(1 point) Let H = span \{(-6, 5, 2), (6, -3, -2), (-3, 2, 1)\}. A basis for the subspace $H \subset \mathbb{R}^3$ is \{ <-6,6,-3>,<0,2,-1/2> \}. vector
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To do this, we can set up a matrix with the given vectors as its columns and row reduce it to see if it has a pivot in every row. If it does, then the vectors are linearly independent. The matrix is: [-6 0] [ 6 2] [-3 -1] Row reducing this matrix, we get: [1 Show more…
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$[\mathbf{M}]$ Let $H=\operatorname{Span}\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\}$ and $\mathcal{B}=\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} .$ Show that $\mathcal{B}$ is a basis for $H$ and $\mathbf{x}$ is in $H,$ and find the $\mathcal{B}$ -coordinate vector of $\mathbf{x},$ for $$ \mathbf{v}_{1}=\left[\begin{array}{r}{-6} \\ {4} \\ {-9} \\ {4}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{8} \\ {-3} \\ {7} \\ {-3}\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{r}{-9} \\ {5} \\ {-8} \\ {3}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{4} \\ {7} \\ {-8} \\ {3}\end{array}\right] $$
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$[\mathbf{M}]$ Let $H=\operatorname{Span}\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\}$ and $B=\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \} .$ Show that $\mathcal{B}$ is a basis for $H$ and $\mathbf{x}$ is in $H,$ and find the $\mathcal{B}$ -coordinate vector of $\mathbf{x},$ when $$ \mathbf{v}_{1}=\left[\begin{array}{r}{-6} \\ {4} \\ {-9} \\ {4}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{8} \\ {-3} \\ {7} \\ {-3}\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{r}{-9} \\ {5} \\ {-8} \\ {3}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{4} \\ {7} \\ {-8} \\ {3}\end{array}\right] $$
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