zero, then the system of equations is inconsistant. If row operations on the augmented matrix result in a row of the form
then you havs shown that one row of the matrix is a linear combination of the other rows and hence the rows are linearly dependent. Row reduce your matrix and see which of the situations you have. Harley  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 