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Solution is on WebWork.
Let $\vw=⟨13,23⟩$, $\vv_1=⟨−1,−2⟩$, $\vv_2=⟨5,10⟩$, $\vv_3=⟨3,5⟩$. If possible, express $\vw$ as a linear combination of the vectors $\vv_1, \vv_2$ and $\vv_3$.
Ans: Form the matrix $A$ with $\vv_i$ as its $i$-th column then the question becomes deciding with $\vb$ is in the column space of $A$. Thus, by Proposition 2 in Notes 14. That amongs to deciding whether the system with augmented matrix $B = [A|\vw]$. Also, any solution of the system $A\vx = \vw$ correspondes to a linear combination for the $\vv_i$'s that sums to $\vw$.
A = column_matrix([(-1,-2),(5,10),(3,5)]); w = column_matrix([13,23]); B = block_matrix([A,w],ncols=2);
B, B.rref()
From the rref of $B$, we see that the system $A\vx = \vx$ is consistent and has 1-free variable ($x_2$). Rename $x_2$ to $s$, then we have $x_1 = -4 +5s, x_2 = s, x_3 = 3$. By fixing $s$ to be any scalar, we get a solution. In particular, if we fix $s$ as $0$, we get the solution $x_1 = -4, x_2 = 0$ and $x_3 = 3$. Therefore, $-4\vv_1 + (0)\vv_2 + 3\vv_3$ is a linear combination of the $\vv_i$'s that sums to $\vw$.
Find a set of vectors $\{\vu,\vv\}$ in $R^4$ that spans the solution set of the equations $w−x−2y−3z=0,w+2x+y+3z=0$.
Ans: To answer this, let's solve the system first. The system in matrix form is: $$\bm{ 1 & -1 & -2 & -3 \\ 1 & 2& 1& 3} \bm{w \\ x\\y \\z} = \bm{0 \\ 0}$$
As usual, we solve this by row reductions on the augmented matrix of the system.
But I want to remark that since the system is homogeneous, instead of the augmented matrix it is enough to look at the coefficient matrix of the system (because no matter what row operations we perform on the augmented matrix, the last column is always zero).
A = matrix([(1,-1,-2,-3),(1,2,1,3)]); B = block_matrix([A,column_matrix((0,0))],ncols=2); B, B.rref()
The free variables in the rref are $y$ and $z$, rename them to $s$ and $t$ respectively, then the system is equivalent to $$ w -s -t = 0 \ \text{and}\ x +s + 2t = 0$$ Thus the solutions are $\bm{w \\x \\ y \\ z} = \bm{s+t \\ -s -2t \\ s \\ t} = s\bm{1 \\ -1 \\ 1 \\ 0} + t\bm{1 \\ -2 \\ 0 \\1}$ where $s$ and $t$ are arbitrary. In this form, we see that we can choose $$ \vu = \bm{1 \\ -1 \\ 1 \\ 0} \ \text{and}\ \vv = \bm{1 \\-2 \\0 \\1}$$ to be the vectors that spans the solution set.