August 8, 2025
Let \(\vec{v} = \begin{bmatrix} 2\\ -1\\ 0\\ 4\end{bmatrix}\) Compute \(3\vec{v}\) and \(-2\vec{v}\).
Let \(\vec{u} = \begin{bmatrix} 4\\ 0\\ -2\end{bmatrix}\) and \(\vec{v} = \begin{bmatrix} -1\\ 3\\ 8\end{bmatrix}\) and compute \(\vec{u} + \vec{v}\).
Let \(\vec{u} = \begin{bmatrix} 1\\ 2\end{bmatrix}\) and \(\vec{v} = \begin{bmatrix} 3\\ -1\end{bmatrix}\) and compute \(2\vec{u} + 3\vec{v}\).
Let \(\vec{v_1} = \begin{bmatrix} 1\\ 0\\ -1\\ 1\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 0\\ -2\\ 0\\ 3\end{bmatrix}\), and \(\vec{v_3} = \begin{bmatrix} 2\\ 0\\ 0\\ 4\end{bmatrix}\). Construct the vector \(\vec{w}\) if \(\vec{w} = 2\vec{v_1} - 3\vec{v_2} + \vec{v_3}\).
Vectors encode both a direction and magnitude
We can multiply vectors by scalars – in this case, each element is multiplied by the scalar
We can add and subtract vectors of the same size
We can move throughout space by scaling and summing vectors together
Goals for Today: Define what is meant by the notion of a linear combination of vectors, identify whether or not a vector can be written as a linear combination of a set of other vectors, and connect linear combinations to what we’ve already learned about linear systems and matrix equations.
Many important properties of linear systems can be described through the lens of vectors.
We encountered vectors and vector operations earlier this semester and the warm-up problems have provided you a reminder of the most important operations we’ll encounter today…
The concept of a linear combination will be extremely important to us in linear algebra. . . .
Definition (Linear Combination of Vectors): The vector \(\vec{y}\) defined by
\[\vec{y} = c_1\vec{v_1} + c_2\vec{v_2} + \cdots + c_p\vec{v_p}\]
is a linear combination of the vectors \(\vec{v_1},~\vec{v_2},~\cdots,~\vec{v_p}\) with weights \(c_1,~c_2,~\cdots,~c_p\).
Example: Let \(\vec{v_1} = \begin{bmatrix} 1\\ 2\\ -1\end{bmatrix}\), \(\vec{v_2} = \begin{bmatrix} 0\\ -3\\ 1\end{bmatrix}\) and \(\vec{v_3} = \begin{bmatrix} -4\\ 1\\ 2\end{bmatrix}\) be vectors. Compute the linear combination \(c_1\vec{v_1} + c_2\vec{v_2} + c_3\vec{v_3}\) where \(c_1 = -2\), \(c_2 = 4\), and \(c_3 = 5\).
The definition of a linear combination leads naturally to the definition of a vector equation.
Definition (Vector Equation): A vector equation is an equation of the form
\[x_1\vec{a_1} + x_2\vec{a_2} + \cdots + x_n\vec{a_n} = \vec{b}\]
Example: Consider the vectors \(\vec{a_1} = \begin{bmatrix} 1\\ 3\\ 1\end{bmatrix}\), \(\vec{a_2} = \begin{bmatrix} -3\\ 2\\ 1\end{bmatrix}\), and \(\vec{b} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\). Determine whether the vector equation \(x_1\begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} + x_2\begin{bmatrix} -3\\ 2\\ 1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\) has solutions (and, if so, what those solutions are).
Example: Determine whether the vector equation \(x_1\begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} + x_2\begin{bmatrix} -3\\ 2\\ 1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\) has solutions (and, if so, what those solutions are).
The vector equation is equivalent to the matrix equation
\[\begin{bmatrix} 1 & -3\\ 3 & 2\\ 1 & 1\end{bmatrix}\begin{bmatrix} x_1\\ x_1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\]
Because of this, we can solve the vector equation in the same way that we solve the matrix equation…
…by row-reducing an augmented matrix
Example: Determine whether the vector equation \(x_1\begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} + x_2\begin{bmatrix} -3\\ 2\\ 1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\) has solutions (and, if so, what those solutions are).
\[\begin{align*} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 3 & 2 & -1\\ 1 & -1 & -7\end{array}\right] \end{align*}\]
Example: Determine whether the vector equation \(x_1\begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} + x_2\begin{bmatrix} -3\\ 2\\ 1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\) has solutions (and, if so, what those solutions are).
\[\begin{align*} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 3 & 2 & -1\\ 1 & -1 & -7\end{array}\right] &\substack{R_2 \leftarrow R_2 + (-3R_1)\\ R_3 \leftarrow R_3 + (-1R_1)\\ \longrightarrow} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 0 & 11 & 44\\ 0 & 2 & 8\end{array}\right] \end{align*}\]
Example: Determine whether the vector equation \(x_1\begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} + x_2\begin{bmatrix} -3\\ 2\\ 1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\) has solutions (and, if so, what those solutions are).
\[\begin{align*} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 3 & 2 & -1\\ 1 & -1 & -7\end{array}\right] &\substack{R_2 \leftarrow R_2 + (-3R_1)\\ R_3 \leftarrow R_3 + (-1R_1)\\ \longrightarrow} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 0 & 11 & 44\\ 0 & 2 & 8\end{array}\right]\\ &\substack{R_2 \leftarrow (1/11)R_2\\ R_3 \leftarrow (1/2)R_3\\ \longrightarrow} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 0 & 1 & 4\\ 0 & 1 & 4\end{array}\right] \end{align*}\]
Example: Determine whether the vector equation \(x_1\begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} + x_2\begin{bmatrix} -3\\ 2\\ 1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\) has solutions (and, if so, what those solutions are).
\[\begin{align*} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 3 & 2 & -1\\ 1 & -1 & -7\end{array}\right] &\substack{R_2 \leftarrow R_2 + (-3R_1)\\ R_3 \leftarrow R_3 + (-1R_1)\\ \longrightarrow} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 0 & 11 & 44\\ 0 & 2 & 8\end{array}\right]\\ &\substack{R_2 \leftarrow (1/11)R_2\\ R_3 \leftarrow (1/2)R_3\\ \longrightarrow} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 0 & 1 & 4\\ 0 & 1 & 4\end{array}\right]\\ &\substack{R_3 \leftarrow R_3 + (-1R_2)\\ \longrightarrow} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 0 & 1 & 4\\ 0 & 0 & 0\end{array}\right] \end{align*}\]
Example: Determine whether the vector equation \(x_1\begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} + x_2\begin{bmatrix} -3\\ 2\\ 1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\) has solutions (and, if so, what those solutions are).
\[\begin{align*} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 3 & 2 & -1\\ 1 & -1 & -7\end{array}\right] &\sim \left[\begin{array}{rr|r} 1 & -3 & -15\\ 0 & 1 & 4\\ 0 & 0 & 0\end{array}\right] \end{align*}\]
Example: Determine whether the vector equation \(x_1\begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} + x_2\begin{bmatrix} -3\\ 2\\ 1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\) has solutions (and, if so, what those solutions are).
\[\begin{align*} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 3 & 2 & -1\\ 1 & -1 & -7\end{array}\right] &\sim \left[\begin{array}{rr|r} 1 & -3 & -15\\ 0 & 1 & 4\\ 0 & 0 & 0\end{array}\right]\\ &\substack{R_1 \leftarrow R_1 + (3R_2)\\ \longrightarrow} \left[\begin{array}{rr|r} 1 & 0 & -3\\ 0 & 1 & 4\\ 0 & 0 & 0\end{array}\right] \end{align*}\]
Example: Determine whether the vector equation \(x_1\begin{bmatrix} 1\\ 3\\ 1\end{bmatrix} + x_2\begin{bmatrix} -3\\ 2\\ 1\end{bmatrix} = \begin{bmatrix} -15\\ -1\\ -7\end{bmatrix}\) has solutions (and, if so, what those solutions are).
\[\begin{align*} \left[\begin{array}{rr|r} 1 & -3 & -15\\ 3 & 2 & -1\\ 1 & -1 & -7\end{array}\right] &\sim \left[\begin{array}{rr|r} 1 & -3 & -15\\ 0 & 1 & 4\\ 0 & 0 & 0\end{array}\right]\\ &\substack{R_1 \leftarrow R_1 + (3R_2)\\ \longrightarrow} \left[\begin{array}{rr|r} 1 & 0 & -3\\ 0 & 1 & 4\\ 0 & 0 & 0\end{array}\right] \end{align*}\]
The reduced-row-echelon matrix above suggests that \(\boxed{~\displaystyle{x_1 = -3 \text{ and } x_2 = 4}~}\) \(_\blacktriangledown\)
In solving the previous example, we noted that the vector equation
\[x_1\vec{a_1} + x_2\vec{a_2} + \cdots + x_n\vec{a_n} = \vec{b}\]
has the same solution set as the augmented coefficient matrix
\[\left[\begin{array}{cccc|c}\vec{a_1} & \vec{a_2} & \cdots & \vec{a_n} & \vec{b}\end{array}\right]\]
That is, the augmented coefficient matrix whose columns are the vectors \(\vec{a_i}\).
This means that vector equations are not new.
For any given vector equation, there is a corresponding…
The tools we’ve used to investigate and find solutions in those scenarios remain directly applicable again here.
Example: Consider the vectors \(\vec{a_1} = \begin{bmatrix} 1\\ 0\\ -2\\ 3\end{bmatrix}\), \(\vec{a_2} = \begin{bmatrix} 3\\ 1\\ 1\\ 0\end{bmatrix}\), \(\vec{a_3} = \begin{bmatrix} 0\\ 2\\ -2\\ -1\end{bmatrix}\), and \(\vec{b} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\).
The vector equation
\[x_1 \begin{bmatrix} 1\\ 0\\ -2\\ 3\end{bmatrix} + x_2 \begin{bmatrix} 3\\ 1\\ 1\\ 0\end{bmatrix} + x_3\begin{bmatrix} 0\\ 2\\ -2\\ -1\end{bmatrix} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\]
Example: Consider the vectors \(\vec{a_1} = \begin{bmatrix} 1\\ 0\\ -2\\ 3\end{bmatrix}\), \(\vec{a_2} = \begin{bmatrix} 3\\ 1\\ 1\\ 0\end{bmatrix}\), \(\vec{a_3} = \begin{bmatrix} 0\\ 2\\ -2\\ -1\end{bmatrix}\), and \(\vec{b} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\).
The vector equation
\[x_1 \begin{bmatrix} 1\\ 0\\ -2\\ 3\end{bmatrix} + x_2 \begin{bmatrix} 3\\ 1\\ 1\\ 0\end{bmatrix} + x_3\begin{bmatrix} 0\\ 2\\ -2\\ -1\end{bmatrix} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\]
is equivalent to the matrix equation
\[\begin{bmatrix} 1 & 3 & 0\\ 0 & 1 & 2\\ -2 & 1 & -2\\ 3 & 0 & -1\end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\]
Example: Consider the vectors \(\vec{a_1} = \begin{bmatrix} 1\\ 0\\ -2\\ 3\end{bmatrix}\), \(\vec{a_2} = \begin{bmatrix} 3\\ 1\\ 1\\ 0\end{bmatrix}\), \(\vec{a_3} = \begin{bmatrix} 0\\ 2\\ -2\\ -1\end{bmatrix}\), and \(\vec{b} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\).
The matrix equation
\[\begin{bmatrix} 1 & 3 & 0\\ 0 & 1 & 2\\ -2 & 1 & -2\\ 3 & 0 & -1\end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\]
Example: Consider the vectors \(\vec{a_1} = \begin{bmatrix} 1\\ 0\\ -2\\ 3\end{bmatrix}\), \(\vec{a_2} = \begin{bmatrix} 3\\ 1\\ 1\\ 0\end{bmatrix}\), \(\vec{a_3} = \begin{bmatrix} 0\\ 2\\ -2\\ -1\end{bmatrix}\), and \(\vec{b} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\).
The matrix equation
\[\begin{bmatrix} 1 & 3 & 0\\ 0 & 1 & 2\\ -2 & 1 & -2\\ 3 & 0 & -1\end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\]
is equivalent to the augmented coefficient matrix
\[\left[\begin{array}{ccc|c} 1 & 3 & 0 & 1\\ 0 & 1 & 2 & 1\\ -2 & 1 & -2 & 2\\ 3 & 0 & 1 & -2\end{array}\right]\]
Example: Consider the vectors \(\vec{a_1} = \begin{bmatrix} 1\\ 0\\ -2\\ 3\end{bmatrix}\), \(\vec{a_2} = \begin{bmatrix} 3\\ 1\\ 1\\ 0\end{bmatrix}\), \(\vec{a_3} = \begin{bmatrix} 0\\ 2\\ -2\\ -1\end{bmatrix}\), and \(\vec{b} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\).
The augmented coefficient matrix
\[\left[\begin{array}{ccc|c} 1 & 3 & 0 & 1\\ 0 & 1 & 2 & 1\\ -2 & 1 & -2 & 2\\ 3 & 0 & 1 & -2\end{array}\right]\]
Example: Consider the vectors \(\vec{a_1} = \begin{bmatrix} 1\\ 0\\ -2\\ 3\end{bmatrix}\), \(\vec{a_2} = \begin{bmatrix} 3\\ 1\\ 1\\ 0\end{bmatrix}\), \(\vec{a_3} = \begin{bmatrix} 0\\ 2\\ -2\\ -1\end{bmatrix}\), and \(\vec{b} = \begin{bmatrix} 1\\ 1\\ 2\\ -2\end{bmatrix}\).
The augmented coefficient matrix
\[\left[\begin{array}{ccc|c} 1 & 3 & 0 & 1\\ 0 & 1 & 2 & 1\\ -2 & 1 & -2 & 2\\ 3 & 0 & 1 & -2\end{array}\right]\]
is equivalent to the system
\[\left\{\begin{array}{rcr} x_1 + 3x_2 & = & 1\\ x_2 + 2x_3 & = & 1\\ -2x_1 + x_2 - 2x_3 & = & 2\\ 3x_1 + x_3 & = & -2\end{array}\right.\]
It’s worth taking a moment to recap the different types of equations and contexts we’ve encountered.
We’ll do that next, with a particular focus on the types of questions we are asking when we write an equation of a particular type.
Vector Equations: \(x_1\vec{v_1} + x_2\vec{v_2} + \cdots x_p\vec{v_p} = \vec{y}\)
Matrix Equations: \(A\vec{x} = \vec{b}\)
Alternatively, we could let \(f: \mathbb{R}^n\to \mathbb{R}^m\) be a function such that \(f\left(\vec{x}\right) = A\vec{x}\).
From this lens, we want to know whether there is some \(\vec{x}\) in the domain of \(f\) such that \(f\left(\vec{x}\right) = \vec{b}\).
Linear Systems: \(\left\{\begin{array}{rcr} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n & = & b_1\\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n & = & b_2\\ & \vdots & \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n & = & b_m\end{array}\right.\)
Example: For the vectors \(\vec{u} = \left[\begin{array}{c}5\\ -3\end{array}\right]\) and \(\vec{v} = \left[\begin{array}{c} -2\\ 1\end{array}\right]\), compute \(\vec{u} + \vec{v}\) and \(\vec{u} - 2\vec{v}\). Draw the vectors \(\vec{u}\), \(\vec{v}\) and the resultant vectors in the plane.
Example: Let \(\vec{u} = \begin{bmatrix} 1\\ 2\end{bmatrix}\), \(\vec{v} = \begin{bmatrix} 3\\ -1\end{bmatrix}\), and \(\vec{w} = \begin{bmatrix} 9\\ 1\end{bmatrix}\). Can \(\vec{w}\) be written as a linear combination of \(\vec{u}\) and \(\vec{v}\)? In order words, do there exist scalars \(c_1\) and \(c_2\) such that \(c_1\vec{u} + c_2\vec{v} = \vec{w}\)?
Example: Find and describe all solutions to the vector equation below, if solutions exist.
\[x_1\begin{bmatrix}1\\ 0\\ 1\\ 0\end{bmatrix} + x_2\begin{bmatrix} 0\\ 1\\ 0\\ 1\end{bmatrix} + x_3\begin{bmatrix} 1\\ 1\\ 1\\ 1\end{bmatrix} = \begin{bmatrix} 3\\ 2\\ 3\\ 2\end{bmatrix}\]
Example: Given the vectors \(\vec{u_1} = \begin{bmatrix} 1\\ 1\\ 0\\ 0\\ 0\end{bmatrix}\), \(\vec{u_2} - \begin{bmatrix} 0\\ 0\\ 1\\ 1\\ 0\end{bmatrix}\), \(\vec{u_3} = \begin{bmatrix} 0\\ 0\\ 0\\ 1\\ 1\end{bmatrix}\), and \(\vec{y} = \begin{bmatrix} 2\\ 2\\ 3\\ 5\\ 2\end{bmatrix}\), find weights that allow you to write \(\vec{y}\) as a linear combination of the vectors \(\vec{u_1}\) , \(\vec{u_2}\), and \(\vec{u_3}\). If such weights do not exist, provide justification for why they do not.
Example: Write a system of linear equations and an augmented coefficient matrix which is equivalent to the vector equation
\[x_1\left[\begin{array}{c}3\\ 0\\ 1\end{array}\right] + x_2\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right] + x_3\left[\begin{array}{c}1\\ 1\\ -2\end{array}\right] = \left[\begin{array}{c}3\\ -5\\ 4\end{array}\right]\]
Use a series of elementary row operations to solve the system.
Try It! 6: Write a vector equation that is equivalent to the system of equations \(\left\{\begin{array}{rcr} 5x_1 + 2x_2 - 4x_3 & = & 20\\ x_1 + 2x_2 + x_3 & = & 6\\ -2x_1 - x_2 + 3x_3 & = & 1\end{array}\right.\).
Use an augmented coefficient matrix and a series of elementary row operations to solve the system and corresponding vector equation.
Example: Determine whether the vector \(\vec{b} = \left[\begin{array}{c} 1\\ 2\\ 5\\ -5\end{array}\right]\) is a linear combination of the vectors \(\vec{v_1} = \left[\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right]\), \(\vec{v_2} = \left[\begin{array}{c}0\\ 1\\ 2\\ -2\end{array}\right]\), and \(\vec{v_3} = \left[\begin{array}{c}0\\ 0\\ -1\\ 1\end{array}\right]\).
To be added…
\[\Huge{\text{Start Homework 4}}\] \[\Huge{\text{on MyOpenMath}}\]
\(\Huge{\text{Matrix Multiplication and}}\)
\(\Huge{\text{Linear Combinations}}\)