Are there any pivot columns in this matrix?

flyingpig
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Homework Statement




Just need help identifying pivots columns

\begin{bmatrix}<br /> 1 &amp; 1 &amp; 1 &amp;1 &amp;1\\ <br /> 0&amp; 0&amp; 0&amp; 0&amp;0\\ <br /> 0&amp; 0&amp; 0&amp; 0&amp;9<br /> \end{bmatrix}

From my understanding, any column with 1s and everything below it 0s are all pivot columns right?

Is the above all pivot columns? Also for a 3 x 5 coefficent matrix that has three pivot columns, is the system consistent?

The book says it is not, but why? The one above is a counterargument given by me. Does anyone know the LaTeX code to make an augmented matrix? Like put a bar before the constants
 
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The last column number should be a 0 not 9
 
remember, the definition of a pivot column is essentially that the column contains a pivot position in reduced echelon form. a pivot is the first nonzero term in the row (which needs to be a 1 for it to be reduced echelon form). how many of your rows in that matrix have a pivot in them?
 
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There can only be one pivot column in every row?
 
There can only be one pivot position in each row. Notice that there is a subtle difference between a pivot column and a pivot itself.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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