Linear Equations in Linear Algebra
Topics: Systems of linear equations (§1.1) · Row reduction and echelon forms (§1.2) · Vector equations (§1.3). Textbook: Lay, Lay & McDonald, Linear Algebra and Its Applications, 6th ed.
By the end of this week you will be able to:
- recognize linear equations and linear systems, and explain what a solution set is;
- write a system in augmented matrix form and apply the three elementary row operations;
- row reduce a matrix to echelon form and reduced echelon form, and locate its pivot positions;
- write the general solution of a consistent system using basic and free variables;
- decide whether a vector is a linear combination of given vectors, and describe a span geometrically.
Day 1 — What This Course Is About, and §1.1: Systems of Linear Equations
Today: the big picture of the whole course, the two words in its name, and our first look at linear systems — what they are, what their solutions look like, and how we start solving them.
The one goal of this course
This entire course is built around a single practical goal: to solve systems of linear equations effectively. That sounds modest, but it is not. We will start this week with systems you could solve in high school, and by the end of the semester the same basic question — “what are all the solutions?” — will lead us to matrices, vector spaces, determinants, and eigenvalues. The equipment gets more powerful because the systems get bigger and the questions get deeper. Keep this in mind whenever a new concept appears: it exists because it helps us understand or solve linear systems better.
Why “linear”, why “algebra”?
Linear comes from the Latin linea, “line”. An equation like \(x_1 + 2x_2 = 4\) describes a line in the plane; with three variables, \(x_1 + 2x_2 - x_3 = 4\) describes a flat plane in space. “Linear” means the unknowns appear in the simplest possible way: multiplied by constants and added. No products of unknowns, no squares, no square roots, no unknowns inside functions like \(\sin\) or \(e^x\). Flat objects only.
Algebra comes from the Arabic al-jabr, a word used by the 9th-century mathematician al-Khwarizmi for the operation of restoring and balancing equations — moving terms across the equals sign by legal steps. Algebra is the art of transforming equations by rules that do not change their solutions.
So “linear algebra” is exactly what it says: a systematic set of legal moves for handling many flat equations at once.
Where this shows up in real life
In textbook exercises our systems will have 2 to 5 unknowns. In applications, they routinely have thousands or millions:
- Economics. In 1949, Wassily Leontief modeled the U.S. economy as a linear system, one equation per economic sector. Harvard's Mark II computer needed 56 hours to solve a 42×42 version of it. That computation helped open the era of large-scale mathematical modeling, and Leontief later received a Nobel Prize for this line of work.
- Engineering and exploration. Ships searching for offshore oil solve thousands of linear systems per day to interpret seismic data.
- Networks. Traffic flow in a road network and current in an electrical circuit are both governed by linear balance equations at each node.
- Computing. Computer graphics, search engine ranking, and the training of machine learning models all reduce, at their core, to enormous linear algebra computations.
None of these are solved by hand. But the method computers use is exactly the one you will learn this week, just automated. The techniques we develop later in the semester exist to answer harder questions about such systems: When is a solution guaranteed to exist? When is it unique? How sensitive is it to errors in the data?
A linear equation in the variables \(x_1,\dots,x_n\) is an equation that can be written in the form \[a_1x_1 + a_2x_2 + \cdots + a_nx_n = b,\] where the coefficients \(a_1,\dots,a_n\) and the number \(b\) are given constants.
A system of linear equations (a linear system) is a collection of one or more linear equations in the same variables.
A solution of the system is a list \((s_1,\dots,s_n)\) of numbers that makes every equation true when substituted for \(x_1,\dots,x_n\). The set of all solutions is the solution set. Two systems are equivalent if they have the same solution set.
Decide whether each equation is linear.
(a) \(4x_1 - 5x_2 + 2 = x_1\)
(b) \(x_2 = 2(\sqrt{6}-x_1)+x_3\)
(c) \(4x_1 - 5x_2 = x_1x_2\)
(d) \(x_2 = 2\sqrt{x_1}-6\)
Solution.
(a) Move \(x_1\) to the left and the constant to the right: \(3x_1 - 5x_2 = -2\). This matches the form \(a_1x_1+a_2x_2=b\), so it is linear. The point: an equation does not have to look like the standard form; it only has to be rearrangeable into it.
(b) Expand and rearrange: \(2x_1 + x_2 - x_3 = 2\sqrt 6\). The \(\sqrt 6\) is fine — it is a constant, not a variable under a root. Linear.
(c) The term \(x_1x_2\) multiplies two unknowns together. No rearrangement can remove it. Not linear.
(d) Here a variable sits under a square root. Not linear.
What can a solution set look like?
With two unknowns, each equation is a line, and solving the system means finding the points common to all the lines. Two lines in the plane can cross at one point, be parallel, or be the same line — Figure 1 shows all three situations. This picture suggests a general fact, which we will actually prove in §1.2:
A linear system has either no solution, exactly one solution, or infinitely many solutions. It can never have, say, exactly two or exactly five. A system with at least one solution is called consistent; a system with no solution is inconsistent.
Solve the system \[\begin{aligned}x_1 - 2x_2 &= -1\\ -x_1 + 3x_2 &= 3\end{aligned}\]
Solution. The strategy: use one equation to eliminate a variable from the other, producing simpler equivalent systems at every step.
Step 1. Add equation 1 to equation 2, and keep equation 1 unchanged. This is legal: any solution of the old pair also satisfies the new pair, and the move is reversible (subtract equation 1 back). The new system is \[\begin{aligned}x_1 - 2x_2 &= -1\\ x_2 &= 2\end{aligned}\]
Step 2. Now the second equation hands us \(x_2 = 2\). Add 2 times equation 2 to equation 1 to eliminate \(x_2\): \(x_1 = 3\).
Step 3 (always do this). Check in the original system: \(3 - 2(2) = -1\) ✓ and \(-3 + 3(2) = 3\) ✓. The solution set is the single point \((3,2)\): the two lines cross exactly there.
Matrix notation and row operations
Notice that in Example 2 the variable names \(x_1, x_2\) did no work — only the coefficients mattered. So we drop the variables and record the numbers in a rectangular array.
For a linear system, the coefficient matrix contains the coefficients of the variables, and the augmented matrix additionally contains the right-hand side as a last column. For Example 2: \[\text{coefficient: }\begin{bmatrix}1 & -2\\ -1 & 3\end{bmatrix},\qquad \text{augmented: }\left[\begin{array}{cc|c}1 & -2 & -1\\ -1 & 3 & 3\end{array}\right]\]
The three elementary row operations are:
- Replacement: add a multiple of one row to another row;
- Interchange: swap two rows;
- Scaling: multiply a row by a nonzero constant.
Two matrices are row equivalent if one can be turned into the other by a sequence of these operations.
Each elementary row operation is reversible, and none of them changes the solution set. Therefore: if the augmented matrices of two systems are row equivalent, the systems have exactly the same solutions. This single fact is the license for everything we do this semester.
Solve \[\begin{aligned}x_1 - 2x_2 + x_3 &= 0\\ 2x_2 - 8x_3 &= 8\\ 5x_1 \phantom{{}-2x_2} - 5x_3 &= 10\end{aligned}\]
Solution. Work on the augmented matrix \(\left[\begin{array}{ccc|c}1&-2&1&0\\0&2&-8&8\\5&0&-5&10\end{array}\right]\).
Step 1. Eliminate the 5 in row 3 using row 1: replace \(R_3\) by \(R_3 - 5R_1\), giving row 3: \(\left[\begin{array}{ccc|c}0&10&-10&10\end{array}\right]\).
Step 2. Scale \(R_2\) by \(\tfrac12\) (nonzero constant, so legal) to get \(\left[\begin{array}{ccc|c}0&1&-4&4\end{array}\right]\), then replace \(R_3\) by \(R_3 - 10R_2\): row 3 becomes \(\left[\begin{array}{ccc|c}0&0&30&-30\end{array}\right]\), i.e. \(30x_3=-30\), so \(x_3 = -1\).
Step 3. Substitute back upward: \(x_2 - 4(-1) = 4\) gives \(x_2 = 0\); then \(x_1 - 2(0) + (-1) = 0\) gives \(x_1 = 1\).
Check in the original system: \(1-0-1=0\) ✓, \(0+8=8\) ✓, \(5+5=10\) ✓. Unique solution \((1,0,-1)\).
Notice the shape of the strategy: create zeros below the diagonal (a “triangular” pattern), then substitute back. In §1.2 we turn this instinct into a precise algorithm that works on any system of any size — including the ones with no solution or infinitely many.
For every linear system we ask:
- Existence: Is the system consistent — does at least one solution exist?
- Uniqueness: If a solution exists, is it the only one?
These two questions will reappear all semester in different forms: this week through row reduction, and later through matrix equations, invertibility, and dimension.
Day 2 — §1.2: Row Reduction and Echelon Forms
Today: the algorithm. We define the target shapes (echelon and reduced echelon form), give the step-by-step procedure that reaches them, and read off complete answers to the existence and uniqueness questions.
(1) Every linear system is recorded by its augmented matrix: coefficients plus the right-hand side as a last column. (2) The three elementary row operations are: add a multiple of one row to another (replacement), swap two rows (interchange), multiply a row by a nonzero constant (scaling). (3) These operations never change the solution set.
Solving systems by improvising with row operations does not scale: a large system needs a procedure that always works and always terminates. That procedure is row reduction (also called Gaussian elimination), and it aims at a specific target shape.
The leading entry of a nonzero row is its leftmost nonzero entry. A matrix is in echelon form if:
- all rows consisting entirely of zeros are at the bottom;
- each leading entry is in a column strictly to the right of the leading entry of the row above it;
- every entry below a leading entry is zero.
It is in reduced echelon form (RREF) if additionally:
- every leading entry is 1;
- each leading 1 is the only nonzero entry in its column.
The word echelon means a staircase formation: the leading entries step down and to the right. With the same pivot pattern, the two forms look like \[\underbrace{\begin{bmatrix} \blacksquare & * & * & *\\ 0 & \blacksquare & * & *\\ 0 & 0 & 0 & \blacksquare\end{bmatrix}}_{\text{echelon form}} \qquad \underbrace{\begin{bmatrix} 1 & 0 & * & 0\\ 0 & 1 & * & 0\\ 0 & 0 & 0 & 1\end{bmatrix}}_{\text{reduced echelon form}}\] where \(\blacksquare\) marks a nonzero leading entry and \(*\) can be anything. Note the difference: in the reduced form every leading entry is 1, and the entries above each leading 1 have also been cleared to 0 — only the \(*\) positions, which sit in non-pivot columns, remain arbitrary.
A matrix may be row reduced to many different echelon forms, depending on the order of your operations. But it is row equivalent to exactly one matrix in reduced echelon form. This is why the RREF of \(A\) is well defined, and why two people who row reduce correctly must end at the same RREF even if their intermediate steps differ.
A pivot position of a matrix \(A\) is a position that holds a leading 1 in the RREF of \(A\). A pivot column is a column containing a pivot position. During row reduction, the nonzero number used in a pivot position to clear out other entries is called a pivot.
Which of these are in echelon form? In reduced echelon form? \[A=\begin{bmatrix}2&1&0\\0&3&4\\0&0&0\end{bmatrix}\quad B=\begin{bmatrix}1&0&5\\0&1&2\\0&0&0\end{bmatrix}\quad C=\begin{bmatrix}0&1&2\\1&0&3\end{bmatrix}\]
Solution.
\(A\): zero row at the bottom ✓, leading entries (2, then 3) step right ✓, zeros below leading entries ✓ — echelon form. But the leading entries are not 1, so not reduced.
\(B\): all echelon conditions hold, leading entries are 1, and each leading 1 is alone in its column — reduced echelon form.
\(C\): the leading entry of row 2 (in column 1) is to the left of the leading entry of row 1 (in column 2) — neither. One row swap would fix it.
The row reduction algorithm
The algorithm has a forward phase (reach echelon form, working left to right) and a backward phase (polish to RREF, working right to left):
- Find the leftmost nonzero column; that is the current pivot column.
- If needed, interchange rows so the pivot position (top of the working part) is nonzero.
- Use replacement operations to create zeros below the pivot.
- Ignore the row containing the pivot (and everything above it), and repeat steps 1–3 on the remaining rows. When no rows remain, the matrix is in echelon form.
- Backward phase: starting from the rightmost pivot, scale each pivot row so the pivot is 1, and create zeros above each pivot.
Row reduce the augmented matrix of a system to RREF and solve: \[\left[\begin{array}{ccc|c}0&1&-4&8\\2&-3&2&1\\4&-8&12&1\end{array}\right]\]
Solution.
Step 1–2. Column 1 is the first pivot column, but its top entry is 0. Interchange \(R_1 \leftrightarrow R_2\) so the pivot is the nonzero 2: \[\left[\begin{array}{ccc|c}2&-3&2&1\\0&1&-4&8\\4&-8&12&1\end{array}\right]\]
Step 3. Clear below the pivot: \(R_3 \to R_3 - 2R_1\): \[\left[\begin{array}{ccc|c}2&-3&2&1\\0&1&-4&8\\0&-2&8&-1\end{array}\right]\]
Step 4. Cover row 1. In the remaining rows, the pivot column is column 2, pivot 1 in row 2. Clear below it: \(R_3 \to R_3 + 2R_2\): \[\left[\begin{array}{ccc|c}2&-3&2&1\\0&1&-4&8\\0&0&0&15\end{array}\right]\]
Read the last row. It says \(0x_1 + 0x_2 + 0x_3 = 15\), i.e. \(0 = 15\). No choice of \(x_1,x_2,x_3\) can make this true, so the system is inconsistent — and we can stop; no backward phase is needed to answer the question. This is typical: the echelon form already answers existence; the RREF is needed when you want to write out the solutions.
Reading solutions from the RREF: basic and free variables
Suppose the augmented matrix of a consistent system has been reduced. A variable is a basic variable if its column is a pivot column, and a free variable otherwise. Free variables can be assigned any value; the basic variables are then determined by the equations.
A system's augmented matrix reduces to \(\left[\begin{array}{ccc|c}1&0&-5&1\\0&1&1&4\\0&0&0&0\end{array}\right]\). Describe all solutions.
Solution. Pivot columns: 1 and 2. So \(x_1, x_2\) are basic and \(x_3\) is free. The nonzero rows say \(x_1 - 5x_3 = 1\) and \(x_2 + x_3 = 4\). Solve each for its basic variable: \[\begin{cases}x_1 = 1 + 5x_3\\ x_2 = 4 - x_3\\ x_3 \text{ free}\end{cases}\] This is the general solution: a recipe producing every solution exactly once. Each value of \(x_3\) gives one solution — for instance \(x_3=0\) gives \((1,4,0)\), and \(x_3=2\) gives \((11,2,2)\). Infinitely many in total.
(1) “Free” does not mean the variable is missing or irrelevant; it means you choose its value, and that choice then pins down the basic variables. (2) A free variable guarantees infinitely many solutions only if the system is consistent. An inconsistent system has no solutions at all, free variables or not.
A linear system is consistent if and only if an echelon form of its augmented matrix has no row of the shape \([\,0\ \cdots\ 0 \mid b\,]\) with \(b \ne 0\). If it is consistent, then:
- no free variables \(\Rightarrow\) exactly one solution;
- at least one free variable \(\Rightarrow\) infinitely many solutions.
This theorem fully justifies Day 1's warning: 0, 1, or infinitely many — nothing else is possible.
Day 3 — §1.3: Vector Equations
Today: a change of viewpoint. The same systems, seen through vectors — linear combinations, the span of a set of vectors, and the exact bridge connecting vector questions back to linear systems.
So far a linear system was a list of equations read row by row. Today we read it column by column, and a new geometric world opens up.
A vector in \(\mathbb{R}^n\) is an ordered list of \(n\) real numbers, written as a column, e.g. \(\mathbf{u}=\begin{bmatrix}1\\-2\end{bmatrix}\in\mathbb{R}^2\). Two vectors are equal only if all corresponding entries are equal. We add vectors entrywise and multiply by a scalar (a real number) entrywise: \[\begin{bmatrix}1\\-2\end{bmatrix}+\begin{bmatrix}2\\5\end{bmatrix}=\begin{bmatrix}3\\3\end{bmatrix},\qquad 3\begin{bmatrix}1\\-2\end{bmatrix}=\begin{bmatrix}3\\-6\end{bmatrix}\] The vector of all zeros is the zero vector \(\mathbf{0}\).
Geometrically, a vector in \(\mathbb{R}^2\) is a point of the plane, or equivalently an arrow from the origin to that point. Addition follows the parallelogram rule (Figure 2): \(\mathbf{u}+\mathbf{v}\) is the fourth corner of the parallelogram with corners \(\mathbf 0\), \(\mathbf u\), \(\mathbf v\). Scalar multiples \(c\mathbf{u}\) stretch, shrink, or (for \(c<0\)) flip the arrow, and all of them lie on one line through the origin. The usual rules of arithmetic hold: \(\mathbf u + \mathbf v = \mathbf v + \mathbf u\), \(c(\mathbf u + \mathbf v) = c\mathbf u + c\mathbf v\), and so on.
Let \(\mathbf u = \begin{bmatrix}2\\1\end{bmatrix}\) and \(\mathbf v = \begin{bmatrix}-1\\3\end{bmatrix}\). Compute \(2\mathbf u - 3\mathbf v\).
Solution. First the scalar multiples: \(2\mathbf u = \begin{bmatrix}4\\2\end{bmatrix}\) and \(3\mathbf v = \begin{bmatrix}-3\\9\end{bmatrix}\) (each entry multiplied). Then subtract entrywise: \[2\mathbf u - 3\mathbf v = \begin{bmatrix}4-(-3)\\2-9\end{bmatrix} = \begin{bmatrix}7\\-7\end{bmatrix}\]
Given vectors \(\mathbf v_1,\dots,\mathbf v_p\) in \(\mathbb{R}^n\) and scalars \(c_1,\dots,c_p\), the vector \[\mathbf y = c_1\mathbf v_1 + \cdots + c_p\mathbf v_p\] is a linear combination of \(\mathbf v_1,\dots,\mathbf v_p\) with weights \(c_1,\dots,c_p\). Weights may be any real numbers, including 0 and negative numbers.
“Linear combination” is the single most important phrase of this course. It is the vector-world counterpart of a linear equation: constants times things, added up.
The vector equation \[x_1\mathbf a_1 + x_2\mathbf a_2 + \cdots + x_n\mathbf a_n = \mathbf b\] has exactly the same solution set as the linear system whose augmented matrix is \[[\,\mathbf a_1\ \ \mathbf a_2\ \cdots\ \mathbf a_n \mid \mathbf b\,].\] In particular: \(\mathbf b\) is a linear combination of \(\mathbf a_1,\dots,\mathbf a_n\) if and only if that system is consistent. So every question about linear combinations is answered by solving a linear system — by hand for small ones, or with row reduction (§1.2) when the system is large.
Let \(\mathbf a_1 = \begin{bmatrix}1\\-2\\-5\end{bmatrix}\), \(\mathbf a_2 = \begin{bmatrix}2\\5\\6\end{bmatrix}\), \(\mathbf b = \begin{bmatrix}7\\4\\-3\end{bmatrix}\). Is \(\mathbf b\) a linear combination of \(\mathbf a_1, \mathbf a_2\)?
Solution. We ask whether weights \(x_1, x_2\) exist with \(x_1\mathbf a_1 + x_2\mathbf a_2 = \mathbf b\). Comparing the three entries on both sides turns the vector equation into an ordinary system: \[\begin{aligned}x_1 + 2x_2 &= 7\\ -2x_1 + 5x_2 &= 4\\ -5x_1 + 6x_2 &= -3\end{aligned}\] Take the first two equations and eliminate \(x_1\): adding 2 times the first equation to the second gives \(9x_2 = 18\), so \(x_2 = 2\), and then the first equation gives \(x_1 = 7 - 2(2) = 3\). These candidate weights must also satisfy the third equation, so check it: \(-5(3) + 6(2) = -3\) ✓. All three entries match, so yes: \(\mathbf b = 3\mathbf a_1 + 2\mathbf a_2\).
Had the third equation failed, the answer would have been no: \(\mathbf b\) would not be a linear combination of \(\mathbf a_1, \mathbf a_2\). And for larger systems, you would organize exactly this computation as the row reduction of the augmented matrix \([\mathbf a_1\ \mathbf a_2 \mid \mathbf b]\) — the answer comes out the same either way.
The set of all linear combinations of \(\mathbf v_1,\dots,\mathbf v_p\) is denoted \[\operatorname{Span}\{\mathbf v_1,\dots,\mathbf v_p\}\] and called the subset of \(\mathbb{R}^n\) spanned by \(\mathbf v_1,\dots,\mathbf v_p\). Asking “is \(\mathbf b\) in \(\operatorname{Span}\{\mathbf v_1,\dots,\mathbf v_p\}\)?” is exactly asking whether \(\mathbf b\) is a linear combination of those vectors.
Describe geometrically, in \(\mathbb{R}^3\):
(a) \(\operatorname{Span}\{\mathbf v\}\) for \(\mathbf v \neq \mathbf 0\);
(b) \(\operatorname{Span}\{\mathbf u, \mathbf v\}\) when neither is a multiple of the other;
(c) \(\operatorname{Span}\{\mathbf u, \mathbf v\}\) when \(\mathbf v = 2\mathbf u\), \(\mathbf u \ne \mathbf 0\).
Solution.
(a) All multiples \(c\mathbf v\): a line through the origin in the direction of \(\mathbf v\).
(b) All \(c_1\mathbf u + c_2\mathbf v\): sliding along two independent directions sweeps out a plane through the origin containing \(\mathbf u\) and \(\mathbf v\).
(c) Every combination \(c_1\mathbf u + c_2(2\mathbf u) = (c_1+2c_2)\mathbf u\) is just a multiple of \(\mathbf u\), so the span collapses to the line through the origin along \(\mathbf u\) — the second vector added nothing new.
Figure 3 shows cases (a) and (b).
Every span contains the zero vector (take all weights 0). So a line or plane not passing through the origin is never a span. Also, do not confuse the set \(\{\mathbf u, \mathbf v\}\) (two vectors) with \(\operatorname{Span}\{\mathbf u, \mathbf v\}\) (usually infinitely many vectors).
Looking ahead: next week we compress the bridge one step further and write the whole system as a single matrix equation \(A\mathbf x = \mathbf b\). Three languages — systems, vector equations, matrix equations — one theory.
Homework 1 — Sections 1.1–1.3 (100 points)
- Write your name and surname in the field above.
- Show your work: for computational questions, the row operations or reasoning must be visible, not just the final answer.
- When finished, click “Save homework as PDF” below, save the PDF, and upload it to Canvas.
- Your answers are saved automatically in this browser only. If you open this file on a different computer or browser, your answers will not appear there.
- Estimated workload: 2–3 hours.
Question 1
Solve the system \[\begin{aligned}x_1 + 2x_2 &= 5\\ 3x_1 - x_2 &= 1\end{aligned}\] by row reducing its augmented matrix, step by step, using the boxes below.
(a) First write the augmented matrix of the system (one number per cell; the double line marks the augmented column). Then click “Add row operation + matrix”: state the operation you apply (e.g. R2 -> R2 - 3R1) and enter the resulting matrix. Repeat until your matrix is reduced enough to read off the solution. You may add or remove steps as needed.
(b) Solution of the system (the values of \(x_1\) and \(x_2\)):
(c) Geometric comment: explain in one or two sentences what your answer means as lines in the plane.
Question 2
Consider the system \[\begin{aligned}x_1 + 3x_2 - 2x_3 &= 4\\ 2x_1 + 6x_2 - 3x_3 &= 9\\ 3x_1 + 9x_2 - 5x_3 &= 13\end{aligned}\]
(a) Enter the reduced echelon form of the augmented matrix (one number per cell):
(b) Identify the basic and free variables, and write the general solution. Show the row operations you used in part (a).
Question 3
Consider the system \[\begin{aligned}x_1 - 3x_2 &= 1\\ 2x_1 - 6x_2 &= h\end{aligned}\] where \(h\) is a constant.
(a) For which value(s) of \(h\) is the system consistent? Justify with a row operation.
(b) For those value(s) of \(h\), describe the solution set completely (how many solutions, and what they are).
Question 4
Let \(\mathbf a_1 = \begin{bmatrix}1\\-2\\0\end{bmatrix}\), \(\mathbf a_2 = \begin{bmatrix}0\\1\\2\end{bmatrix}\), \(\mathbf a_3 = \begin{bmatrix}1\\0\\1\end{bmatrix}\), \(\mathbf b = \begin{bmatrix}2\\-1\\6\end{bmatrix}\).
Determine whether \(\mathbf b\) is a linear combination of \(\mathbf a_1, \mathbf a_2, \mathbf a_3\). If it is, find weights that work and verify your answer by direct substitution. Show your row reduction.
Question 5
Let \(\mathbf v_1 = \begin{bmatrix}1\\2\\3\end{bmatrix}\), \(\mathbf v_2 = \begin{bmatrix}2\\4\\6\end{bmatrix}\), \(\mathbf v_3 = \begin{bmatrix}0\\1\\1\end{bmatrix}\).
(a) Describe \(\operatorname{Span}\{\mathbf v_1, \mathbf v_2\}\) geometrically, and explain why.
(b) Describe \(\operatorname{Span}\{\mathbf v_1, \mathbf v_3\}\) geometrically, and explain why.
(c) Is \(\begin{bmatrix}3\\6\\9\end{bmatrix}\) in \(\operatorname{Span}\{\mathbf v_1, \mathbf v_2\}\)? Justify.
Question 6
True or false? For each statement write T or F and justify your answer in one or two sentences. (2 points each; no credit without justification.)
(a) A system of linear equations can have exactly two solutions.
(b) Elementary row operations on an augmented matrix never change the solution set of the corresponding system.
(c) Every matrix is row equivalent to exactly one matrix in echelon form.
(d) If a linear system has a free variable, then it has infinitely many solutions.
(e) The vector \(\mathbf b\) is a linear combination of \(\mathbf a_1,\dots,\mathbf a_n\) if and only if the linear system with augmented matrix \([\mathbf a_1 \cdots \mathbf a_n \mid \mathbf b]\) is consistent.