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Question

In each case, write x as a linear combination of the orthogonal basis of the subspace U. (a)

x=(13,20,15)U=span{(1,2,3),(1,1,1)}\begin{array} { l } { \mathbf { x } = ( 13 , - 20,15 ) } \\ { U = \operatorname { span } \{ ( 1 , - 2,3 ) , ( - 1,1,1 ) \} } \end{array}

, (b)

x=(14,1,8,5)U=span{(2,1,0,3),(2,1,2,1)}\begin{array} { l } { \mathbf { x } = ( 14,1 , - 8,5 ) } \\ { U = \operatorname { span } \{ ( 2 , - 1,0,3 ) , ( 2,1 , - 2 , - 1 ) \} } \end{array}

Solution

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Step 1
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(a):\text{\color{#19804f}(a):} We need to find scalars α,βR\alpha,\beta\in R, such that

α[123]+β[111]=[132015]\alpha\begin{bmatrix}1\\-2\\3\end{bmatrix}+\beta\begin{bmatrix}-1\\1\\1\end{bmatrix}=\begin{bmatrix}13\\-20\\15\end{bmatrix}

.

This gives us system of equations:

αβ=132α+β=203α+β=15\begin{align*} \alpha&-\beta=13\\ -2\alpha&+\beta=-20\\ 3\alpha&+\beta=15 \end{align*}

If we add first and second equation, we get

α=7    α=7-\alpha=-7\implies\boxed{\alpha=7}.

If we substitute this value back into the third equation of the system, we get

21+β=15    β=621+\beta=15\implies\boxed{\beta=-6}.

Therefore,

[132015]=7[123]+(6)[111]{\color{#4257b2}\begin{bmatrix}13\\-20\\15\end{bmatrix}=7\begin{bmatrix}1\\-2\\3\end{bmatrix}+(-6)\begin{bmatrix}-1\\1\\1\end{bmatrix}}

.

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