Enter linear systems and read the determinant, ranks and solution in the solver, and predict — before you read it off the screen — the determinant of a system, whether an inconsistent system has no solution, whether a dependent one has infinitely many, and the exact solution of a system with a unique answer.
For a square system Ax = b, the determinant det(A) decides the easy case: det ≠ 0 means one unique solution. When det(A) = 0 the verdict turns on the ranks (Rouché–Capelli): if rank([A|b]) > rank(A) the system is inconsistent — no solution; if the ranks are equal but below the number of unknowns, there are free variables — infinitely many solutions. Only the unique case has a single vector to report.
Open the solver. Each step names the coefficient matrix A and the right-hand side b to enter; type them into the inputs and read the determinant, rank and solution readouts.
A x = b (enter the matrix A and vector b each step names)
Read: det(A), rank(A), rank([A|b]), solution vector
Watch the determinant, rank and solution readouts in the main panel — every number you predict is printed there.
Enter A = [[1, 1], [1, -1]] and b = (3, 1) — the system x + y = 3, x - y = 1. The determinant is det(A) = a11 a22 - a12 a21. Predict det(A), then read the readout.
The readout shows det(A) = 1·(-1) - 1·1 = -2. Because det ≠ 0, the system has one unique solution — here x = 2, y = 1.
Enter A = [[1, 1], [1, 1]] and b = (1, 2) — the system x + y = 1, x + y = 2. Predict the rank gap rank([A|b]) - rank(A), then read the ranks.
The augmented rank exceeds the coefficient rank by 1: rank(A) = 1 but rank([A|b]) = 2, a "0 = 1" row. The system is inconsistent — no solution.
Enter A = [[1, 1], [2, 2]] and b = (1, 2) — the system x + y = 1, 2x + 2y = 2 (the second row is twice the first). Predict the number of free variables, then read the readout.
The ranks are equal — rank(A) = rank([A|b]) = 1 — but below the 2 unknowns, so there is 1 free variable. The system is consistent and under-determined — infinitely many solutions along a line.
Enter A = [[2, 1, -1], [-3, -1, 2], [-2, 1, 2]] and b = (8, -11, -3). Predict the third coordinate z of the solution, then read the solution readout.
Gaussian elimination gives the unique solution (x, y, z) = (2, 3, -1), so the third coordinate is z = -1. Cramer's rule inside the demo agrees, an independent second route to the same answer.
Everything above is waiting in the solver. Enter any square system, watch the elimination steps march to echelon form, and see the determinant, the ranks and the one/none/infinite verdict update at once — with the two lines of a 2×2 drawn so you can see them cross, run parallel, or coincide.
- Read a system's determinant, det(A) = -2, and saw det ≠ 0 means one unique solution
- Saw an inconsistent system's rank gap rank([A|b]) - rank(A) = 1 — no solution
- Saw a dependent system with 1 free variable — infinitely many solutions
- Solved a 3×3 to the unique (2, 3, -1), with third coordinate z = -1
- Every value you predicted is the demo's own Gaussian elimination, not a picture