# Decide a number $a, b, c$ (expressed by $\lambda$) so that $a\mathbf{u}_1 + b\mathbf{u}_2 + c\mathbf{u}_3 = \mathbf{x}$

See the following vectors in $$\mathbb{R}^3$$:

$$\mathbf{u}_1= \begin{pmatrix} 1 \\ -1\\ 1 \\ 1 \end{pmatrix}$$, $$\mathbf{u}_2 = \begin{pmatrix} 1 \\ 2 \\ 0 \\ 2 \end{pmatrix}$$, $$\mathbf{u}_3 = \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix}$$

and

$$\mathbf{v}_1= \begin{pmatrix} 5 \\ -3\\ -4 \\ -1 \end{pmatrix}$$, $$\mathbf{v}_2 = \begin{pmatrix} 6 \\ 4 \\ 1 \\ 8 \end{pmatrix}$$, $$\mathbf{v}_3 = \begin{pmatrix} 7 \\ 2 \\ -1 \\ 6 \end{pmatrix}$$

We are told that $$\mathcal{B} = (\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3)$$ and $$\mathcal{C} = (\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3)$$ both are bases for the same subspace $$\mathcal{U}$$.

Let $$\mathcal{\lambda}$$ be an unknown number og see the vector $$\mathbf{x} = \mathbf{v}_1 + \lambda\mathbf{v}_2 - \mathbf{v}_3$$.

Decide a number $$a, b, c$$ (expressed by $$\lambda$$) so that $$a\mathbf{u}_1 + b\mathbf{u}_2 + c\mathbf{u}_3 = \mathbf{x}$$

1. Find some $$\alpha_i$$ such that $$v_1=\alpha_1 u_1+\alpha_2u_2+\alpha_3u_3$$
2. Find some $$\beta_i$$ such that $$v_2=\beta_1 u_1+\beta_2u_2+\beta_3u_3$$
3. Find some $$\gamma_i$$ such that $$v_3=\gamma_1 u_1+\gamma_2u_2+\gamma_3u_3$$
4. Simplify the expression $$x=v_1+\lambda v_2-v_3=\alpha_1 u_1+\alpha_2u_2+\alpha_3u_3+\lambda(\beta_1 u_1+\beta_2u_2+\beta_3u_3)-(\gamma_1 u_1+\gamma_2u_2+\gamma_3u_3)$$
• Thanks 5xum! When I reduce it, I get $\begin{pmatrix} 8 \\ 3\\ -9 \\ 1 \end{pmatrix} + a\begin{pmatrix} 2 \\ -2\\ 1 \\ 1 \end{pmatrix} + b\begin{pmatrix} 3 \\ 6\\ 0 \\ 6 \end{pmatrix} + c\begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix}$ What do I do from here? – jubibanna May 15 '19 at 13:56