From Three-Dimensional Targets to Coefficients
A column-combination question asks whether a target vector can be assembled from a fixed list of vectors. In \(\mathbb{R}^3\), the target has three coordinates, but the number of available columns can be two, three, or another finite number. The calculation is to find coefficients that make all three coordinates agree at once. A match in only one or two coordinates is not enough.
Let the given columns be \(\mathbf{a}_1,\dots,\mathbf{a}_p\in\mathbb{R}^3\), and let \(\mathbf{b}\in\mathbb{R}^3\) be a target. A representation of \(\mathbf{b}\) using these columns is a choice of real coefficients \(x_1,\dots,x_p\) such that \(\mathbf{b}=x_1\mathbf{a}_1+\cdots+x_p\mathbf{a}_p\). Put the columns side by side to form \(\mathbf{A}=\begin{bmatrix}\mathbf{a}_1&\cdots&\mathbf{a}_p\end{bmatrix}\), and put the coefficients into the column vector \(\mathbf{x}=(x_1,\dots,x_p)^\top\).
This equivalence follows directly from matrix multiplication: the product \(\mathbf{A}\mathbf{x}\) is the sum of the columns weighted by the entries of \(\mathbf{x}\). For example, with three columns, the equations in coordinates are \(x_1a_{11}+x_2a_{12}+x_3a_{13}=b_1\), \(x_1a_{21}+x_2a_{22}+x_3a_{23}=b_2\), and \(x_1a_{31}+x_2a_{32}+x_3a_{33}=b_3\). All equations use the same coefficients. The main task is therefore to solve the coefficient system, not to guess separate coefficients for separate coordinates.
Proof. Write \(\mathbf{x}=(x_1,\dots,x_p)^\top\). By the definition of the matrix-vector product, \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_p\mathbf{a}_p\). If \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for some \(\mathbf{x}\), this equality expresses \(\mathbf{b}\) as a combination of the columns. Conversely, if \(\mathbf{b}=x_1\mathbf{a}_1+\cdots+x_p\mathbf{a}_p\), putting those coefficients into \(\mathbf{x}\) gives \(\mathbf{A}\mathbf{x}=\mathbf{b}\). Thus each condition implies the other. \(\square\)
Solving a Target System in \(\mathbb{R}^3\)
Worked Example: A Target That Is a Combination
Determine whether \(\mathbf{b}=(2,-1,1)^\top\) is a combination of \(\mathbf{a}_1=(1,0,1)^\top\) and \(\mathbf{a}_2=(0,1,1)^\top\). If it is, find the coefficients.
Set \(x_1\mathbf{a}_1+x_2\mathbf{a}_2=\mathbf{b}\). Matching coordinates gives
The first two equations require \(x_1=2\) and \(x_2=-1\). Substitution into the third gives \(2+(-1)=1\), so all three equations hold. Therefore the target is a combination. In matrix form, the coefficients are \(\mathbf{x}=(2,-1)^\top\).
Check: Multiplying the coefficients by the columns gives \(2(1,0,1)^\top-1(0,1,1)^\top=(2,0,2)^\top-(0,1,1)^\top=(2,-1,1)^\top\), which is the target.
When a system does not have a solution, no choice of coefficients can represent the target. Row reduction makes the obstruction visible. The coefficient columns must be retained in their original order, and the target is appended as the final column of the augmented matrix. Row operations simplify the equations of the system; they do not change which coefficient vectors solve it.
For the last sentence, choose any values for variables without pivots, such as zero, then solve the pivot equations from the bottom upward. If no row requires \(0=c\) with \(c\ne0\), each remaining equation either determines a pivot variable or is an all-zero equation \(0=0\). This gives a solution.
Worked Example: A Target That Is Not a Combination
Use the same two columns as in the previous example, but take the target \(\mathbf{c}=(1,1,3)^\top\). Determine whether \(\mathbf{c}\) is a combination of the columns.
The coefficient equations are \(x_1=1\), \(x_2=1\), and \(x_1+x_2=3\). The first two equations force \(x_1+x_2=2\), which contradicts the third equation. Row reduction displays the contradiction:
Here the final column is the augmented right-hand side. The last row means \(0=1\), so there is no solution and \(\mathbf{c}\) is not a combination of the two columns.
Check: The first coordinate of a combination \(x_1\mathbf{a}_1+x_2\mathbf{a}_2\) must equal \(x_1\), and the second must equal \(x_2\). To match \(\mathbf{c}\), these would have to be \(1\) and \(1\); then the third coordinate would be \(1+1=2\), not \(3\). This confirms the inconsistency using the original columns.
When a Consistent Target Has More Than One Representation
A consistent coefficient system can have one solution or several. The distinction is determined by the variables after reduction, but consistency must be checked first. A variable without a pivot is called a free variable: it can be chosen freely in a consistent system, and the pivot variables are then obtained from the pivot equations.
Proof. Assume the system is consistent. If every variable has a pivot, each pivot equation determines its variable from variables in later columns; because there are no free variables, solving from the last pivot upward determines every variable uniquely. Hence there is exactly one solution. If at least one variable is free, assign arbitrary values to the free variables and use the pivot equations to determine the pivot variables. Since the system is consistent, this produces solutions. Choose two different values for one free variable while keeping the other free variables fixed; the resulting coefficient vectors differ in that coordinate, so they are distinct solutions. Thus there is more than one solution. \(\square\)
Worked Example: A Target with Multiple Representations
Let \(\mathbf{a}_1=(1,0,1)^\top\), \(\mathbf{a}_2=(0,1,1)^\top\), and \(\mathbf{a}_3=(1,1,2)^\top\). Find all coefficient vectors representing \(\mathbf{b}=(2,3,5)^\top\).
The system is \(x_1+x_3=2\), \(x_2+x_3=3\), and \(x_1+x_2+2x_3=5\). Its augmented matrix reduces as follows:
The system is consistent, and \(x_3\) is free. Set \(x_3=t\), where \(t\in\mathbb{R}\). The first two rows give \(x_1=2-t\) and \(x_2=3-t\). Thus all representations are given by \((x_1,x_2,x_3)=(2-t,3-t,t)\).
Check: For any real \(t\), the first coordinate of the combination is \((2-t)+t=2\), the second is \((3-t)+t=3\), and the third is \((2-t)+(3-t)+2t=5\). For instance, \(t=0\) gives coefficients \((2,3,0)\), while \(t=1\) gives \((1,2,1)\); both represent \((2,3,5)^\top\).
A Three-Column System with No Solution
Having three columns in \(\mathbb{R}^3\) does not guarantee that every target can be represented. The coefficient equations still have to be mutually consistent. The next example uses the same columns as the previous one, so the change in outcome comes from the target, not from the available columns.
Worked Example: An Inconsistent Target with Three Columns
Using \(\mathbf{a}_1=(1,0,1)^\top\), \(\mathbf{a}_2=(0,1,1)^\top\), and \(\mathbf{a}_3=(1,1,2)^\top\), test whether \(\mathbf{d}=(2,3,6)^\top\) is a combination.
The coefficient system is \(x_1+x_3=2\), \(x_2+x_3=3\), and \(x_1+x_2+2x_3=6\). Augment and row reduce:
The final row is \(0=1\), so the system is inconsistent. Therefore \(\mathbf{d}\) is not a combination of the columns. The first two equations would force the third coordinate of the combination to be \((x_1+x_3)+(x_2+x_3)=2+3=5\), whereas the target has third coordinate \(6\).
Why the Row-Reduction Test Works
The contradiction in the examples is not an artifact of a particular sequence of operations. Each elementary row operation preserves the solution set of the augmented system. Swapping two rows only changes the order of equations. Multiplying one row by a nonzero scalar replaces an equation by an equivalent one. Adding a multiple of one row to another replaces an equation by a consequence of the two equations and can be reversed by subtracting that multiple. Consequently, a contradiction reached by row reduction was already an obstruction in the original coefficient system.
Proof. Each elementary row operation is reversible and preserves the solutions, so the original system and the echelon system have the same solutions. If the echelon system contains a row \([0\ \cdots\ 0\mid c]\) with \(c\ne0\), that row asserts \(0=c\), which no coefficient vector can satisfy; therefore neither system has a solution. Conversely, suppose there is no such row. In echelon form, every row with a nonzero coefficient has a leading variable, and any row with all zero coefficients must have zero on the right as well. Assign zero to each free variable and solve the leading-variable equations from bottom to top. Each such equation determines its leading variable, and the all-zero rows impose no further condition. This constructs a solution. Hence the absence of a contradictory row implies consistency, proving both directions. \(\square
This result gives a reliable routine for column-combination problems in \(\mathbb{R}^3\): arrange the given vectors as columns, append the target, reduce the augmented matrix, and inspect the equations. If the system is inconsistent, stop: there is no representation. If it is consistent, solve for coefficients and substitute them back into the original columns. Only after confirming consistency should free variables be interpreted as producing multiple representations.
Common Pitfall: A Free Variable Does Not Repair Inconsistency
Consider again the inconsistent system from the three-column example. Its echelon form contains \(x_1+x_3=2\), \(x_2+x_3=3\), and \(0=1\). A wrong conclusion would be: “\(x_3\) is free, so there are multiple representations.” That reasoning ignores the last equation. Setting \(x_3=0\) gives \(x_1=2\) and \(x_2=3\), but the third coordinate of the resulting combination is \(2+3=5\), not the target coordinate \(6\); the equation \(0=1\) remains unsatisfied. No value of \(x_3\) can fix it.
The correct statement includes consistency: if the system is consistent, then it has a unique solution when every variable is determined, and multiple solutions when at least one variable is free. An inconsistent system has no representations, whether or not its coefficient part has free variables.
Check Your Understanding
For each question, write the coefficient system before drawing a conclusion. Verify any proposed representation by substituting into the original columns.
- Is \((3,2,5)^\top\) a combination of \((1,0,1)^\top\) and \((0,1,1)^\top\)? If so, find the coefficients.
- For the same two columns, explain why \((1,1,4)^\top\) is not a combination.
- Using columns \((1,0,1)^\top\), \((0,1,1)^\top\), and \((1,1,2)^\top\), find all representations of \((1,2,3)^\top\).
- What does a row \([0\ 0\ 0\mid -2]\) in an augmented echelon form tell you about the target?
- Explain why the presence of a free variable alone does not prove that a target has multiple representations.
