How to generate a 3×3 linear system with a unique integer solution

Use this method when you want students to solve a system of three linear equations in three variables and ensure that the system has a unique solution with integer values.

Following this guide, you will generate a 3×3 linear system whose determinant is non-zero and whose solution is predefined and guaranteed to be composed of integers.

See it in action: Watch how this logic is implemented inside Nubric:

Before you begin

Requirements

• Basic knowledge of how to create a Nubric question
• Familiarity with adding an algorithm to generate random variables
• Basic understanding of matrices and determinants

Steps

Define a reusable random function.

r() := random([-5..5]/[0])

This helper function generates random non-zero integers.

Generate a valid coefficient matrix.

repeat
    A = [ [r(), r(), r()],
           [r(), r(), r()],
           [r(), r(), r()] ]
until determinant(A) != 0

This guarantees that the system has a unique solution.

Define a known integer solution.

sx = r()
sy = r()
sz = r()
solution_vect = [ sx, sy, sz ]

This sets the exact solution the system will have.

Compute the right-hand side.

b_vect = A * solution_vect

This ensures that the generated system corresponds to the predefined solution.

Extract coefficients.

a11 = A subindex_operator(1,1)
a12 = A subindex_operator(1,2)
a13 = A subindex_operator(1,3)
a21 = A subindex_operator(2,1)
a22 = A subindex_operator(2,2)
a23 = A subindex_operator(2,3)
a31 = A subindex_operator(3,1)
a32 = A subindex_operator(3,2)
a33 = A subindex_operator(3,3)

b1 = b_vect subindex_operator(1)
b2 = b_vect subindex_operator(2)
b3 = b_vect subindex_operator(3)

These variables can now be used to display the system:

Verify it worked

• Preview the question multiple times.
• Confirm that the determinant of matrix A is never 0.
• Solve the generated system manually.
• Verify that the solution is always (sx, sy, sz)

Full algorithm (copy-paste version)

Use the complete version below if you want to copy the logic directly into your question algorithm:

# Define a function to generate random integers, excluding zero
r() := random([-5..5]/[0]) 

# Generate a coefficient matrix A with a non-zero determinant
repeat
    A = [ [r(), r(), r()],
           [r(), r(), r()],
           [r(), r(), r()] ]
until determinant(A) != 0

# Generate a unique integer solution vector
sx = r()
sy = r()
sz = r()
solution_vect = [ sx, sy, sz ]

# Calculate the right-hand side vector
b_vect = A * solution_vect

# Extract coefficients
a11 = A subindex_operator(1,1)
a12 = A subindex_operator(1,2)
a13 = A subindex_operator(1,3)
a21 = A subindex_operator(2,1) 
a22 = A subindex_operator(2,2)
a23 = A subindex_operator(2,3)
a31 = A subindex_operator(3,1)
a32 = A subindex_operator(3,2)
a33 = A subindex_operator(3,3)

b1 = b_vect subindex_operator(1)
b2 = b_vect subindex_operator(2)
b3 = b_vect subindex_operator(3)

Options and variations

• If you want larger systems, you can increase the interval in r()
• If you want rational solutions instead of integers, you can allow fractions in the r() function.
• If you want more challenging systems, you can use larger coefficient ranges to increase computational difficulty.

Common errors

The system has no unique solution.
Ensure the condition determinant(A) != 0 is correctly applied

Infinite regeneration loop.
Broaden the interval if the determinant condition is too restrictive

The displayed coefficients don't match the solution.
Confirm that the right-hand side vector is computed as A * solution_vect

Related

• Understanding Advanced Logic in Nubric
• Common Patterns and Best Practices
• Glossary of Commands