# Systems of Linear Equations

Business I · Mathematics I · https://tryals.app/learn/business-i/systems-of-linear-equations

## Many Equations, One Verdict

A **system of linear equations** is business modeling's native language, written compactly as $Ax = b$. Three fates govern every system, compared by the **rank** of $A$, the augmented matrix $(A|b)$, and unknowns $n$.

| Comparison | Verdict | Geometry |
|---|---|---|
| $rank(A) = rank(A|b) = n$ | **Unique solution** | Planes meet at one point |
| $rank(A) = rank(A|b) < n$ | **Infinitely many** | Overlapping equations |
| $rank(A) < rank(A|b)$ | **No solution** | Parallel contradiction |

> **Common pitfall:** A zero row like $0 = 0$ means redundancy (infinite solutions), while $0 = c$ where $c \ne 0$ means no solution.

## Solving Methods

**Gaussian elimination** uses row operations (swapping, scaling, adding rows) to transform the matrix into **triangular** shape, then back-substitutes from the bottom up. Row operations preserve the exact solution set.

**Cramer's rule** solves square systems when $\det A \ne 0$ via the formula:

$$x_i = \frac{\det A_i}{\det A}$$

Here, $A_i$ is $A$ with column $i$ replaced by $b$. It shines for $2 \times 2$ and $3 \times 3$ systems, but Gaussian elimination wins for anything larger.

**The geometry to keep:** last lesson's determinant is your early warning. $\det A \ne 0$ guarantees the unique solution case immediately.

## Practice questions

7 of this lesson's 14 practice questions, with answers. The full set is in the app.

### 1. Order the steps of solving $Ax = b$ by Gaussian elimination.

**Answer:**

1. Write the augmented matrix $(A|b)$
2. Use row operations to zero out below the diagonal
3. Reach triangular form
4. Solve the last equation, one unknown
5. Back-substitute upward to recover the rest

**Why:** Augment → sweep → triangle → bottom equation → climb back up. Elimination turns a tangle of equations into a staircase you descend once and ascend once.

Page: https://tryals.app/practice/business-i/systems-of-linear-equations/order-the-steps-of-solving-ax-b-by-gaussian-elimination

### 2. If a square system $Ax = b$ has two different solutions, it has infinitely many. Why can there be no system with exactly two?

A. Because square systems always have either one solution or none
B. Because the difference of two solutions solves Ax = 0
C. Because two solutions would make the matrix non-square
D. Because the determinant would have to take two different values at once

**Answer:** B. Because the difference of two solutions solves Ax = 0

**Why:** Solution sets are affine: a particular solution plus the null space. A null space with any nonzero vector in it is infinite, so "exactly two" is impossible.

Page: https://tryals.app/practice/business-i/systems-of-linear-equations/if-a-square-system-ax-b-has-two-different-solutions-it-has

### 3. Complete the classification.

**Answer:** If $rank(A)$ equals $rank(A|b)$ and both equal the number of unknowns, the system has a **unique** solution; if the augmented rank is larger, the system is **incompatible**.

**Why:** Rouché–Frobenius in one sentence: matching full ranks = one point; a rank jump in the augmented matrix = contradiction.

Page: https://tryals.app/practice/business-i/systems-of-linear-equations/complete-the-classification

### 4. Which moves are legal row operations in Gaussian elimination?

**Answer:**

- Legal row operation: Swap row 1 and row 3, Multiply a row by 5, Add twice row 1 to row 2
- Illegal move: Multiply a row by 0, Delete an inconvenient equation, Square both sides of one equation

**Why:** Legal operations are reversible, they preserve the solution set exactly. Multiplying by zero erases information; deleting or squaring changes what the system says.

Page: https://tryals.app/practice/business-i/systems-of-linear-equations/which-moves-are-legal-row-operations-in-gaussian-elimination

### 5. Match each Rouché–Frobenius situation to its verdict.

**Answer:**

- $rank(A) = rank(A|b) = n$ → Unique solution
- $rank(A) = rank(A|b) < n$ → Infinitely many solutions
- $rank(A) < rank(A|b)$ → No solution
- $\det A \ne 0$ (square system) → Unique solution guaranteed

**Why:** Ranks equal and full: one point. Ranks equal but short: freedom remains. Augmented rank higher: contradiction. And the determinant is the square-system shortcut to case one.

Page: https://tryals.app/practice/business-i/systems-of-linear-equations/match-each-rouche-frobenius-situation-to-its-verdict

### 6. A factory models output with $Ax = b$ and finds the system incompatible. Which statements are true?

A. Doubling all capacities ($2b$) automatically restores a solution
B. Geometrically, some constraint lines/planes never meet
C. $rank(A|b)$ exceeds $rank(A)$
D. The constraints contradict each other, no production plan satisfies all of them

**Answer:** B. Geometrically, some constraint lines/planes never meet; C. $rank(A|b)$ exceeds $rank(A)$; D. The constraints contradict each other, no production plan satisfies all of them

**Why:** Incompatibility is a *structural* contradiction between constraints. Rescaling $b$ may or may not cure it, the fix is renegotiating constraints, not multiplying wishes.

Page: https://tryals.app/practice/business-i/systems-of-linear-equations/a-factory-models-output-with-ax-b-and-finds-the-system

### 7. A production planner writes four constraints in three unknowns and the system has no solution. What has the model actually discovered?

A. That there are too many equations, which always makes a system unsolvable
B. That the solution exists but requires more unknowns to express
C. That the requirements are mutually inconsistent
D. That the coefficient matrix must be square before solving

**Answer:** C. That the requirements are mutually inconsistent

**Why:** Inconsistency is a finding rather than an error: the constraints as stated cannot all hold, so the negotiation moves from the maths to the business. Overdetermined systems are often perfectly solvable, which is why option 2's counting rule is not the test.

Page: https://tryals.app/practice/business-i/systems-of-linear-equations/a-production-planner-writes-four-constraints-in-three-unknowns-and