Square System of Equations
Square System of Equations
A linear system of equations in which the number of variables equals the number of equations. The coefficient matrix of a square system is a square matrix.
Example:
2x + 3y = 4
5x + 6y = 7
Key Formula
Ax=b
Where:
- A = The n × n coefficient matrix (square because the system has n equations and n unknowns)
- x = The n × 1 column vector of unknown variables
- b = The n × 1 column vector of constants on the right-hand side
Worked Example
Problem: Solve the square system of equations:
2x + 3y = 12
4x − y = 2
Step 1: Confirm the system is square. There are 2 equations and 2 unknowns (x and y), so this is a 2 × 2 square system.
Step 2: Write the coefficient matrix and check its determinant to verify a unique solution exists.
A=(243−1),det(A)=(2)(−1)−(3)(4)=−2−12=−14=0
Step 3: Since the determinant is nonzero, a unique solution exists. Use elimination: multiply the first equation by 2 and subtract from the second equation, or solve directly. From the second equation, solve for y in terms of x.
4x−y=2⟹y=4x−2
Step 4: Substitute into the first equation and solve for x.
2x+3(4x−2)=12⟹2x+12x−6=12⟹14x=18⟹x=1418=79
Step 5: Substitute back to find y.
y=4(79)−2=736−714=722
Answer: The unique solution is x = 9/7 and y = 22/7.
Another Example
This example extends to a 3 × 3 square system, showing that the same determinant-based check and elimination approach generalizes beyond 2 × 2 systems.
Problem: Determine whether the following 3 × 3 square system has a unique solution, and if so, find it:
x + y + z = 6
2x − y + z = 3
x + 2y − z = 3
Step 1: Confirm the system is square: 3 equations, 3 unknowns (x, y, z). Write the coefficient matrix.
A=1211−1211−1
Step 2: Compute the determinant to check for a unique solution.
det(A)=1((−1)(−1)−(1)(2))−1((2)(−1)−(1)(1))+1((2)(2)−(−1)(1))=1(1−2)−1(−2−1)+1(4+1)=−1+3+5=7=0
Step 3: Use elimination. Add equations 1 and 3 to eliminate z: (x + y + z) + (x + 2y − z) = 6 + 3.
2x+3y=9...(i)
Step 4: Add equations 2 and 3 to eliminate z: (2x − y + z) + (x + 2y − z) = 3 + 3.
3x+y=6...(ii)
Step 5: Solve equations (i) and (ii). From (ii): y = 6 − 3x. Substitute into (i): 2x + 3(6 − 3x) = 9, giving 2x + 18 − 9x = 9, so −7x = −9, thus x = 9/7. Then y = 6 − 27/7 = 15/7, and from equation 1: z = 6 − 9/7 − 15/7 = 18/7.
x=79,y=715,z=718
Answer: The unique solution is x = 9/7, y = 15/7, z = 18/7.
Frequently Asked Questions
Does a square system of equations always have exactly one solution?
No. A square system has exactly one solution only when its coefficient matrix is invertible (i.e., its determinant is nonzero). If the determinant is zero, the system is either inconsistent (no solution) or dependent (infinitely many solutions). Being square is necessary but not sufficient for a unique solution.
What is the difference between a square system and an overdetermined or underdetermined system?
A square system has the same number of equations as unknowns. An overdetermined system has more equations than unknowns, while an underdetermined system has fewer equations than unknowns. Only square systems can be written in the form Ax = b with A being a square matrix, which allows you to use the determinant or matrix inverse to analyze and solve them directly.
How do you solve a square system using the inverse matrix?
If the coefficient matrix A is invertible, you can find the solution by computing x = A⁻¹b. This works because multiplying both sides of Ax = b on the left by A⁻¹ isolates x. For large systems, however, methods like Gaussian elimination or Cramer's rule are often more practical than computing the full inverse.
Square System vs. Non-Square System
| Square System | Non-Square System | |
|---|---|---|
| Definition | Number of equations equals number of unknowns (n × n) | Number of equations differs from number of unknowns (m × n, m ≠ n) |
| Coefficient matrix shape | Square matrix (n × n) | Rectangular matrix (m × n) |
| Unique solution possible? | Yes, when det(A) ≠ 0 | Overdetermined: possible but rare; Underdetermined: generally no unique solution |
| Solvable via matrix inverse? | Yes, when A is invertible: x = A⁻¹b | No — A⁻¹ does not exist for non-square matrices |
| Typical solution method | Gaussian elimination, Cramer's rule, or inverse matrix | Gaussian elimination, least squares (overdetermined), or parametric solutions (underdetermined) |
Why It Matters
Square systems appear throughout algebra and linear algebra courses whenever you set up equations to find unknown quantities—from solving two-variable word problems in algebra to analyzing circuits and structural loads in engineering. Recognizing that a system is square tells you immediately that a determinant test can reveal whether a unique solution exists. Many standard solution techniques, including Cramer's rule and the inverse-matrix method, apply only to square systems.
Common Mistakes
Mistake: Assuming a square system always has a unique solution.
Correction: A square system can still have no solution or infinitely many solutions if the determinant of the coefficient matrix is zero. Always check det(A) ≠ 0 before concluding a unique solution exists.
Mistake: Confusing the coefficient matrix with the augmented matrix when checking if a system is square.
Correction: The coefficient matrix contains only the coefficients of the variables (n × n for a square system). The augmented matrix includes the constants column and is n × (n + 1). Use the coefficient matrix to determine whether the system is square.
Related Terms
- Linear System of Equations — The broader category that includes square systems
- Variable — The unknowns whose count defines a square system
- Equation — Each equation contributes one row to the system
- Coefficient Matrix — The square matrix formed from the coefficients
- Square Matrix — The shape of the coefficient matrix in a square system
- Determinant — Used to test whether a square system has a unique solution
- Cramer's Rule — A solution method that applies only to square systems
- Inverse of a Matrix — If A is invertible, x = A⁻¹b solves the system
