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Linear algebra question

Mathematics
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\[\left[\begin{matrix}1 & 0&-4 \\0 & 3&-2 \\ -2 & 6& 3\end{matrix}\right]\]
Let ^ be A, b= \[\left(\begin{matrix}4 \\ 1\\ -4\end{matrix}\right)\]
columns of A is a1,a2,a3 and W= span{a1,a2,a3}

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Is b in W? how many vectors are in {a1,a2,a3}
If W is in span of A, means C1a1+C2a2+C3a3=W?
Looks like you're just checking to see if \(b\) can be written as a linear combination of \(a_1,a_2,a_3\), which amounts to solving for \(c_1,c_2,c_3\) such that \[c_1a_1+c_2a_2+c_3a_3=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}\] which means you'll be solving a system of 3 equations with 3 unknowns.
In other words, \[\begin{cases} c_1-2c_3=4\\ 3c_2+6c_3=1\\ -4c_1-2c_2+3c_3=-4 \end{cases}\]
True, I found it to be unique. Which means b is in the span of W. How many vectors are in W?
Well, I'd say there's an infinite number of vectors. \(a_1,a_2,a_3\) are linearly independent, so \(W\) forms a basis of \(\mathbb{R}^3\), which contains an infinite number of vectors.
I see, how would be interpret if the question was Is W in b?
I don't think that question would make sense. \(W\) is a set/space of vectors, while \(b\) is just a vector. A set can't belong to an element.
Okay, you said "number of vectors. a1,a2,a3 are linearly independent" is it because they are not scalar multiplies of each other?
Yes, the only way to get \(k_1a_1+k_2a_2+k_3a_3=\begin{pmatrix}0\\0\\0\end{pmatrix}\) is if all the \(k=0\). It's more accurate to say "linear combination" in place of "scalar multiples" here. (The second term is a special case of the first term.)
thank you so much :-)

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