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Determine the dimensions of Nul $A$ and $\mathrm{Col} A$ for the matrices shown in Exercises $13-18$ .$$A=\left[\begin{array}{llll}{1} & {0} & {9} & {5} \\ {0} & {0} & {1} & {-4}\end{array}\right]$$

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Calculus 3

Chapter 4

Vector Spaces

Section 5

The Dimension of a Vector Space

Vectors

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the question as to determine the dimensions of null space. A in column space A for the Matrix shown below as a reminder of the dimensions of column space A is basically the dimensions in which that subspace fans or the number of the nearly independent columns in the subspace or the number of vectors in its bases, on the other hand, the dimensions of null space. A is just the number of free variable columns and is a reminder. Um, no Space A is a set of solutions of the homogeneous equation. A X equals zero. So once we reduce this to reduce echelon form and right into its parametric form, the columns, the solutions well, the number of columns in thes solutions set will be equal to the number of free variable columns. So going back to our problem, first thing we could do is reduce this to this echelon form. But lucky for us, it's already and it's special in form, so we can just start counting the number of pivot columns. So, looking at our first column, there is our first pivot column going down into the right. This is our second pivot column, so therefore first column and and the third column, um, our our pivot columns and the's are linearly independent, so the dimensions of column space A is the number of linearly independent columns in the subspace, which is to now for the dimensions of no space. A A. We just need to find We just need toe. Find the number of free variable columns, and in this case, the free variable columns are in the second column and in third column, So the dimensions of null space A is the number of free variable columns, which is just to

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