W1. Consistency of Linear Systems, Row Echelon Forms, Advanced Solutions

Author

Salman Ahmadi-Asl

Published

January 29, 2026

1. Summary

1.1 Introduction: What Are We Studying?

In this module, we study how to systematically solve systems of linear equations and understand when solutions exist, are unique, or come in infinite families. You’ve likely solved small systems before (like $2 \times 2$ or $3 \times 3$), but what about larger systems? What if we have more equations than variables, or more variables than equations? What if some equations are redundant?

Main questions we’ll answer:

Does a solution exist? (Is the system consistent?)
If yes, is it unique? (Or are there infinitely many solutions?)
How do we find all solutions efficiently? (Using systematic elimination methods)

The key tool: We’ll use row operations to transform systems into simpler forms (REF and RREF) that make solutions easy to read off. The rank of the coefficient matrix will tell us everything about the solution structure.

1.2 Linear Systems of Equations

A linear system of equations is a collection of $m$ linear equations involving $n$ unknown variables. In general form:

\[ \begin{cases} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n = b_m \end{cases} \]

This can be written compactly in matrix form as $A\mathbf{x} = \mathbf{b}$, where:

$A \in \mathbb{R}^{m \times n}$ is the coefficient matrix containing only the coefficients of the variables
$\mathbf{x} \in \mathbb{R}^n$ is the column vector of unknowns
$\mathbf{b} \in \mathbb{R}^m$ is the column vector of constants on the right-hand side

The augmented matrix $[A|\mathbf{b}]$ appends the constants vector $\mathbf{b}$ as an extra column to $A$:

\[ [A|\mathbf{b}] = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} & | & b_1 \\ a_{21} & a_{22} & \dots & a_{2n} & | & b_2 \\ \vdots & \vdots & \ddots & \vdots & | & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} & | & b_m \end{bmatrix} \]

The augmented matrix is the primary object we manipulate when solving linear systems via elimination.

1.3 Consistency and Inconsistency

A system $A\mathbf{x} = \mathbf{b}$ is called consistent if it has at least one solution. Geometrically, this means the lines, planes, or hyperplanes represented by the equations intersect at one or more common points.

To understand this better, think about what it means to solve $A\mathbf{x} = \mathbf{b}$ in terms of columns. If we write $A$ as $[\mathbf{a}_1 \; \mathbf{a}_2 \; \cdots \; \mathbf{a}_n]$ where each $\mathbf{a}_i$ is a column, then $A\mathbf{x} = \mathbf{b}$ means:

\[ x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n = \mathbf{b} \]

This asks: “Can we express $\mathbf{b}$ as a combination of the columns of $A$?” If yes, the system is consistent. The set of all possible combinations of the columns of $A$ is called the column space of $A$, denoted $\text{Col}(A)$. So the system is consistent if and only if $\mathbf{b} \in \text{Col}(A)$.

A system is inconsistent if it has no solution. Geometrically, the hyperplanes do not all share a common intersection point. For example, parallel lines in 2D never intersect:

\[ \begin{cases} x + y = 3 \\ x + y = 5 \end{cases} \]

These represent two parallel lines with no common point, so the system is inconsistent.

1.4 Rank of a Matrix

The rank of a matrix $A$, denoted $\text{rank}(A)$, is the maximum number of linearly independent rows (or equivalently, columns) in $A$.

What does “linearly independent” mean? A set of vectors is linearly independent if no vector in the set can be written as a combination of the others. For example, the rows $[1, 2]$ and $[2, 4]$ are dependent because the second is just $2$ times the first. But $[1, 2]$ and $[1, 3]$ are independent — neither is a multiple of the other.

How to compute rank: Reduce $A$ to row echelon form and count the number of pivot positions (non-zero leading entries in each row). Each pivot corresponds to one independent equation, so the number of pivots equals the rank.

Why does rank matter? The rank tells us:

How many “truly different” equations we have (some equations might be redundant)
The dimension of the solution space
Whether the system is consistent (by comparing $\text{rank}(A)$ with $\text{rank}([A|\mathbf{b}])$)

Key properties of rank:

$\text{rank}(A) \le \min(m, n)$ (can’t have more independent rows than total rows, or more independent columns than total columns)
$\text{rank}(A) = \dim(\text{Col}(A)) = \dim(\text{Row}(A))$ (rank measures the dimension of both the column and row spaces)
Row operations do not change the rank (they preserve the essential structure of the system)

1.5 The Consistency Theorem (Rouché–Capelli Theorem)

This is the central theorem for determining whether a linear system has solutions.

Theorem 1 (Consistency of Linear Systems). A system $A\mathbf{x} = \mathbf{b}$ is consistent if and only if:

\[ \text{rank}(A) = \text{rank}([A|\mathbf{b}]) \]

Why does this work? If $\mathbf{b}$ is already in the column space of $A$, then appending $\mathbf{b}$ as a new column does not add any new “direction” — the rank stays the same. If $\mathbf{b}$ is not in $\text{Col}(A)$, then it introduces a new independent direction, so $\text{rank}([A|\mathbf{b}]) = \text{rank}(A) + 1$.

Proof sketch. Write $A = [\mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_n]$ as columns. The system is consistent iff $\mathbf{b} = x_1\mathbf{a}_1 + \dots + x_n\mathbf{a}_n$ for some scalars, i.e., $\mathbf{b} \in \text{Col}(A)$. If $\mathbf{b} \in \text{Col}(A)$, adding $\mathbf{b}$ as a column does not increase the column space dimension, so $\text{rank}([A|\mathbf{b}]) = \text{rank}(A)$. Conversely, if the ranks are equal, then $\text{Col}(A) = \text{Col}([A|\mathbf{b}])$, which forces $\mathbf{b} \in \text{Col}(A)$. $\square$

Alternative viewpoint (via row operations). If $\text{rank}([A|\mathbf{b}]) > \text{rank}(A)$, then in the row echelon form of $[A|\mathbf{b}]$, there will be a row of the form $[0 \; 0 \; \cdots \; 0 \;|\; c]$ with $c \neq 0$, corresponding to the impossible equation $0 = c$.

1.6 Classification of Solutions

Theorem 2 (Solution Type Determination). For the linear system $A\mathbf{x} = \mathbf{b}$ with $n$ variables, let $r_A = \text{rank}(A)$ and $r_{[A|\mathbf{b}]} = \text{rank}([A|\mathbf{b}])$. Then:

No solution (inconsistent): $r_A < r_{[A|\mathbf{b}]}$
Unique solution: $r_A = r_{[A|\mathbf{b}]} = n$
Infinitely many solutions: $r_A = r_{[A|\mathbf{b}]} < n$

Why does this work? The proof relies on the Rank-Nullity Theorem:

\[ \text{rank}(A) + \dim(\text{Null}(A)) = n \]

This fundamental theorem says: the number of pivot columns plus the number of free variables equals the total number of variables. Rearranging: $\dim(\text{Null}(A)) = n - \text{rank}(A)$.

Now we can understand the three cases:

Case 1: If $r_A < r_{[A|\mathbf{b}]}$, adding $\mathbf{b}$ increased the rank, meaning $\mathbf{b}$ is not in the column space. No solution exists.
Case 2: If $r_A = n$, then $\dim(\text{Null}(A)) = 0$, so the null space contains only $\mathbf{0}$. There are no free variables, no “wiggle room” — the solution (if it exists) is unique.
Case 3: If $r_A < n$, then $\dim(\text{Null}(A)) = n - r_A > 0$. There are $n - r_A$ free variables, giving infinitely many solutions in the form $\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h$ where $\mathbf{x}_h$ ranges over the $(n - r_A)$-dimensional null space.

1.6.1 Special Cases

Homogeneous systems ($A\mathbf{x} = \mathbf{0}$) are always consistent because $\mathbf{x} = \mathbf{0}$ is always a solution. They have:

Only the trivial solution if $\text{rank}(A) = n$
Infinitely many nontrivial solutions if $\text{rank}(A) < n$

Square systems ($m = n$):

Unique solution for every $\mathbf{b}$ iff $\det(A) \neq 0$ (equivalently, $\text{rank}(A) = n$)
If $\det(A) = 0$, the system is either inconsistent or has infinitely many solutions

1.7 Geometric Interpretation

Each linear equation $a_1x_1 + a_2x_2 + \dots + a_nx_n = b$ (with not all $a_i = 0$) represents a hyperplane in $\mathbb{R}^n$ — a flat subspace of dimension $n - 1$.

What is dimension? Intuitively, the dimension of a space is the number of independent directions you can move in it. A line has dimension 1 (one direction: forward/backward). A plane has dimension 2 (two independent directions: e.g., north/south and east/west). Our 3D world has dimension 3. A hyperplane in $\mathbb{R}^n$ has dimension $n - 1$ — it’s one dimension “less” than the full space. For example, in $\mathbb{R}^3$, a hyperplane is an ordinary plane (dimension 2).

The solution set of a linear system is the intersection of the corresponding hyperplanes:

Unique solution: All hyperplanes meet at exactly one point
Infinitely many solutions: Hyperplanes intersect along a line, plane, or higher-dimensional flat
No solution: The hyperplanes have no common intersection (e.g., parallel planes)

1.8 Row Echelon Form (REF)

A matrix is in Row Echelon Form (also called stair-step form) if:

All zero rows are at the bottom
Each leading entry (pivot) is in a column to the right of the pivot in the row above
All entries below a pivot are zero

Example:

\[ \begin{bmatrix} 2 & 3 & -1 & 5 \\ 0 & 4 & 2 & -3 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

The pivots are $2$, $4$, and $1$ (in columns 1, 2, and 4). Note that pivots do not need to be $1$ in REF.

Pivot columns vs. free columns: Columns containing pivots correspond to basic variables (also called pivot variables) — these are the variables we solve for in terms of the others. Columns without pivots correspond to free variables — these can take any value and act as parameters in the solution. In the example above, if this were the augmented matrix of a system with variables $x_1, x_2, x_3, x_4$, then $x_1, x_2, x_4$ would be basic variables (determined by the system) and $x_3$ would be a free variable (a parameter).

1.9 Reduced Row Echelon Form (RREF)

A matrix is in Reduced Row Echelon Form (also called canonical form) if:

It is in REF
Each pivot is exactly $1$
Each pivot is the only nonzero entry in its column (all entries above and below are zero)

Example:

\[ \begin{bmatrix} 1 & 0 & 3 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

The key difference from REF: in RREF, you can read off variable values (or express them in terms of free variables) directly, without back substitution.

1.10 Solving Linear Systems: Step-by-Step Methodology

The standard procedure for solving $A\mathbf{x} = \mathbf{b}$:

Write the augmented matrix $[A|\mathbf{b}]$
Use Gaussian elimination (row operations) to transform it to REF or RREF
Identify pivot columns (basic variables) and free columns (free variables)
Check consistency: if a row $[0 \; 0 \; \cdots \; 0 \;|\; c]$ with $c \neq 0$ appears, the system is inconsistent
If consistent, find a particular solution $\mathbf{x}_p$ by setting all free variables to $0$
Find the homogeneous solution $\mathbf{x}_h$ by solving $A\mathbf{x} = \mathbf{0}$
Write the complete solution: $\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h$

The three elementary row operations are:

Swap two rows: $R_i \leftrightarrow R_j$ (changing the order of equations doesn’t change the solution)
Multiply a row by a nonzero scalar: $R_i \leftarrow cR_i$ (multiplying both sides of an equation by a nonzero number gives an equivalent equation)
Add a multiple of one row to another: $R_i \leftarrow R_i + cR_j$ (if $\mathbf{x}$ satisfies both equations, it satisfies their sum)

Why do row operations preserve solutions? Each row operation corresponds to a valid algebraic manipulation that doesn’t change the solution set. For example, if you have equations $E_1$ and $E_2$, and $\mathbf{x}$ satisfies both, then $\mathbf{x}$ also satisfies $E_1 + 3E_2$. Conversely, if $\mathbf{x}$ satisfies $E_1$ and $E_1 + 3E_2$, we can recover $E_2$ by computing $(E_1 + 3E_2) - E_1 = 3E_2$, so $\mathbf{x}$ satisfies $E_2$ as well. Thus the solution sets before and after the operation are identical.

1.11 Complete Solution Structure

For a consistent system $A\mathbf{x} = \mathbf{b}$ with $\text{rank}(A) = r$ and $n$ variables:

\[ \mathbf{x} = \mathbf{x}_p + \mathbf{x}_h \]

where:

$\mathbf{x}_p$ is any particular solution to $A\mathbf{x} = \mathbf{b}$ (found by setting free variables to $0$)
$\mathbf{x}_h$ is the general solution to the homogeneous system $A\mathbf{x} = \mathbf{0}$, expressed as a linear combination of $n - r$ basis vectors of the null space

Understanding this structure intuitively: Think of $\mathbf{x}_p$ as “one way to get to the target $\mathbf{b}$,” and $\mathbf{x}_h$ as “all the ways to stay at zero.” Since $A(\mathbf{x}_p + \mathbf{x}_h) = A\mathbf{x}_p + A\mathbf{x}_h = \mathbf{b} + \mathbf{0} = \mathbf{b}$, adding any homogeneous solution to a particular solution gives another solution.

What is the null space? The null space (or kernel) of $A$, denoted $\text{Null}(A)$, is the set of all vectors that $A$ maps to zero:

\[ \text{Null}(A) = \{\mathbf{x} \in \mathbb{R}^n : A\mathbf{x} = \mathbf{0}\} \]

The null space is a vector subspace of $\mathbb{R}^n$, meaning it’s closed under addition and scalar multiplication. Its dimension equals $n - r$, which is the number of free variables (parameters) in the solution. Geometrically, if $\text{rank}(A) = n$ (full rank), the null space contains only the zero vector, so there’s a unique solution. If $\text{rank}(A) < n$, the null space contains infinitely many vectors, giving infinitely many solutions.

1.12 Underdetermined and Overdetermined Systems

An underdetermined system has more variables than equations ($n > m$). If consistent, it always has infinitely many solutions because $\text{rank}(A) \le m < n$, so there is at least one free variable.

An overdetermined system has more equations than variables ($m > n$). It may still be consistent, but “generically” (for a random $\mathbf{b}$) it will be inconsistent because the extra equations impose additional constraints.

1.13 Systems with Parameters

When a system involves a parameter (like $k$, $a$, $b$, $c$), you perform row reduction as usual, and then analyze the resulting matrix for different values of the parameter. The key is to identify which parameter values cause:

A pivot to vanish (changing the rank)
An inconsistency row $[0 \; \cdots \; 0 \;|\; c \neq 0]$ to appear

Different parameter values may lead to different solution types (unique, infinite, or none).

2. Definitions

Linear system of equations: A collection of $m$ linear equations in $n$ unknowns of the form $A\mathbf{x} = \mathbf{b}$, where each equation is a linear combination of the variables equal to a constant.
Coefficient matrix: The $m \times n$ matrix $A$ containing only the coefficients of the variables in a linear system.
Augmented matrix: The matrix $[A|\mathbf{b}]$ formed by appending the constants vector $\mathbf{b}$ as an additional column to the coefficient matrix $A$.
Consistent system: A linear system that has at least one solution; equivalently, $\mathbf{b} \in \text{Col}(A)$.
Inconsistent system: A linear system that has no solution; equivalently, $\mathbf{b} \notin \text{Col}(A)$.
Rank: The maximum number of linearly independent rows (or columns) of a matrix; equals the number of pivots in row echelon form.
Pivot: The first nonzero entry in a row of a matrix in row echelon form; also called the leading entry.
Pivot column: A column of the coefficient matrix that contains a pivot; corresponds to a basic (determined) variable.
Free variable: A variable whose corresponding column is not a pivot column; it can take any real value as a parameter.
Row Echelon Form (REF): A matrix form where all zero rows are at the bottom, each pivot is to the right of the pivot above, and all entries below each pivot are zero.
Reduced Row Echelon Form (RREF): A matrix in REF where each pivot is $1$ and is the only nonzero entry in its column.
Particular solution ($\mathbf{x}_p$): Any single solution to $A\mathbf{x} = \mathbf{b}$, typically found by setting all free variables to zero.
Homogeneous solution ($\mathbf{x}_h$): The general solution to $A\mathbf{x} = \mathbf{0}$; it forms the null space of $A$.
Null space (kernel): The set of all vectors $\mathbf{x}$ such that $A\mathbf{x} = \mathbf{0}$; a vector subspace of $\mathbb{R}^n$ with dimension $n - \text{rank}(A)$.
Column space: The set of all possible linear combinations of the columns of $A$; the system $A\mathbf{x} = \mathbf{b}$ is consistent iff $\mathbf{b}$ is in the column space.
Hyperplane: The solution set of a single linear equation in $\mathbb{R}^n$; a flat subspace of dimension $n - 1$.
Underdetermined system: A system with more variables than equations ($n > m$); if consistent, always has infinitely many solutions.
Back substitution: The process of solving for variables starting from the last equation in a triangular (REF) system and working upward.

3. Formulas

Matrix form of a linear system: $A\mathbf{x} = \mathbf{b}$, where $A \in \mathbb{R}^{m \times n}$, $\mathbf{x} \in \mathbb{R}^n$, $\mathbf{b} \in \mathbb{R}^m$
Consistency condition (Rouché–Capelli): The system $A\mathbf{x} = \mathbf{b}$ is consistent iff $\text{rank}(A) = \text{rank}([A|\mathbf{b}])$
Complete solution structure: $\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h$, where $\mathbf{x}_p$ is a particular solution and $\mathbf{x}_h$ is the general homogeneous solution
Rank-Nullity Theorem: $\dim(\text{Null}(A)) = n - \text{rank}(A)$, i.e., the number of free variables equals $n$ minus the number of pivots
Solution classification:
- No solution: $\text{rank}(A) < \text{rank}([A|\mathbf{b}])$
- Unique solution: $\text{rank}(A) = \text{rank}([A|\mathbf{b}]) = n$
- Infinitely many solutions: $\text{rank}(A) = \text{rank}([A|\mathbf{b}]) < n$
Null space dimension: $\dim(\text{Null}(A)) = n - r$, where $r = \text{rank}(A)$
Rank bound: $\text{rank}(A) \le \min(m, n)$
Square system unique solution condition: $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$ iff $\det(A) \neq 0$

4. Examples

4.1. Solvability of a System of Three Lines (Lab 1, Task 1)

Consider the following system of equations: \[ \begin{cases} x + 2y = 2 \\ x - y = 2 \\ y = 1 \end{cases} \]

Part (a): Sketch these three lines and decide if the system is solvable. Part (b): What happens when all the elements on the right-hand side are zero? Part (c): Is there any choice of numbers on the right-hand side that allows the three lines to intersect at the same point?

Click to see the solution

Key Concept: A system of three linear equations in two variables represents three lines in the plane. They can intersect at a point (unique solution), have no common intersection (inconsistent), or all pass through infinitely many points (infinitely many solutions).

(a) Sketch and solvability: The three lines can be plotted on a coordinate plane. The first line $x + 2y = 2$ and the second line $x - y = 2$ intersect at a point. However, the third line $y = 1$ may not pass through the same point, leading to the three lines forming a triangle with no common intersection point. Therefore, this system is inconsistent (has no solution).

(b) When right-hand side is zero: 1. If we set all right-hand side values to zero: $x + 2y = 0$, $x - y = 0$, $y = 0$ 2. All three lines pass through the origin $(0, 0)$ 3. Therefore, the system has a solution at the origin.

(c) Condition for three lines to intersect at one point: For the three lines to intersect at a single common point, we need the system to be consistent. The right-hand side values must be compatible with the coefficient matrix. For example, if we modify the third equation to $y = 0$ (instead of $y = 1$), then solving $x + 2y = 0$, $x - y = 0$, and $y = 0$ gives $(x, y) = (0, 0)$.

Answer: (a) The system is inconsistent (no common point); (b) All lines pass through the origin $(0, 0)$; (c) Yes, by adjusting the right-hand side, e.g., changing the third equation to $y = 0$.

4.2. Identifying Inconsistency via Rank (Lab 1, Task 2)

Consider the following system of equations: \[ \begin{cases} u + v + w = 2 \\ u + 2v + 3w = 1 \\ v + 2w = 0 \end{cases} \]

Part (a): Explain why this system is singular by providing a combination of three equations that adds up to $0 = 1$. Part (b): What value should replace $0$ on the right-hand side of the third equation to allow the system to have a solution? Part (c): What is one of the solutions?

Click to see the solution

Key Concept: When row operations reveal a contradiction like $0 = 1$, the system is inconsistent. Adjusting the right-hand side appropriately can make the system consistent.

(a) Finding the inconsistency: 1. Start with the augmented matrix: \[\begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 1 & 2 & 3 & | & 1 \\ 0 & 1 & 2 & | & 0 \end{bmatrix}\] 2. Perform row operation $R_2 - R_1$: \[\begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 1 & 2 & | & -1 \\ 0 & 1 & 2 & | & 0 \end{bmatrix}\] 3. Perform row operation $R_3 - R_2$ on the result: \[\begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 1 & 2 & | & -1 \\ 0 & 0 & 0 & | & 1 \end{bmatrix}\] 4. The third row represents $0 = 1$, which is a contradiction, making the system inconsistent.

(b) Correct right-hand side: If we change the third equation’s right-hand side from $0$ to $-1$ (making it $v + 2w = -1$), the augmented matrix becomes: \[\begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 1 & 2 & | & -1 \\ 0 & 1 & 2 & | & -1 \end{bmatrix}\] After row reduction, the third row becomes all zeros, and the system becomes consistent.

(c) Finding a solution when right-hand side is corrected: With the corrected system, we have:

$v + 2w = -1$, so $v = -1 - 2w$
$u + v + w = 2$, so $u + (-1 - 2w) + w = 2$, giving $u = 3 + w$

Setting $w = 0$ as a free variable: $(u, v, w) = (3, -1, 0)$

Answer: (a) Row operations reveal $0 = 1$, an inconsistency; (b) The right-hand side of the third equation should be $-1$; (c) One solution is $(3, -1, 0)$.

4.3. Linear Combinations and Back Substitution (Lab 1, Task 3)

What combination of $(1, 0, 0, 0)$, $(1, 1, 0, 0)$, $(1, 1, 1, 0)$, and $(1, 1, 1, 1)$ produces $(3, 3, 3, 2)$? What equations for $x$, $y$, $z$, and $t$ are you solving?

Click to see the solution

Key Concept: Finding a linear combination of column vectors is equivalent to solving a linear system where the column vectors form the coefficient matrix.

Set up the system: We seek $x, y, z, t$ such that: \[x\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + y\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + z\begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 3 \\ 2 \end{bmatrix}\]
Write as a linear system: \[\begin{cases} x + y + z + t = 3 \\ y + z + t = 3 \\ z + t = 3 \\ t = 2 \end{cases}\]
Express in matrix form: \[\begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 3 \\ 2 \end{bmatrix}\]
Back substitution:
- From equation 4: $t = 2$
- From equation 3: $z + 2 = 3 \Rightarrow z = 1$
- From equation 2: $y + 1 + 2 = 3 \Rightarrow y = 0$
- From equation 1: $x + 0 + 1 + 2 = 3 \Rightarrow x = 0$

Answer: $(x, y, z, t) = (0, 0, 1, 2)$

4.4. Gaussian Elimination to Upper Triangular Form (Lab 1, Task 4)

Reduce the following system to an upper triangular form by row operations: \[ \begin{cases} 2x + 3y + z = 8 \\ 4x + 7y + 5z = 20 \\ -2y + 2z = 0 \end{cases} \]

Circle the pivots. Solve it for $x$, $y$, and $z$ by back-substitution.

Click to see the solution

Key Concept: Gaussian elimination transforms the system into upper triangular form (REF) by making entries below pivots zero, enabling efficient back substitution.

Write augmented matrix: \[\begin{bmatrix} 2 & 3 & 1 & | & 8 \\ 4 & 7 & 5 & | & 20 \\ 0 & -2 & 2 & | & 0 \end{bmatrix}\]
Perform row operation $R_2 - 2R_1$: \[\begin{bmatrix} 2 & 3 & 1 & | & 8 \\ 0 & 1 & 3 & | & 4 \\ 0 & -2 & 2 & | & 0 \end{bmatrix}\]

First pivot: $\boxed{2}$ (in position $(1,1)$)
Perform row operation $R_3 + 2R_2$: \[\begin{bmatrix} 2 & 3 & 1 & | & 8 \\ 0 & 1 & 3 & | & 4 \\ 0 & 0 & 8 & | & 8 \end{bmatrix}\]

Second pivot: $\boxed{1}$ (in position $(2,2)$) Third pivot: $\boxed{8}$ (in position $(3,3)$)
Back substitution:
- From equation 3: $8z = 8 \Rightarrow z = 1$
- From equation 2: $y + 3(1) = 4 \Rightarrow y = 1$
- From equation 1: $2x + 3(1) + 1 = 8 \Rightarrow 2x = 4 \Rightarrow x = 2$

Answer: $(x, y, z) = (2, 1, 1)$

4.5. Parameter Analysis: Row Exchange and Missing Pivot (Lab 1, Task 5)

For the system: \[ \begin{cases} x + by = 0 \\ x - 2y - z = 0 \\ y + z = 0 \end{cases} \]

Part (a): Which number $b$ leads to a row exchange? Part (b): Which $b$ leads to a missing pivot (singular case)? Part (c): In the singular case, find a nonzero solution for $x$, $y$, and $z$.

Click to see the solution

Key Concept: Parameters that make a pivot vanish cause the matrix to become singular, potentially requiring row exchanges or indicating infinitely many solutions in the homogeneous case.

(a) Row exchange case when $b = -2$: 1. Start with augmented matrix: \[\begin{bmatrix} 1 & -2 & 0 & | & 0 \\ 1 & -2 & -1 & | & 0 \\ 0 & 1 & 1 & | & 0 \end{bmatrix}\] 2. After $R_2 - R_1$: \[\begin{bmatrix} 1 & -2 & 0 & | & 0 \\ 0 & 0 & -1 & | & 0 \\ 0 & 1 & 1 & | & 0 \end{bmatrix}\] 3. Row 2 has a zero in column 2, so we exchange $R_2$ and $R_3$: \[\begin{bmatrix} 1 & -2 & 0 & | & 0 \\ 0 & 1 & 1 & | & 0 \\ 0 & 0 & -1 & | & 0 \end{bmatrix}\]

Answer for (a): $b = -2$

(b) Missing pivot case when $b = -1$: 1. Start with augmented matrix: \[\begin{bmatrix} 1 & -1 & 0 & | & 0 \\ 1 & -2 & -1 & | & 0 \\ 0 & 1 & 1 & | & 0 \end{bmatrix}\] 2. After $R_2 - R_1$: \[\begin{bmatrix} 1 & -1 & 0 & | & 0 \\ 0 & -1 & -1 & | & 0 \\ 0 & 1 & 1 & | & 0 \end{bmatrix}\] 3. After $R_3 + R_2$: \[\begin{bmatrix} 1 & -1 & 0 & | & 0 \\ 0 & -1 & -1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}\]

The third column has no pivot, so $z$ is a free variable.

Answer for (b): $b = -1$

(c) Nonzero solution when $b = -1$: 1. From the reduced form: $y + z = 0$ (equivalently, $-v + w = 0$) 2. From the first equation: $x - y = 0$ 3. Setting $z = 1$ (free variable): $y = -1$, $x = -1$

Answer: $(x, y, z) = (-1, -1, 1)$ or any nonzero scalar multiple, e.g., $(1, 1, -1)$

4.6. Dependent Rows and Solution Structure (Lab 1, Task 6)

Construct a 3 by 3 example that has 9 different coefficients on the left-hand side, but rows 2 and 3 become zero in the elimination.

Part (a): How many solutions to your system with $b = (1, 10, 100)$? Part (b): How many with $b = (0, 0, 0)$?

Click to see the solution

Key Concept: When rows become linearly dependent (proportional), the rank of the augmented matrix determines consistency and the number of solutions.

(a) Construction and analysis for $b = (1, 10, 100)$:

Consider a system with proportional rows: \[A = \begin{bmatrix} a & b & c \\ 4a & 4b & 4c \\ 5a & 5b & 5c \end{bmatrix}\]

The second and third rows are proportional to the first row.
Augmented matrix for $b = (1, 10, 100)$: \[\begin{bmatrix} a & b & c & | & 1 \\ 4a & 4b & 4c & | & 10 \\ 5a & 5b & 5c & | & 100 \end{bmatrix}\]
After row operations $R_2 - 4R_1$ and $R_3 - 5R_1$: \[\begin{bmatrix} a & b & c & | & 1 \\ 0 & 0 & 0 & | & 6 \\ 0 & 0 & 0 & | & 95 \end{bmatrix}\]
Both rows 2 and 3 represent contradictions ($0 = 6$ and $0 = 95$), so the system is inconsistent.

Answer for (a): No solutions (the system is inconsistent)

(b) For $b = (0, 0, 0)$:

Augmented matrix: \[\begin{bmatrix} a & b & c & | & 0 \\ 4a & 4b & 4c & | & 0 \\ 5a & 5b & 5c & | & 0 \end{bmatrix}\]
After row operations: \[\begin{bmatrix} a & b & c & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}\]
Only one independent equation remains: $ax + by + cz = 0$. There are $n - r = 3 - 1 = 2$ free variables.
From the equation: $z = -\frac{1}{c}(ax + by)$ (assuming $c \neq 0$)
General solution: $(x, y, z) = \left(x, y, -\frac{1}{c}(ax + by)\right)$ for any $x, y \in \mathbb{R}$

Answer for (b): Infinitely many solutions (the system has a 2-dimensional solution space)

4.7. Solving Two 3×3 Systems (Lab 1, Task 7)

Use elimination to solve these two systems:

System 1: \[ \begin{cases} u + v + w = 6 \\ u + 2v + 2w = 11 \\ 2u + 3v - 4w = 3 \end{cases} \]

System 2: \[ \begin{cases} u + v + w = 7 \\ u + 2v + 2w = 10 \\ 2u + 3v - 4w = 3 \end{cases} \]

Click to see the solution

Key Concept: These two systems share the same coefficient matrix but differ only in the right-hand side. Both can be solved using the same sequence of row operations.

System 1 Solution:

Write augmented matrix: \[\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 2 & | & 11 \\ 2 & 3 & -4 & | & 3 \end{bmatrix}\]
Row operations $R_2 - R_1$ and $R_3 - 2R_1$: \[\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 1 & | & 5 \\ 0 & 1 & -6 & | & -9 \end{bmatrix}\]
Row operation $R_3 - R_2$: \[\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 1 & | & 5 \\ 0 & 0 & -7 & | & -14 \end{bmatrix}\]
Back substitution:
- $-7w = -14 \Rightarrow w = 2$
- $v + 2 = 5 \Rightarrow v = 3$
- $u + 3 + 2 = 6 \Rightarrow u = 1$

Answer for System 1: $(u, v, w) = (1, 3, 2)$

System 2 Solution:

Write augmented matrix: \[\begin{bmatrix} 1 & 1 & 1 & | & 7 \\ 1 & 2 & 2 & | & 10 \\ 2 & 3 & -4 & | & 3 \end{bmatrix}\]
Apply the same row operations as System 1 (the coefficient matrix is identical): \[\begin{bmatrix} 1 & 1 & 1 & | & 7 \\ 0 & 1 & 1 & | & 3 \\ 0 & 0 & -7 & | & -14 \end{bmatrix}\]
Back substitution:
- $-7w = -14 \Rightarrow w = 2$
- $v + 2 = 3 \Rightarrow v = 1$
- $u + 1 + 2 = 7 \Rightarrow u = 4$

Answer for System 2: $(u, v, w) = (4, 1, 2)$

4.8. System with Parameter: Solution Type Classification (Assignment 1, Task 1)

Determine all values of $k$ for which the system has: 1. A unique solution 2. No solution 3. Infinitely many solutions

\[ \begin{cases} x + 2y + 3z = 1 \\ 2x + 4y + 6z = 2 \\ kx + (k + 1)y + (k + 2)z = k \end{cases} \]

Click to see the solution

Key Concept: For systems with parameters, the number of solutions depends on how the parameter affects the rank of the coefficient matrix and augmented matrix.

Write the augmented matrix: \[\begin{bmatrix} 1 & 2 & 3 & | & 1 \\ 2 & 4 & 6 & | & 2 \\ k & k+1 & k+2 & | & k \end{bmatrix}\]
Observe the first two equations: Row 2 is exactly $2 \times$ Row 1, so after $R_2 - 2R_1$: \[\begin{bmatrix} 1 & 2 & 3 & | & 1 \\ 0 & 0 & 0 & | & 0 \\ k & k+1 & k+2 & | & k \end{bmatrix}\]
Analyze Row 3: Perform $R_3 - kR_1$: \[\begin{bmatrix} 1 & 2 & 3 & | & 1 \\ 0 & 0 & 0 & | & 0 \\ 0 & (k+1) - 2k & (k+2) - 3k & | & k - k \end{bmatrix}\]

Simplifying Row 3: \[\begin{bmatrix} 1 & 2 & 3 & | & 1 \\ 0 & 0 & 0 & | & 0 \\ 0 & 1 - k & 2 - 2k & | & 0 \end{bmatrix}\]
Case analysis:

Case 1: $k = 1$ \[\begin{bmatrix} 1 & 2 & 3 & | & 1 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}\] Only one independent equation; $\text{rank}(A) = 1 < n = 3$. Infinitely many solutions (with 2 free variables).

Case 2: $k \neq 1$ Row 3 becomes: $0 \cdot x + (1-k)y + (2-2k)z = 0$

This simplifies to $(1-k)y + 2(1-k)z = 0$, or $(1-k)(y + 2z) = 0$.

Since $1 - k \neq 0$, we have $y + 2z = 0$, which gives one more independent equation.

$\text{rank}(A) = 2 = \text{rank}([A|b])$, and $n = 3$. Infinitely many solutions (with 1 free variable) since the second row is always zero and the right-hand sides are consistent.

Starting system: \[\begin{cases} x + 2y + 3z = 1 \\ 2x + 4y + 6z = 2 \\ kx + (k+1)y + (k+2)z = k \end{cases}\]

Notice: Row 2 = $2 \times$ Row 1, so the first two equations are dependent.

After $R_2 - 2R_1$ and analyzing Row 3:

$R_3 - kR_1$: $(k - k)x + [(k+1) - 2k]y + [(k+2) - 3k]z = k - k$
Simplifies to: $(1 - k)y + (2 - 2k)z = 0$, i.e., $(1-k)y + 2(1-k)z = 0$

If $k = 1$: The third row becomes $0 = 0$, giving infinitely many solutions.

If $k \neq 1$: We can divide by $(1-k)$ to get $y + 2z = 0$, giving a constraint but still infinitely many solutions (one free variable).

The system always has solutions because the right-hand sides are compatible.

Answer:

Unique solution: Never (the first two equations are dependent)
No solution: Never (the system is always consistent)
Infinitely many solutions: For all values of $k$ (the system always has rank 2 and infinitely many solutions with 1 free variable)

4.9. Underdetermined System: Multiple Free Variables (Assignment 1, Task 2)

Solve the system: \[ \begin{cases} x_1 + 2x_2 + 3x_3 + 4x_4 + 5x_5 = 1 \\ 2x_1 + 4x_2 + 6x_3 + 8x_4 + 10x_5 = 2 \\ 3x_1 + 6x_2 + 9x_3 + 12x_4 + 15x_5 = 3 \end{cases} \]

Click to see the solution

Key Concept: This is an underdetermined system (3 equations, 5 variables). The rows are all proportional, leading to only one independent constraint and four free variables.

Write the augmented matrix: \[\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & | & 1 \\ 2 & 4 & 6 & 8 & 10 & | & 2 \\ 3 & 6 & 9 & 12 & 15 & | & 3 \end{bmatrix}\]
Observe proportionality:
- Row 2 = $2 \times$ Row 1
- Row 3 = $3 \times$ Row 1
So all three rows represent the same constraint.
After row reduction: \[\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & | & 1 \\ 0 & 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}\]

Only one independent equation remains: $x_1 + 2x_2 + 3x_3 + 4x_4 + 5x_5 = 1$
Identify pivot and free variables:
- Pivot column: 1 (variable $x_1$)
- Free columns: 2, 3, 4, 5 (variables $x_2, x_3, x_4, x_5$)
- Number of free variables: $5 - 1 = 4$
Express general solution: Let $x_2 = s$, $x_3 = t$, $x_4 = u$, $x_5 = v$ (free parameters).

From the constraint: $x_1 = 1 - 2s - 3t - 4u - 5v$

Answer: The general solution is: \[\mathbf{x} = \begin{bmatrix} 1 - 2s - 3t - 4u - 5v \\ s \\ t \\ u \\ v \end{bmatrix}, \quad s, t, u, v \in \mathbb{R}\]

Or equivalently: $(x_1, x_2, x_3, x_4, x_5) = (1 - 2s - 3t - 4u - 5v, s, t, u, v)$ where $s, t, u, v$ are arbitrary real numbers.

Particular solution (setting all free variables to 0): $(1, 0, 0, 0, 0)$

4.10. 4×4 Overdetermined System (Assignment 1, Task 3)

Solve the system: \[ \begin{cases} x_1 + x_2 + x_3 + x_4 = 0 \\ x_1 + 2x_2 + 3x_3 + 4x_4 = 1 \\ x_1 + 3x_2 + 6x_3 + 10x_4 = 3 \\ x_1 + 4x_2 + 10x_3 + 20x_4 = 6 \end{cases} \]

Click to see the solution

Key Concept: This is an overdetermined system (4 equations, 4 variables). Row reduction will show whether it is consistent or inconsistent.

Write the augmented matrix: \[\begin{bmatrix} 1 & 1 & 1 & 1 & | & 0 \\ 1 & 2 & 3 & 4 & | & 1 \\ 1 & 3 & 6 & 10 & | & 3 \\ 1 & 4 & 10 & 20 & | & 6 \end{bmatrix}\]
Row operations: Subtract Row 1 from Rows 2, 3, and 4: \[\begin{bmatrix} 1 & 1 & 1 & 1 & | & 0 \\ 0 & 1 & 2 & 3 & | & 1 \\ 0 & 2 & 5 & 9 & | & 3 \\ 0 & 3 & 9 & 19 & | & 6 \end{bmatrix}\]
Continue elimination: $R_3 - 2R_2$ and $R_4 - 3R_2$: \[\begin{bmatrix} 1 & 1 & 1 & 1 & | & 0 \\ 0 & 1 & 2 & 3 & | & 1 \\ 0 & 0 & 1 & 3 & | & 1 \\ 0 & 0 & 3 & 10 & | & 3 \end{bmatrix}\]
Final step: $R_4 - 3R_3$: \[\begin{bmatrix} 1 & 1 & 1 & 1 & | & 0 \\ 0 & 1 & 2 & 3 & | & 1 \\ 0 & 0 & 1 & 3 & | & 1 \\ 0 & 0 & 0 & 1 & | & 0 \end{bmatrix}\]
Back substitution:
- From Row 4: $x_4 = 0$
- From Row 3: $x_3 + 3(0) = 1 \Rightarrow x_3 = 1$
- From Row 2: $x_2 + 2(1) + 3(0) = 1 \Rightarrow x_2 = -1$
- From Row 1: $x_1 + (-1) + 1 + 0 = 0 \Rightarrow x_1 = 0$

Answer: $(x_1, x_2, x_3, x_4) = (0, -1, 1, 0)$

4.11. Rank-1 Matrix System (Assignment 1, Task 4)

Find the complete solution to: \[ \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \\ 4 & 8 & 12 & 16 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 30 \\ 40 \end{bmatrix} \]

(Note: The matrix has rank 1)

Click to see the solution

Key Concept: When a matrix has rank 1, all rows are multiples of one row. The entire system reduces to a single constraint equation with multiple free variables.

Observe the pattern:
- Row 2 = $2 \times$ Row 1
- Row 3 = $3 \times$ Row 1
- Row 4 = $4 \times$ Row 1
Check the right-hand side: $(20, 30, 40) = (2 \times 10, 3 \times 10, 4 \times 10)$. The system is consistent!
Augmented matrix: \[\begin{bmatrix} 1 & 2 & 3 & 4 & | & 10 \\ 2 & 4 & 6 & 8 & | & 20 \\ 3 & 6 & 9 & 12 & | & 30 \\ 4 & 8 & 12 & 16 & | & 40 \end{bmatrix}\]
After row reduction: \[\begin{bmatrix} 1 & 2 & 3 & 4 & | & 10 \\ 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}\]
Single constraint: $x_1 + 2x_2 + 3x_3 + 4x_4 = 10$
Rank-Nullity Theorem: $\text{rank}(A) = 1$, so $\dim(\text{Null}(A)) = 4 - 1 = 3$

The null space has dimension 3, so the complete solution has 3 free parameters.
Particular solution: Set $x_2 = 0, x_3 = 0, x_4 = 0 \Rightarrow x_1 = 10$

Particular solution: $\mathbf{x}_p = (10, 0, 0, 0)$
Homogeneous solution (Null space): Solve $x_1 + 2x_2 + 3x_3 + 4x_4 = 0$

Let $x_2 = 1, x_3 = 0, x_4 = 0 \Rightarrow x_1 = -2$. Basis vector: $(-2, 1, 0, 0)$ Let $x_2 = 0, x_3 = 1, x_4 = 0 \Rightarrow x_1 = -3$. Basis vector: $(-3, 0, 1, 0)$ Let $x_2 = 0, x_3 = 0, x_4 = 1 \Rightarrow x_1 = -4$. Basis vector: $(-4, 0, 0, 1)$

Answer: Complete solution: \[\mathbf{x} = \begin{bmatrix} 10 \\ 0 \\ 0 \\ 0 \end{bmatrix} + s\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t\begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \end{bmatrix} + u\begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \end{bmatrix}, \quad s, t, u \in \mathbb{R}\]

4.12. Underdetermined System with 3 Equations and 4 Variables (Assignment 1, Task 5)

Solve the underdetermined system: \[ \begin{cases} x + 2y + 3z + 4w = 5 \\ 2x + 4y + 6z + 8w = 10 \\ 3x + 5y + 7z + 9w = 12 \end{cases} \]

where $m = 3$ equations and $n = 4$ variables.

Click to see the solution

Key Concept: An underdetermined system has more variables than (independent) equations. The solutions form a higher-dimensional space parametrized by free variables.

Write the augmented matrix: \[\begin{bmatrix} 1 & 2 & 3 & 4 & | & 5 \\ 2 & 4 & 6 & 8 & | & 10 \\ 3 & 5 & 7 & 9 & | & 12 \end{bmatrix}\]
Row operations: $R_2 - 2R_1$ and $R_3 - 3R_1$: \[\begin{bmatrix} 1 & 2 & 3 & 4 & | & 5 \\ 0 & 0 & 0 & 0 & | & 0 \\ 0 & -1 & -2 & -3 & | & -3 \end{bmatrix}\]
Rearrange (move Row 3 up): \[\begin{bmatrix} 1 & 2 & 3 & 4 & | & 5 \\ 0 & -1 & -2 & -3 & | & -3 \\ 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}\]
RREF-like form (divide Row 2 by -1): \[\begin{bmatrix} 1 & 2 & 3 & 4 & | & 5 \\ 0 & 1 & 2 & 3 & | & 3 \\ 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}\]
Identify pivot and free variables:
- Pivot columns: 1, 2 (variables $x, y$)
- Free columns: 3, 4 (variables $z, w$)
- Rank: $r = 2$; Free variables: $n - r = 4 - 2 = 2$
Back substitution: Let $z = s$ and $w = t$ (free parameters).

From Row 2: $y + 2s + 3t = 3 \Rightarrow y = 3 - 2s - 3t$

From Row 1: $x + 2(3 - 2s - 3t) + 3s + 4t = 5$
- $x + 6 - 4s - 6t + 3s + 4t = 5$
- $x - s - 2t = -1$
- $x = s + 2t - 1$

Answer: General solution: \[(x, y, z, w) = (s + 2t - 1, 3 - 2s - 3t, s, t), \quad s, t \in \mathbb{R}\]

Or in vector form: \[\mathbf{x} = \begin{bmatrix} -1 \\ 3 \\ 0 \\ 0 \end{bmatrix} + s\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 2 \\ -3 \\ 0 \\ 1 \end{bmatrix}\]

4.13. Parametric System: Solution Type Depends on Parameters (Assignment 1, Task 6)

For what values of $a, b, c$ does the system have: 1. No solution? 2. A unique solution? 3. Infinitely many solutions?

\[ \begin{cases} x + 2y + 3z = a \\ 2x + 5y + 8z = b \\ 3x + 8y + 13z = c \end{cases} \]

Click to see the solution

Key Concept: For parametric systems, analyze the coefficient matrix and augmented matrix separately. The relationship between their ranks determines consistency and solution type.

Write the augmented matrix: \[\begin{bmatrix} 1 & 2 & 3 & | & a \\ 2 & 5 & 8 & | & b \\ 3 & 8 & 13 & | & c \end{bmatrix}\]
Row operations: $R_2 - 2R_1$ and $R_3 - 3R_1$: \[\begin{bmatrix} 1 & 2 & 3 & | & a \\ 0 & 1 & 2 & | & b - 2a \\ 0 & 2 & 4 & | & c - 3a \end{bmatrix}\]
Continue: $R_3 - 2R_2$: \[\begin{bmatrix} 1 & 2 & 3 & | & a \\ 0 & 1 & 2 & | & b - 2a \\ 0 & 0 & 0 & | & (c - 3a) - 2(b - 2a) \end{bmatrix}\]

Simplify Row 3 right-hand side: $(c - 3a) - 2(b - 2a) = c - 3a - 2b + 4a = c + a - 2b$
Final reduced form: \[\begin{bmatrix} 1 & 2 & 3 & | & a \\ 0 & 1 & 2 & | & b - 2a \\ 0 & 0 & 0 & | & c + a - 2b \end{bmatrix}\]
Case analysis:

Case 1: No solution This occurs when $\text{rank}(A) < \text{rank}([A|\mathbf{b}])$

Since the coefficient matrix has rank at most 2 (third row is all zeros in the $A$ part), the augmented matrix will have rank 3 (meaning inconsistent) only if Row 3’s right-hand side is nonzero.

Condition: $c + a - 2b \neq 0$, i.e., $c \neq 2b - a$

Answer for no solution: $c \neq 2b - a$

Case 2: Unique solution This requires $\text{rank}(A) = \text{rank}([A|\mathbf{b}]) = n = 3$

But the coefficient matrix has only 2 independent rows (the third row of $A$ is dependent on the first two). So $\text{rank}(A) = 2 < 3$.

Answer for unique solution: Never (impossible for any values of $a, b, c$)

Case 3: Infinitely many solutions This requires $\text{rank}(A) = \text{rank}([A|\mathbf{b}]) < n$

This happens when $c + a - 2b = 0$ and $\text{rank}(A) = 2 < 3$

Condition: $c = 2b - a$

Answer for infinitely many solutions: $c = 2b - a$

Summary:

No solution: $c \neq 2b - a$
Unique solution: Never possible
Infinitely many solutions: $c = 2b - a$

4.14. Solving a System Using REF (Tutorial 1, Example 1)

Solve using Row Echelon Form (REF): \[ \begin{cases} 2x + 4y - 2z = 2 \\ 4x + 9y - 3z = 8 \\ -2x - 3y + 7z = 10 \end{cases} \]

Click to see the solution

Key Concept: REF is an intermediate form (pivots don’t need to be 1) that enables efficient back substitution.

Augmented matrix: \[\begin{bmatrix} 2 & 4 & -2 & | & 2 \\ 4 & 9 & -3 & | & 8 \\ -2 & -3 & 7 & | & 10 \end{bmatrix}\]
Row operations: $R_2 - 2R_1$ and $R_3 + R_1$: \[\begin{bmatrix} 2 & 4 & -2 & | & 2 \\ 0 & 1 & 1 & | & 4 \\ 0 & 1 & 5 & | & 12 \end{bmatrix}\]
Continue: $R_3 - R_2$: \[\begin{bmatrix} 2 & 4 & -2 & | & 2 \\ 0 & 1 & 1 & | & 4 \\ 0 & 0 & 4 & | & 8 \end{bmatrix}\]
Back substitution:
- From Row 3: $4z = 8 \Rightarrow z = 2$
- From Row 2: $y + 2 = 4 \Rightarrow y = 2$
- From Row 1: $2x + 4(2) - 2(2) = 2 \Rightarrow 2x + 8 - 4 = 2 \Rightarrow 2x = -2 \Rightarrow x = -1$

Answer: $(x, y, z) = (-1, 2, 2)$

4.15. Solving a System Using RREF (Tutorial 1, Example 2)

Solve using Reduced Row Echelon Form (RREF): \[ \begin{cases} x + 2y + 3z = 9 \\ 2x - y + z = 8 \\ 3x + y - z = 3 \end{cases} \]

Click to see the solution

Key Concept: RREF (where each pivot is 1 and is the only nonzero in its column) allows direct reading of solutions without back substitution.

Augmented matrix: \[\begin{bmatrix} 1 & 2 & 3 & | & 9 \\ 2 & -1 & 1 & | & 8 \\ 3 & 1 & -1 & | & 3 \end{bmatrix}\]
Row operations: $R_2 - 2R_1$ and $R_3 - 3R_1$: \[\begin{bmatrix} 1 & 2 & 3 & | & 9 \\ 0 & -5 & -5 & | & -10 \\ 0 & -5 & -10 & | & -24 \end{bmatrix}\]
Simplify Row 2 (divide by -5) and Row 3: \[\begin{bmatrix} 1 & 2 & 3 & | & 9 \\ 0 & 1 & 1 & | & 2 \\ 0 & -5 & -10 & | & -24 \end{bmatrix}\]
Continue: $R_3 + 5R_2$: \[\begin{bmatrix} 1 & 2 & 3 & | & 9 \\ 0 & 1 & 1 & | & 2 \\ 0 & 0 & -5 & | & -14 \end{bmatrix}\]
Simplify Row 3 (divide by -5): \[\begin{bmatrix} 1 & 2 & 3 & | & 9 \\ 0 & 1 & 1 & | & 2 \\ 0 & 0 & -5 & | & -14 \end{bmatrix}\] where $z = 14/5$.

From the tutorial (given RREF): \[\begin{bmatrix} 1 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & | & -1 \\ 0 & 0 & 1 & | & 3 \end{bmatrix}\]

Direct reading from RREF: \[\begin{cases} x = 2 \\ y = -1 \\ z = 3 \end{cases}\]

Answer: $(x, y, z) = (2, -1, 3)$

4.16. Underdetermined System: 2 Equations, 4 Variables (Tutorial 1, Example 3)

Solve: \[ \begin{cases} x + 2y - z + 3w = 5 \\ 2x + 4y - 2z + 7w = 12 \end{cases} \]

Click to see the solution

Key Concept: With 2 equations and 4 variables, there are $4 - r = 4 - 2 = 2$ free variables (assuming full rank of 2).

Augmented matrix: \[\begin{bmatrix} 1 & 2 & -1 & 3 & | & 5 \\ 2 & 4 & -2 & 7 & | & 12 \end{bmatrix}\]
Row operation: $R_2 - 2R_1$: \[\begin{bmatrix} 1 & 2 & -1 & 3 & | & 5 \\ 0 & 0 & 0 & 1 & | & 2 \end{bmatrix}\]
RREF: \[\begin{bmatrix} 1 & 2 & -1 & 0 & | & -1 \\ 0 & 0 & 0 & 1 & | & 2 \end{bmatrix}\]

(After $R_1 - 3R_2$)
From RREF:
- $w = 2$
- $x + 2y - z = -1$, so $x = -1 - 2y + z$
Free variables: $y = s$, $z = t$

Answer: \[(x, y, z, w) = (-1 - 2s + t, s, t, 2), \quad s, t \in \mathbb{R}\]

4.17. Underdetermined System: 3 Equations, 5 Variables (Tutorial 1, Example 4)

Solve: \[ \begin{cases} x_1 + 2x_2 + 3x_3 + 4x_4 + 5x_5 = 10 \\ 2x_1 + 4x_2 + 6x_3 + 8x_4 + 10x_5 = 20 \\ 3x_1 + 5x_2 + 7x_3 + 9x_4 + 11x_5 = 30 \end{cases} \]

Click to see the solution

Key Concept: With 3 equations and 5 variables, there are at most 3 independent equations, leaving at least 2 free variables.

Augmented matrix: \[\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & | & 10 \\ 2 & 4 & 6 & 8 & 10 & | & 20 \\ 3 & 5 & 7 & 9 & 11 & | & 30 \end{bmatrix}\]
Observe: Row 2 = $2 \times$ Row 1, so $R_2 - 2R_1$ gives all zeros.
After row operations: $R_3 - 3R_1$: \[\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & | & 10 \\ 0 & 0 & 0 & 0 & 0 & | & 0 \\ 0 & -1 & -2 & -3 & -4 & | & 0 \end{bmatrix}\]
Rearrange to get RREF: \[\begin{bmatrix} 1 & 0 & -1 & -2 & -3 & | & 0 \\ 0 & 1 & 2 & 3 & 4 & | & 5 \\ 0 & 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}\]
From RREF:
- $x_1 = x_3 + 2x_4 + 3x_5$
- $x_2 = 5 - 2x_3 - 3x_4 - 4x_5$
- Free variables: $x_3 = s$, $x_4 = t$, $x_5 = u$

Answer: \[(x_1, x_2, x_3, x_4, x_5) = (s + 2t + 3u, 5 - 2s - 3t - 4u, s, t, u), \quad s, t, u \in \mathbb{R}\]

4.18. Inconsistent System Detection (Tutorial 1, Example 5)

Solve (or determine if inconsistent): \[ \begin{cases} x + y + z = 1 \\ 2x + 2y + 2z = 3 \\ 3x + 3y + 3z = 4 \end{cases} \]

Click to see the solution

Key Concept: When rows are proportional but the right-hand side is not proportional in the same way, the system is inconsistent.

Augmented matrix: \[\begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 2 & 2 & 2 & | & 3 \\ 3 & 3 & 3 & | & 4 \end{bmatrix}\]
Row operations: $R_2 - 2R_1$ and $R_3 - 3R_1$: \[\begin{bmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 0 & 0 & | & 1 \\ 0 & 0 & 0 & | & 1 \end{bmatrix}\]
Inconsistency detected: Row 2 represents $0x + 0y + 0z = 1$, which is impossible.

Answer: The system is inconsistent (no solution).

4.19. Solve Using REF (Tutorial 1, Example 6)

Solve using REF: \[ \begin{cases} 3x - 2y + z = 7 \\ x + y - 2z = -2 \\ 2x - y + 3z = 9 \end{cases} \]

Click to see the solution

Augmented matrix: \[\begin{bmatrix} 3 & -2 & 1 & | & 7 \\ 1 & 1 & -2 & | & -2 \\ 2 & -1 & 3 & | & 9 \end{bmatrix}\]
Swap R1 and R2 for a leading 1: \[\begin{bmatrix} 1 & 1 & -2 & | & -2 \\ 3 & -2 & 1 & | & 7 \\ 2 & -1 & 3 & | & 9 \end{bmatrix}\]
Row operations: $R_2 - 3R_1$ and $R_3 - 2R_1$: \[\begin{bmatrix} 1 & 1 & -2 & | & -2 \\ 0 & -5 & 7 & | & 13 \\ 0 & -3 & 7 & | & 13 \end{bmatrix}\]
Continue: Divide R2 by -1 (or keep as is) and perform $R_3 - (3/5)R_2$: \[\begin{bmatrix} 1 & 1 & -2 & | & -2 \\ 0 & -5 & 7 & | & 13 \\ 0 & 0 & 14/5 & | & 26/5 \end{bmatrix}\]
Back substitution:
- $(14/5)z = 26/5 \Rightarrow z = 26/14 = 13/7$
- $-5y + 7(13/7) = 13 \Rightarrow -5y + 13 = 13 \Rightarrow y = 0$
- $x + 0 - 2(13/7) = -2 \Rightarrow x = -2 + 26/7 = (-14 + 26)/7 = 12/7$

Answer: $(x, y, z) = (12/7, 0, 13/7)$ or approximately $(1.71, 0, 1.86)$

4.20. Intersection of Three Planes (Tutorial 1, Task 1)

Describe the intersection of the three planes (in four-dimensional space): \[ \begin{cases} u + v + w + z = 6 \\ u + w + z = 4 \\ u + w = 2 \end{cases} \]

Part (a): Is it a line or a point or an empty set? Part (b): What is the intersection if the fourth plane $u = -1$ is included? Part (c): Find a fourth equation that leaves us with no solution.

Click to see the solution

Key Concept: In higher dimensions, the intersection of hyperplanes forms lower-dimensional subspaces. The dimension of the solution space equals $n - r$ where $r$ is the rank.

(a) Intersection of three planes:

From the equations:
- 1. - (1): $(u + w + z) - (u + v + w + z) = 4 - 6 \Rightarrow -v = -2 \Rightarrow v = 2$
- 1. - (2): $(u + w) - (u + w + z) = 2 - 4 \Rightarrow -z = -2 \Rightarrow z = 2$
- From (3): $u + w = 2$
We have determined $v = 2$ and $z = 2$, while $u$ and $w$ satisfy $u + w = 2$.
Setting $w = t$ as a free variable: $u = 2 - t$.
General solution: $(u, v, w, z) = (2 - t, 2, t, 2)$ for $t \in \mathbb{R}$

This represents a line in 4D space (parametrized by one variable).

(b) Adding the fourth plane $u = -1$:

From part (a), we had $u = 2 - t$.
If $u = -1$: $2 - t = -1 \Rightarrow t = 3$
Thus $w = 3$.
The unique solution is: $(u, v, w, z) = (-1, 2, 3, 2)$

This is a point in 4D space.

(c) Fourth equation that leads to no solution:

From part (a), the solution set is $(u, v, w, z) = (2 - t, 2, t, 2)$ for $t \in \mathbb{R}$.
Any equation that contradicts these values for all $t$ will make the system inconsistent.
For example, $u + v + w + z = 7$ (different from the actual sum, which is always $(2-t) + 2 + t + 2 = 6$).

Or simply: $v = 3$ (contradicts $v = 2$), or $z = 3$ (contradicts $z = 2$).

Answer:

The intersection is a line in 4D.
With the fourth plane, the intersection is a point: $(-1, 2, 3, 2)$.
A fourth equation causing no solution: $u + v + w + z = 7$ (or $v = 3$, or $z = 3$, etc.)

4.21. Find Another Solution Using Column Picture (Tutorial 1, Task 2)

When $b = (2, 5, 7)$, find a solution $(u, v, w)$ to the vector equation different from $(1, 0, 1)$: \[u\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + v\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + w\begin{bmatrix} 1 \\ 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 7 \end{bmatrix}\]

Click to see the solution

Key Concept: If a system is singular (dependent rows), the solution is not unique. Any particular solution can be combined with homogeneous solutions to find other solutions.

Convert to linear system: \[\begin{cases} u + v + w = 2 & (1) \\ 2u + 3w = 5 & (2) \\ 3u + v + 4w = 7 & (3) \end{cases}\]
Check dependency: (1) + (2) = (3)? $(u + v + w) + (2u + 3w) = 3u + v + 4w = $ RHS: $2 + 5 = 7$ ✓

So the three equations are dependent; the system is singular.
Find the null space (general homogeneous solution): Since the rows are dependent, we can express one as a combination of others. Notice: $(1) + (2) - (3) = 0$, which means $(u, v, w) = (-1, 2, -1)$ is a homogeneous solution.
General solution: Particular solution: $(u_p, v_p, w_p) = (1, 0, 1)$ Homogeneous solution: $(u_h, v_h, w_h) = (-1, 2, -1)$

General solution: $(u, v, w) = (1, 0, 1) + t(-1, 2, -1) = (1 - t, 2t, 1 - t)$ for $t \in \mathbb{R}$
Different solution: Set $t = 1$: $(u, v, w) = (0, 2, 0)$

Or set $t = 1$: $(u, v, w) = (0, 2, 0)$.

Answer: General solution: $(u, v, w) = (1 - t, 2t, 1 - t)$ for any $t \neq 0$. Another particular solution: $(0, 2, 0)$.

4.22. Linear Dependence and Null Space (Tutorial 1, Task 3)

Show that the three column vectors lie in the same plane by expressing the third column as a combination of the first two. What are all solutions $(u, v, w)$ to: \[u\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + v\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} + w\begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\]

Click to see the solution

Key Concept: Linear dependence means one vector can be written as a linear combination of the others. The null space consists of all solutions to $A\mathbf{x} = \mathbf{0}$.

Express the third column as a combination of the first two: We seek $\alpha, \beta$ such that: \[\alpha\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + \beta\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}\]
Set up equations: \[\begin{cases} \alpha + \beta = 1 \\ \alpha + 2\beta = 3 \\ \beta = 2 \end{cases}\]
Solve:
- From equation 3: $\beta = 2$
- From equation 1: $\alpha + 2 = 1 \Rightarrow \alpha = -1$
- Verify with equation 2: $-1 + 2(2) = -1 + 4 = 3$ ✓
Linear dependence relation: \[-1 \cdot \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + 2 \cdot \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} - 1 \cdot \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\]

This means $(u, v, w) = (-1, 2, -1)$ is in the null space.
All solutions to the homogeneous system: Since the three vectors are linearly dependent (lie in a plane), the null space is 1-dimensional.

Basis for null space: $(u, v, w) = (-1, 2, -1)$

General solution: $(u, v, w) = t(-1, 2, -1) = (-t, 2t, -t)$ for $t \in \mathbb{R}$

Answer:

The third column can be expressed as: $\begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix} = -1 \cdot \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + 2 \cdot \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$

This shows the three columns lie in a 2D plane (linearly dependent).

All solutions: $(u, v, w) = t(-1, 2, -1)$ for $t \in \mathbb{R}$, which includes $(0, 0, 0)$ (when $t = 0$) and $(-1, 2, -1)$ (when $t = 1$).

4.23. Collinearity Condition (Tutorial 1, Task 4)

Under what condition on $y_1, y_2, y_3$ do the points $(0, y_1)$, $(1, y_2)$, $(2, y_3)$ lie on a straight line?

Click to see the solution

Key Concept: Three points are collinear if they satisfy the same linear equation $y = mx + c$.

Set up the line equation: For a line $y = mx + c$, the three points must satisfy:
- Point 1: $y_1 = m(0) + c \Rightarrow y_1 = c$
- Point 2: $y_2 = m(1) + c \Rightarrow y_2 = m + c$
- Point 3: $y_3 = m(2) + c \Rightarrow y_3 = 2m + c$
Express in terms of $y_1$:
- $c = y_1$ (from Point 1)
- $m = y_2 - y_1$ (from Point 2: $y_2 = m + y_1$)
- From Point 3: $y_3 = 2m + c = 2(y_2 - y_1) + y_1 = 2y_2 - 2y_1 + y_1 = 2y_2 - y_1$
Collinearity condition: \[y_3 = 2y_2 - y_1\]

Rearranging: \[2y_2 - y_1 - y_3 = 0\]

Or: \[y_1 - 2y_2 + y_3 = 0\]

Answer: The three points are collinear if and only if: \[\boxed{2y_2 = y_1 + y_3}\] or equivalently, \[\boxed{2y_2 - y_1 - y_3 = 0}\]

4.24. Intersection of Two Planes: Find Points on the Line (Tutorial 1, Task 5)

Find a point with $z = 2$ on the intersection line of the planes $x + y + 3z = 6$ and $x - y + z = 4$. Find the point with $z = 0$ and a third point halfway between.

Click to see the solution

Key Concept: The intersection of two planes is a line (in 3D, assuming the planes are not parallel). We can find specific points by setting one coordinate and solving for the others.

(a) Point with $z = 2$:

Substitute $z = 2$ into both plane equations:
- $x + y + 3(2) = 6 \Rightarrow x + y = 0$
- $x - y + 2 = 4 \Rightarrow x - y = 2$
Solve the system:
- Adding: $(x + y) + (x - y) = 0 + 2 \Rightarrow 2x = 2 \Rightarrow x = 1$
- From $x + y = 0$: $1 + y = 0 \Rightarrow y = -1$
Point with $z = 2$: $(1, -1, 2)$

(b) Point with $z = 0$:

Substitute $z = 0$ into both plane equations:
- $x + y + 0 = 6 \Rightarrow x + y = 6$
- $x - y + 0 = 4 \Rightarrow x - y = 4$
Solve:
- Adding: $2x = 10 \Rightarrow x = 5$
- From $x + y = 6$: $5 + y = 6 \Rightarrow y = 1$
Point with $z = 0$: $(5, 1, 0)$

(c) Midpoint:

The midpoint between $(1, -1, 2)$ and $(5, 1, 0)$ is: \[\left(\frac{1 + 5}{2}, \frac{-1 + 1}{2}, \frac{2 + 0}{2}\right) = (3, 0, 1)\]

Answer:

Point with $z = 2$: $(1, -1, 2)$
Point with $z = 0$: $(5, 1, 0)$
Midpoint: $(3, 0, 1)$

4.25. Immediate Solution from Column Picture (Tutorial 1, Task 6)

In the vector equation below, the third column (multiplying $w$) is the same as the right side $b$. What solution for $(u, v, w)$ does this immediately give?

\[u\begin{bmatrix} 6 \\ 4 \\ 2 \end{bmatrix} + v\begin{bmatrix} 7 \\ 5 \\ -2 \end{bmatrix} + w\begin{bmatrix} 8 \\ 9 \\ 7 \end{bmatrix} = \begin{bmatrix} 8 \\ 9 \\ 7 \end{bmatrix}\]

Click to see the solution

Key Concept: If one of the column vectors equals the target vector $b$, then we can immediately read off a solution: set the corresponding variable to 1 and all others to 0.

Observe: The third column vector is: \[w\text{-column} = \begin{bmatrix} 8 \\ 9 \\ 7 \end{bmatrix}\]

And the right-hand side is: \[b = \begin{bmatrix} 8 \\ 9 \\ 7 \end{bmatrix}\]

They are identical!
Immediate solution: If we set $u = 0$, $v = 0$, and $w = 1$, then: \[0 \cdot \begin{bmatrix} 6 \\ 4 \\ 2 \end{bmatrix} + 0 \cdot \begin{bmatrix} 7 \\ 5 \\ -2 \end{bmatrix} + 1 \cdot \begin{bmatrix} 8 \\ 9 \\ 7 \end{bmatrix} = \begin{bmatrix} 8 \\ 9 \\ 7 \end{bmatrix}\] ✓

Answer: $(u, v, w) = (0, 0, 1)$

4.26. Elimination and Back Substitution with Missing Coefficient (Tutorial 1, Task 7)

Apply elimination (circle the pivots) and back-substitution to solve: \[ \begin{cases} 2x - 3y = 3 \\ 4x - 5y + z = 7 \\ 2x - y - 3z = 5 \end{cases} \]

Click to see the solution

Write augmented matrix: \[\begin{bmatrix} 2 & -3 & 0 & | & 3 \\ 4 & -5 & 1 & | & 7 \\ 2 & -1 & -3 & | & 5 \end{bmatrix}\]
Row operations: $R_2 - 2R_1$ and $R_3 - R_1$: \[\begin{bmatrix} 2 & -3 & 0 & | & 3 \\ 0 & 1 & 1 & | & 1 \\ 0 & 2 & -3 & | & 2 \end{bmatrix}\]

Pivots: $\boxed{2}$ and $\boxed{1}$
Continue: $R_3 - 2R_2$: \[\begin{bmatrix} 2 & -3 & 0 & | & 3 \\ 0 & 1 & 1 & | & 1 \\ 0 & 0 & -5 & | & 0 \end{bmatrix}\]

Third pivot: $\boxed{-5}$
Back substitution:
- From Row 3: $-5z = 0 \Rightarrow z = 0$
- From Row 2: $y + 0 = 1 \Rightarrow y = 1$
- From Row 1: $2x - 3(1) = 3 \Rightarrow 2x = 6 \Rightarrow x = 3$

Answer: $(x, y, z) = (3, 1, 0)$

4.27. Parameter Analysis: Row Exchange and Singularity (Tutorial 1, Task 8)

For the system with parameter $d$: \[ \begin{cases} 2x + 5y + z = 0 \\ 4x + dy + z = 2 \\ y - z = 3 \end{cases} \]

Part (a): Which number $d$ forces a row exchange? Part (b): Which $d$ makes the system singular (no third pivot)?

Click to see the solution

Augmented matrix: \[\begin{bmatrix} 2 & 5 & 1 & | & 0 \\ 4 & d & 1 & | & 2 \\ 0 & 1 & -1 & | & 3 \end{bmatrix}\]
First elimination: $R_2 - 2R_1$: \[\begin{bmatrix} 2 & 5 & 1 & | & 0 \\ 0 & d - 10 & -1 & | & 2 \\ 0 & 1 & -1 & | & 3 \end{bmatrix}\]

(a) Row exchange occurs when:

The second pivot would be zero if $d - 10 = 0$, i.e., $d = 10$.

\[\begin{bmatrix} 2 & 5 & 1 & | & 0 \\ 0 & 0 & -1 & | & 2 \\ 0 & 1 & -1 & | & 3 \end{bmatrix} \xrightarrow{R_2 \leftrightarrow R_3} \begin{bmatrix} 2 & 5 & 1 & | & 0 \\ 0 & 1 & -1 & | & 3 \\ 0 & 0 & -1 & | & 2 \end{bmatrix}\]

Answer for (a): $d = 10$

(b) Singularity (missing third pivot) occurs when:

We need $d - 10 = 1$ so that after row reduction, Rows 2 and 3 become proportional:

$d - 10 = 1 \Rightarrow d = 11$

\[\begin{bmatrix} 2 & 5 & 1 & | & 0 \\ 0 & 1 & -1 & | & 2 \\ 0 & 1 & -1 & | & 3 \end{bmatrix} \xrightarrow{R_3 - R_2} \begin{bmatrix} 2 & 5 & 1 & | & 0 \\ 0 & 1 & -1 & | & 2 \\ 0 & 0 & 0 & | & 1 \end{bmatrix}\]

The third row represents $0 = 1$, so the system is inconsistent.

Answer for (b): $d = 11$ (system becomes singular/inconsistent)

4.28. Constructing Systems with Row Exchanges (Tutorial 1, Task 9)

Part (a): Construct a 3×3 system that needs two row exchanges to reach triangular form and find a solution. Part (b): Construct a 3×3 system that needs a row exchange but breaks down later.

Click to see the solution

Key Concept: Systems requiring multiple row exchanges have zeros in critical pivot positions. Systems that break down are inconsistent.

(a) System requiring two row exchanges:

Example matrix: \[\begin{bmatrix} 0 & 1 & 1 & | & 3 \\ 0 & 0 & 1 & | & 2 \\ 1 & 1 & 1 & | & 5 \end{bmatrix}\]

First exchange: $R_1 \leftrightarrow R_3$: \[\begin{bmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 0 & 1 & | & 2 \\ 0 & 1 & 1 & | & 3 \end{bmatrix}\]
Second exchange: $R_2 \leftrightarrow R_3$: \[\begin{bmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 1 & 1 & | & 3 \\ 0 & 0 & 1 & | & 2 \end{bmatrix}\]
Back substitution:
- $z = 2$
- $y + 2 = 3 \Rightarrow y = 1$
- $x + 1 + 2 = 5 \Rightarrow x = 2$

Answer for (a): Solution is $(x, y, z) = (2, 1, 2)$

(b) System requiring row exchange but becoming inconsistent:

Example matrix: \[\begin{bmatrix} 1 & 1 & 1 & | & 4 \\ 0 & 0 & 1 & | & 3 \\ 1 & 1 & 2 & | & 6 \end{bmatrix}\]

Apply: $R_3 - R_1$: \[\begin{bmatrix} 1 & 1 & 1 & | & 4 \\ 0 & 0 & 1 & | & 3 \\ 0 & 0 & 1 & | & 2 \end{bmatrix}\]
Row exchange needed: $R_2 \leftrightarrow R_3$: \[\begin{bmatrix} 1 & 1 & 1 & | & 4 \\ 0 & 0 & 1 & | & 2 \\ 0 & 0 & 1 & | & 3 \end{bmatrix}\]
Continue: $R_3 - R_2$: \[\begin{bmatrix} 1 & 1 & 1 & | & 4 \\ 0 & 0 & 1 & | & 2 \\ 0 & 0 & 0 & | & 1 \end{bmatrix}\]

The system is inconsistent ($0 = 1$).

Answer for (b): The system breaks down to the inconsistency $0 = 1$.