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For each subspace in Exercises 1–8, (a) find a basis, and (b) state the dimension.$$\left\{\left[\begin{array}{c}{3 a+6 b-c} \\ {6 a-2 b-2 c} \\ {-9 a+5 b+3 c} \\ {-3 a+b+c}\end{array}\right] : a, b, c \text { in } \mathbb{R}\right\}$$

a. See solutionb. 2

Calculus 3

Chapter 4

Vector Spaces

Section 5

The Dimension of a Vector Space

Vectors

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were given a linear system which is, uh, three a US six. B minus C and six a minus to be minus to see to see. And also we have negative nine a us five e and plus three say and elected three a plus B plus C where ABC are all real numbers. Okay, So to find the basis off this system, um, we first need to separate those, um, the scale er's ABC. So to do that, we have first scaler a and we can't those those turns that contain that those come that contain eight. So that's 36 negative. 89 and 10 to 3. So that's three six connective 9 to 3. And also for Ms Cater be We checked the coefficients. That's Ah, six negative 25 and one six connected to buy one. And for skaters. See you. We have a negative 12 and three and one. So that's next one Naked too. And three and one. So that's for skater. See? So the vectors generated by, uh, the system is will be 36 elective nigh and elective, three for the one say, and to be two, we have six negative too. Five and a one. And for the three, we have negative one. Next, two and three and one. Okay, so the first thing we need to do is to write here. We need to check the win your independence off these three vectors. So, um, one thing to observe dead is be one is scuse me is negative. Three times B three. Right? Because if we times to be three by a skater, negative three, then the first term will be three. And second term will be six. The third term will be negative. Die and fourth term is negative. Three. So that means a V one is dependent on the B three. So the only reason you're independent vectors in this system is vey to MP three. There are only too linearly independent vectors B two and B three. Sorry. Oh, revised is you too. And B three linearly independent independent. Okay, so that means that dimension of this you mention off this subspace is two. And this is the basis which is formed by 3 to 1 of these three

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