One System, Three Questions
Purpose: a short summary of Week 1. The same computation (row reduction of the augmented matrix) answers every question we asked this week about a linear system. This note shows that on one example, then works one more example in the style of Homework Question 3.
1. A 3×2 linear system
Start with three linear equations in two unknowns:
\[\begin{aligned} x_1 + 2x_2 &= 5\\ 2x_1 - x_2 &= 0\\ x_1 - 3x_2 &= -5 \end{aligned}\]Its augmented matrix is
\[\left[\begin{array}{cc|c} 1 & 2 & 5\\ 2 & -1 & 0\\ 1 & -3 & -5 \end{array}\right].\]Read column by column, the same system is the vector equation
\[x_1\underbrace{\begin{bmatrix}1\\2\\1\end{bmatrix}}_{\mathbf a_1} + x_2\underbrace{\begin{bmatrix}2\\-1\\-3\end{bmatrix}}_{\mathbf a_2} = \underbrace{\begin{bmatrix}5\\0\\-5\end{bmatrix}}_{\mathbf b}.\]The columns of the coefficient matrix are the coefficient vectors \(\mathbf a_1, \mathbf a_2\), and the right-hand side is the constant vector \(\mathbf b\).
2. The questions we can ask about it
This week the same system appeared in three different phrasings. Each one is a question you may see on a quiz or an exam.
Does the system have a solution? If yes, how many: exactly one, or infinitely many? What are they?
Is \(\mathbf b\) a linear combination of \(\mathbf a_1\) and \(\mathbf a_2\)? That is, do weights \(x_1, x_2\) exist with \(x_1\mathbf a_1 + x_2\mathbf a_2 = \mathbf b\)? If yes, find them.
Is \(\mathbf b\) in \(\operatorname{Span}\{\mathbf a_1, \mathbf a_2\}\)?
By definition, \(\operatorname{Span}\{\mathbf a_1,\mathbf a_2\}\) is the set of all linear combinations of \(\mathbf a_1,\mathbf a_2\). So Question C is Question B in different words. And the vector equation \(x_1\mathbf a_1 + x_2\mathbf a_2 = \mathbf b\) has exactly the same solutions as the system with augmented matrix \([\mathbf a_1\ \mathbf a_2 \mid \mathbf b]\). So Question B is Question A in different words. Therefore:
system is consistent \(\Longleftrightarrow\) \(\mathbf b\) is a linear combination of \(\mathbf a_1,\mathbf a_2\) \(\Longleftrightarrow\) \(\mathbf b \in \operatorname{Span}\{\mathbf a_1,\mathbf a_2\}\)
All three are answered by one computation: row reduce the augmented matrix.
3. The one computation: row reduce
Forward phase. Pivot in row 1, column 1. Clear below it:
\(R_2 \to R_2 - 2R_1\) and \(R_3 \to R_3 - R_1\):
\[\left[\begin{array}{cc|c} 1 & 2 & 5\\ 0 & -5 & -10\\ 0 & -5 & -10 \end{array}\right]\]Next pivot in row 2, column 2. Clear below it: \(R_3 \to R_3 - R_2\):
\[\left[\begin{array}{cc|c} 1 & 2 & 5\\ 0 & -5 & -10\\ 0 & 0 & 0 \end{array}\right]\]This is an echelon form. Backward phase: scale \(R_2 \to -\tfrac15 R_2\), then clear above the pivot with \(R_1 \to R_1 - 2R_2\):
\[\left[\begin{array}{cc|c} 1 & 0 & 1\\ 0 & 1 & 2\\ 0 & 0 & 0 \end{array}\right]\]This is the reduced echelon form. Pivot columns: 1 and 2. No row of the form \([\,0\ 0 \mid b\,]\) with \(b \ne 0\).
4. Reading the answers from the RREF
There is no row \([\,0\ 0 \mid b\,]\) with \(b\neq 0\), so the system is consistent. Both variables are basic (both columns are pivot columns), so there is no free variable and the solution is unique: \(x_1 = 1\), \(x_2 = 2\).
Check in the original system: \(1 + 4 = 5\) ✓, \(2 - 2 = 0\) ✓, \(1 - 6 = -5\) ✓.
Geometrically: three lines in the plane pass through the same point \((1,2)\). The zero row \([\,0\ 0 \mid 0\,]\) tells us the third equation gave no new information; it is a combination of the first two.
Yes. The weights are the solution we just found: \(x_1 = 1\), \(x_2 = 2\), so
\[\mathbf b = 1\cdot\mathbf a_1 + 2\cdot\mathbf a_2, \qquad \text{check: } \begin{bmatrix}1\\2\\1\end{bmatrix} + 2\begin{bmatrix}2\\-1\\-3\end{bmatrix} = \begin{bmatrix}5\\0\\-5\end{bmatrix} ✓\]Because the solution is unique, these are the only weights that work.
Yes, \(\mathbf b \in \operatorname{Span}\{\mathbf a_1,\mathbf a_2\}\), because it is a linear combination of them.
Geometrically: \(\mathbf a_1\) and \(\mathbf a_2\) are not multiples of each other, so their span is a plane through the origin in \(\mathbb{R}^3\). The vector \(\mathbf b\) lies on that plane.
If the RREF had contained a row \([\,0\ 0 \mid b\,]\) with \(b \ne 0\), then all three answers would flip at once: the system would be inconsistent, \(\mathbf b\) would not be a linear combination of \(\mathbf a_1,\mathbf a_2\), and \(\mathbf b\) would not be in their span. There is no situation where one answer is yes and another is no.
5. Worked example: a system with parameters
This example is in the spirit of Homework Question 3, but with two parameters instead of one. This makes all three outcomes possible: no solution, exactly one solution, infinitely many solutions.
Consider the system \[\begin{aligned} x_1 + h\,x_2 &= 2\\ 3x_1 + 6x_2 &= k \end{aligned}\] where \(h\) and \(k\) are constants.
(a) For which values of \(h\) and \(k\) does the system have no solution?
(b) For which values does it have exactly one solution?
(c) For which values does it have infinitely many solutions?
Solution. One row operation answers all three parts. The augmented matrix is \(\left[\begin{array}{cc|c} 1 & h & 2\\ 3 & 6 & k \end{array}\right]\). Apply \(R_2 \to R_2 - 3R_1\):
\[\left[\begin{array}{cc|c} 1 & h & 2\\ 0 & 6 - 3h & k - 6 \end{array}\right]\]This is an echelon form as long as we know whether \(6 - 3h\) is zero or not. So the analysis splits on \(h = 2\).
Case \(h \ne 2\). Then \(6 - 3h \ne 0\), so it is a pivot. Both columns are pivot columns, there is no free variable, and the last row is not of the form \([\,0\ 0 \mid b\,]\). The system is consistent with exactly one solution, for every value of \(k\). Solving: from row 2, \(x_2 = \dfrac{k-6}{6-3h}\), and from row 1, \(x_1 = 2 - h\,x_2\).
Case \(h = 2\). The second row becomes \([\,0\ 0 \mid k - 6\,]\), which says \(0 = k - 6\).
- If \(k \ne 6\), this is false. The system has no solution.
- If \(k = 6\), the row is \([\,0\ 0 \mid 0\,]\), always true. The only pivot is in column 1, so \(x_2\) is free and the system has infinitely many solutions: \(x_1 = 2 - 2x_2\), \(x_2\) free.
Answers.
(a) No solution: \(h = 2\) and \(k \ne 6\).
(b) Exactly one solution: \(h \ne 2\) (any \(k\)).
(c) Infinitely many solutions: \(h = 2\) and \(k = 6\).