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Determine the dimensions of Nul $A$ and $\mathrm{Col} A$ for the matrices shown in Exercises $13-18$ .$$A=\left[\begin{array}{rr}{3} & {4} \\ {-6} & {10}\end{array}\right]$$

There are two pivot columns, so the dimension of $\mathrm{Col} A$ is 2 .There are no columns without pivots.So the equation $A \overrightarrow{\mathrm{x}}=0$ has only the trivial solution $\overrightarrow{0}$ .Thus Nul$A=\{\overrightarrow{0}\}$ and so the dimension of $\mathrm{Nul} A$ is $0 .$

Calculus 3

Chapter 4

Vector Spaces

Section 5

The Dimension of a Vector Space

Vectors

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a queue of matrix is two by two matrix, which is three four connective, 6 10 Okay, so, in order to determine the dimensions, um, first transformed this matrix thio reduced ashen of form. So this will be three. I'm sorry. Three or and we use second role. Plus, why sell the first room? So that would be zero. I saw a bit of personal eight. 18. So that gives X two. It's cereal and X one could be ex want us or ex too, is you? So that means everything is cereal. Because we have X to zero and plugging into this equation, we have x one will be zero to so no A it is zero. What about a column? Obey. This dimension will turn out to be, uh, too, because we have two columns, have two columns with the people positions. So we have two people columns so that that is too

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