The Timpson Lyceum · Linear Algebra (2026)

1.3 Exercises

Problems with worked solutions · 20 points · 4 extra credit

Vector Equations

1. 1. (2 pts) Let $\mathbf{u} = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$ and $\mathbf{v} = \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}$. Compute $3\mathbf{u} - 2\mathbf{v}$.

   Solution

   1. $3\mathbf{u} - 2\mathbf{v} = \begin{bmatrix} 6 \\ -3 \\ 9 \end{bmatrix} - \begin{bmatrix} 2 \\ 0 \\ -4 \end{bmatrix} = \begin{bmatrix} 4 \\ -3 \\ 13 \end{bmatrix}.$

2. 2. (3 pts) Determine whether $\mathbf{b} = \begin{bmatrix} 5 \\ 3 \end{bmatrix}$ is a linear combination of $\mathbf{a}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\mathbf{a}_2 = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$. If so, give the weights.

   Solution

   2. Solve $x_1\begin{bmatrix}1\\2\end{bmatrix} + x_2\begin{bmatrix}3\\-1\end{bmatrix} = \begin{bmatrix}5\\3\end{bmatrix}$, i.e.

   $\begin{aligned} x_1 + 3x_2 &= 5 \\ 2x_1 - x_2 &= 3. \end{aligned}$

   From the second equation $x_2 = 2x_1 - 3$; substituting, $x_1 + 3(2x_1 - 3) = 5 \Rightarrow 7x_1 = 14 \Rightarrow x_1 = 2$, so $x_2 = 1$. Thus $\mathbf{b} = 2\mathbf{a}_1 + \mathbf{a}_2$; yes, $\mathbf{b}$ is a linear combination, with weights $2$ and $1$.

3. 3. (3 pts) Determine whether $\mathbf{b} = \begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix}$ lies in $\operatorname{Span}\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3\}$, where

   $\mathbf{a}_1 = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix},\quad \mathbf{a}_2 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix},\quad \mathbf{a}_3 = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}.$

   Solution

   3. Row reduce the augmented matrix $[\,\mathbf{a}_1\;\mathbf{a}_2\;\mathbf{a}_3\mid\mathbf{b}\,]$:

   $\left[\begin{array}{ccc|c} 1 & 2 & 0 & 3 \\ 0 & 1 & 1 & 2 \\ -1 & 0 & 2 & -1 \end{array}\right] \xrightarrow{R_3 + R_1} \left[\begin{array}{ccc|c} 1 & 2 & 0 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 2 & 2 & 2 \end{array}\right] \xrightarrow{R_3 - 2R_2} \left[\begin{array}{ccc|c} 1 & 2 & 0 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & -2 \end{array}\right].$

   Row 3 reads $0 = -2$, so the system is inconsistent. Therefore $\mathbf{b} \notin \operatorname{Span}\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3\}$. (In fact these three vectors span only a plane through the origin, and $\mathbf{b}$ does not lie on that plane.)

4. 4. (3 pts) Describe geometrically, in $\mathbb{R}^3$:

   (a) $\operatorname{Span}\left\{\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}\right\}$

   (b) $\operatorname{Span}\left\{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\right\}$

   Solution

   4.

   (a) A single nonzero vector spans all its scalar multiples, so $\operatorname{Span}\left\{(1,2,-1)\right\}$ is the line through the origin in the direction $(1, 2, -1)$.

   (b) All combinations $x_1(1,0,0) + x_2(0,1,0) = (x_1, x_2, 0)$ fill the $x_1x_2$-plane (all vectors with third coordinate $0$), a plane through the origin.

5. 5. (3 pts) State whether each of the following is true or false, and justify each answer.

   (a) The vector equation $x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + x_3\mathbf{a}_3 = \mathbf{b}$ has the same solution set as the linear system whose augmented matrix is $[\,\mathbf{a}_1 \; \mathbf{a}_2 \; \mathbf{a}_3 \mid \mathbf{b}\,]$.

   (b) The span of a single nonzero vector in $\mathbb{R}^3$ is a plane.

   (c) Asking whether $\mathbf{b} \in \operatorname{Span}\{\mathbf{a}_1, \dots, \mathbf{a}_p\}$ is the same as asking whether the system $[\,\mathbf{a}_1 \; \cdots \; \mathbf{a}_p \mid \mathbf{b}\,]$ is consistent.

   Solution

   5.

   (a) True. By definition, $x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + x_3\mathbf{a}_3 = \mathbf{b}$ is exactly the system whose augmented matrix is $[\,\mathbf{a}_1\;\mathbf{a}_2\;\mathbf{a}_3\mid\mathbf{b}\,]$; the weights are the unknowns.

   (b) False. The span of one nonzero vector is a line through the origin (all its scalar multiples), not a plane.

   (c) True. $\mathbf{b} \in \operatorname{Span}\{\mathbf{a}_1,\dots,\mathbf{a}_p\}$ means $\mathbf{b}$ is a linear combination of the $\mathbf{a}_i$, i.e. the equation $[\,\mathbf{a}_1\;\cdots\;\mathbf{a}_p\mid\mathbf{b}\,]$ has a solution — precisely consistency.

6. 6. (3 pts) Let $\mathbf{a}_1 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$ and $\mathbf{a}_2 = \begin{bmatrix} -3 \\ 6 \end{bmatrix}$. Describe exactly which vectors $\mathbf{b} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$ belong to $\operatorname{Span}\{\mathbf{a}_1, \mathbf{a}_2\}$, and give the defining equation of that set.

   Solution

   6. Note $\mathbf{a}_2 = -3\mathbf{a}_1$, so $\operatorname{Span}\{\mathbf{a}_1, \mathbf{a}_2\} = \operatorname{Span}\{\mathbf{a}_1\}$: the line through the origin in the direction $(1, -2)$. A vector $\mathbf{b}$ lies on this line iff it is a scalar multiple of $(1,-2)$, i.e. iff $\dfrac{b_1}{1} = \dfrac{b_2}{-2}$. The defining equation is $2b_1 + b_2 = 0$.

7. 7. (3 pts) Explain why $\mathbf{0}$ belongs to $\operatorname{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$ for any vectors $\mathbf{v}_1, \dots, \mathbf{v}_p$. Then explain, for two vectors $\mathbf{v}_1, \mathbf{v}_2$ in $\mathbb{R}^3$, the geometric possibilities for $\operatorname{Span}\{\mathbf{v}_1, \mathbf{v}_2\}$.

   Solution

   7. Choosing all weights equal to $0$ gives $0\mathbf{v}_1 + \cdots + 0\mathbf{v}_p = \mathbf{0}$, so $\mathbf{0}$ is always a linear combination of the vectors and hence lies in their span. For two vectors in $\mathbb{R}^3$: if both are $\mathbf{0}$, the span is $\{\mathbf{0}\}$; if exactly one direction is present (one is zero, or one is a scalar multiple of the other), the span is a line through the origin; if neither is a multiple of the other, the span is a plane through the origin.

8. 8. (4 pts ••) Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\mathbb{R}^n$. Prove that $\operatorname{Span}\{\mathbf{u}, \mathbf{v}\} = \operatorname{Span}\{\mathbf{u}, \mathbf{u} + \mathbf{v}\}$.

   Solution

   8. Write $S = \operatorname{Span}\{\mathbf{u}, \mathbf{v}\}$ and $T = \operatorname{Span}\{\mathbf{u}, \mathbf{u} + \mathbf{v}\}$.

   ($T \subseteq S$) Both $\mathbf{u}$ and $\mathbf{u} + \mathbf{v}$ are linear combinations of $\mathbf{u}, \mathbf{v}$ (namely $\mathbf{u}$ and $1\mathbf{u} + 1\mathbf{v}$), so any combination $a\mathbf{u} + b(\mathbf{u}+\mathbf{v}) = (a+b)\mathbf{u} + b\mathbf{v}$ lies in $S$. Hence $T \subseteq S$.

   ($S \subseteq T$) Conversely $\mathbf{u} \in T$, and $\mathbf{v} = (\mathbf{u}+\mathbf{v}) - \mathbf{u}$ is a combination of $\mathbf{u}$ and $\mathbf{u}+\mathbf{v}$, so $\mathbf{v} \in T$. Then any $c\mathbf{u} + d\mathbf{v} \in T$, giving $S \subseteq T$.

   Therefore $S = T$. $\blacksquare$