ksaimouli
  • ksaimouli
Linear algebra question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ksaimouli
  • ksaimouli
\[\left[\begin{matrix}1 & 0&-4 \\0 & 3&-2 \\ -2 & 6& 3\end{matrix}\right]\]
ksaimouli
  • ksaimouli
Let ^ be A, b= \[\left(\begin{matrix}4 \\ 1\\ -4\end{matrix}\right)\]
ksaimouli
  • ksaimouli
columns of A is a1,a2,a3 and W= span{a1,a2,a3}

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ksaimouli
  • ksaimouli
Is b in W? how many vectors are in {a1,a2,a3}
ksaimouli
  • ksaimouli
If W is in span of A, means C1a1+C2a2+C3a3=W?
anonymous
  • anonymous
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.
anonymous
  • anonymous
In other words, \[\begin{cases} c_1-2c_3=4\\ 3c_2+6c_3=1\\ -4c_1-2c_2+3c_3=-4 \end{cases}\]
ksaimouli
  • ksaimouli
True, I found it to be unique. Which means b is in the span of W. How many vectors are in W?
anonymous
  • anonymous
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.
ksaimouli
  • ksaimouli
I see, how would be interpret if the question was Is W in b?
anonymous
  • anonymous
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.
ksaimouli
  • ksaimouli
Okay, you said "number of vectors. a1,a2,a3 are linearly independent" is it because they are not scalar multiplies of each other?
anonymous
  • anonymous
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.)
ksaimouli
  • ksaimouli
thank you so much :-)

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