Definition. Given vectors \( \bbm v_1, \bbm v_2, \ldots, \bbm v_p \in \mathbb R^n \), a linear combination of these vectors is any vector of the form \( c_1 \bbm v_1 + c_2 \bbm v_2 + \cdots + c_p \bbm v_p, \) where \( c_1, c_2, \ldots, c_p \) are scalars.
Example 1. If \( \bbm u = \vecthree 0 {-3} 5 \) and \( \bbm v = \vecthree 1 4 7 \), then \( 2\bbm u + 3\bbm v = \vecthree 3 6 {31} \) is a linear combination of \( \bbm u \) and \( \bbm v \).
Example 2. If \( \bbm p = \vectwo 1 0 \), \( \bbm q = \vectwo {-2} {-2} \), and \( \bbm r = \vectwo {-4} 8 \), then \( -1\bbm p + 0\bbm q + \frac 1 2 \bbm r = \vectwo {-3} 4 \) is a linear combination of \( \bbm p \), \( \bbm q \), and \( \bbm r \).
Note that any or all of the scalars in the linear combination are allowed to be zero. This means that the zero vector \( \bbm 0 \) is always a linear combination of any given collection of vectors.
Definition. A vector equation has the form \( x_1 \bbm v_1 + x_2 \bbm v_2 + \cdots + x_p \bbm v_p = \bbm b \), where \( \bbm v_1, \bbm v_2, \ldots, \bbm v_p\) and \( \bbm b\) are vectors in \( \mathbb R^n \). A solution of this vector equation is a value for each variable \( x_1, x_2, \ldots, x_p \) that makes the left-hand side equal the right-hand side.
Example 3. Solve the vector equation \[ x_1 \vecthree 2 {-1} 0 + x_2 \vecthree 1 0 {-3} = \vecthree 4 {-1} {-6}. \]
Working out the vector operations on the left-hand sides gives the equation \( \vecthree {2x_1 + x_2} {-x_1} {-3x_2} = \vecthree 4 {-1} {-6} \), which can be thought of as a system of three linear equations. Solving this system in the normal way gives the solution \( x_1 = 1 \) and \( x_2 = 2 \). \( \Box \)
Example 3 illustrates that we can set up an augmented matrix to solve a vector equation \( x_1 \bbm v_1 + x_2 \bbm v_2 + \cdots + x_p \bbm v_p = \bbm b \). The columns of this augmented matrix are \( \bbm v_1, \bbm v_2, \ldots, \bbm v_p \) and \( \bbm b \).
Consider again the vector equation in Example 3. We have two equivalent questions that arise given this vector equation:
In general, given any vector equation \( x_1 \bbm v_1 + x_2 \bbm v_2 + \cdots + x_p \bbm v_p = \bbm b \), the two equivalent questions are:
Example 4. Let \( \bbm v_1 = \vecthree 1 {-2} {-5} \), \( \bbm v_2 = \vecthree 2 5 6 \), and \( \bbm b = \vecthree 7 4 {-3} \). Is \( \bbm b\) a linear combination of \( \bbm v_1 \) and \( \bbm v_2 \)?
To solve this problem, we first consider the equivalent question: Does the equation \( x_1 \bbm v_1 + x_2 \bbm v_2 = \bbm b \) have a solution? As we saw in Example 3, we can construct an augmented matrix for this question whose columns are \( \bbm v_1 \), \( \bbm v_2 \), and \( \bbm b \): \[ \begin{bmatrix} 1 & 2 & 7 \\ -2 & 5 & 4 \\ -5 & 6 & -3 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \]
The row-reduced augmented matrix shows us that, since there is no pivot in the last column, the vector equation is consistent. This tells us that, yes, \( \bbm b \) is a linear combination of \( \bbm v_1 \) and \( \bbm v_2 \). \( \Box \)
Note that, in our solution to Example 4, we actually had more information: we discovered exactly how to construct \( \bbm b \) as a linear combination of \( \bbm v_1 \) and \( \bbm v_2 \). However, the question did not ask us for that information. This illustrates an important idea: Always answer the question you are being asked.
Example 5. Let \( \bbm w_1 = \vecthree 1 {-2} 0 \), \( \bbm w_2 = \vecthree 0 1 1 \), \( \bbm w_3 = \vecthree 2 {-7} {-3} \), and \( \bbm b = \vecthree {-1} 0 3 \). Is \( \bbm b \) a linear combination of \( \bbm w_1 \), \( \bbm w_2 \), and \( \bbm w_3 \)?
To answer this question, we consider whether the equation \( x_1 \bbm w_1 + x_2 \bbm w_2 + x_3 \bbm w_3 = \bbm b \) is consistent. To solve this equation, we set up the corresponding augmented matrix and row-reduce: \[ \begin{bmatrix} 1 & 0 & 2 & -1 \\ -2 & 1 & -7 & 0 \\ 0 & 1 & -3 & 3 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
Since there is a pivot in the last column of our augmented matrix, the vector equation has no solutions. This means that, no, \( \bbm b \) is not a linear combination of \( \bbm w_1 \), \( \bbm w_2 \), and \( \bbm w_3 \). \( \Box \)
The previous examples illustrate a good format for your solutions that you should emulate in your own writing. Specifically, make sure to include the following elements in your written solutions:
Many students, in their initial attempts at these kinds of solutions, will omit items (1) and (3) above and only show a row-reduction process. It is very important to explain what that matrix represents, and how the result of the row-reduction provides information about the question being considered. Look back over the examples you have seen and adapt your writing style to make sure that you include these essential elements.
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