Math  /  Algebra

Question(1 point)
Let WW be the set of all vectors of the form [2s+3t5s+2t4st]\left[\begin{array}{c}2 s+3 t \\ 5 s+2 t \\ 4 s-t\end{array}\right]. Find vectors w~\tilde{w} and zz in R3\mathbb{R}^{3} such that W=span{z~undefined,}\mathbb{W}=\operatorname{span}\{\overrightarrow{\tilde{z}}, \vec{\nabla}\}. u=[]v=[]\vec{u}=\left[\begin{array}{l} \square \\ \square \\ \square \end{array}\right] \cdot \vec{v}=\left[\begin{array}{c} \square \\ \square \\ \square \end{array}\right]

Studdy Solution
Verify that the span of {w~,z}\{\tilde{w}, z\} is indeed the set WW.
The set WW is the span of {w~,z}\{\tilde{w}, z\} because any vector in WW can be written as a linear combination of w~\tilde{w} and zz, as shown in the expression:
[2s+3t5s+2t4st]=s[254]+t[321]\begin{bmatrix} 2s + 3t \\ 5s + 2t \\ 4s - t \end{bmatrix} = s \begin{bmatrix} 2 \\ 5 \\ 4 \end{bmatrix} + t \begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix}
Thus, the vectors w~\tilde{w} and zz are:
u=[254],v=[321]\vec{u} = \begin{bmatrix} 2 \\ 5 \\ 4 \end{bmatrix}, \quad \vec{v} = \begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix}

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