Question

Linear algebra question

Answer 1

\[\left[\begin{matrix}1 & 0&-4 \\0 & 3&-2 \\ -2 & 6& 3\end{matrix}\right]\]

Answer 2

Let ^ be A, b= \[\left(\begin{matrix}4 \\ 1\\ -4\end{matrix}\right)\]

Answer 3

columns of A is a1,a2,a3 and W= span{a1,a2,a3}

Answer 4

Is b in W? how many vectors are in {a1,a2,a3}

Answer 5

If W is in span of A, means C1a1+C2a2+C3a3=W?

Answer 6

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.

Answer 7

In other words,
\[\begin{cases}
c_1-2c_3=4\\
3c_2+6c_3=1\\
-4c_1-2c_2+3c_3=-4
\end{cases}\]

Answer 8

True, I found it to be unique. Which means b is in the span of W. How many vectors are in W?

Answer 9

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.

Answer 10

I see, how would be interpret if the question was Is W in b?

Answer 11

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.

Answer 12

Okay, you said "number of vectors. a1,a2,a3 are linearly independent" is it because they are not scalar multiplies of each other?

Answer 13

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.)

Answer 14

thank you so much :-)