Matrix Inverse Method for Linear Systems
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Direct answer
A nonsingular coefficient matrix converts AX = B into X = A⁻¹B — compute |A|, form adj A, and the unique solution falls out of one multiplication. If |A| = 0 the inverse route dies: the system then has either no solution or infinitely many, decided by comparing rank(A) with rank([A | B]). Rectangular systems, such as two equations in three unknowns, have no inverse at all; row reduction plus one free parameter finishes them. Pre-multiplication is the whole game — X = A⁻¹B, never BA⁻¹.
What you must remember
- Square method (NCERT Class 12): |A| ≠ 0 ⇒ X = A⁻¹B = (adj A/|A|)B, unique; the method is a four-mark staple presented as "system of equations by matrix method".
- 2×2 inverse shortcut: [[a, b], [c, d]]⁻¹ = (1/(ad − bc))[[d, −b], [−c, a]] — swap the diagonal, negate the off-diagonal, divide by the determinant.
- Singular square case: |A| = 0 ⇒ the system is either inconsistent or has infinitely many solutions; it can never have exactly one.
- Rectangular systems: 2 equations in 3 unknowns carry rank at most 2 < 3, so if consistent they hold infinitely many solutions, parametrised with 3 − rank free variables.
- Homogeneous systems: AX = 0 with |A| ≠ 0 forces X = 0; nontrivial solutions require |A| = 0 — Advanced asks for parameter values that make this happen.
- Verification habit: substitute the answer back into all original equations; an adjoint arithmetic slip is the single most common error here.
- Cramer connection: x = |A₁|/|A| is the same computation organised column-by-column — cofactors either way.
One inverse computation, start to verified finish
Solve x + y + z = 6, x − y + z = 2, 2x + y − z = 1. Here A = [[1, 1, 1], [1, −1, 1], [2, 1, −1]] and |A| = 1(1 − 1) − 1(−1 − 2) + 1(1 + 2) = 6. The cofactor matrix works out to [[0, 3, 3], [2, −3, 1], [2, 0, −2]], so adj A — its transpose — is [[0, 2, 2], [3, −3, 0], [3, 1, −2]]. Multiplying: X = (1/6)[[0, 2, 2], [3, −3, 0], [3, 1, −2]]·(6, 2, 1)ᵀ = (1/6)(6, 12, 18)ᵀ = (1, 2, 3). Check: 1 + 2 + 3 = 6, 1 − 2 + 3 = 2, 2 + 2 − 3 = 1 — all three hold. Now the rectangular contrast: x + y + z = 6 with 2x − y + z = 3 alone. Subtracting gives x + 2y = 3; set z = t, then y = t − 3 and x = 9 − 2t — a one-parameter family, exactly as the rank count predicts. Two computations, two regimes, one page.
Where students slip
Forgetting to transpose the cofactor matrix is the most frequent adjoint error — adj A is the transpose of cofactors, not the cofactor matrix itself. Dividing by a miscomputed |A| ranks second: one wrong 2×2 minor poisons every component of X, which is why the substitution check is not optional. Conceptually, students declare a singular system "inconsistent" — but |A| = 0 allows infinitely many solutions too, and only rank(A) versus rank([A | B]) separates them. Finally, order slips: X = A⁻¹B requires pre-multiplying by A⁻¹; post-multiplying (BA⁻¹) is undefined dimensionally or plain wrong, and examiners include such distractor answers. For the rectangular case, the trap is expecting a unique solution at all — with more unknowns than equations, uniqueness is structurally impossible.
Frequently asked questions
When can a system be solved as X = A⁻¹B?
Exactly when the coefficient matrix is square and nonsingular; the solution is then unique and equals (adj A/|A|)B.
What happens when |A| = 0?
The inverse does not exist; the system is either inconsistent or has infinitely many solutions, decided by comparing rank(A) with rank([A | B]).
How many solutions can two equations in three unknowns have?
If consistent, infinitely many — one free parameter — since rank can never reach 3; inconsistency is also possible when the equations contradict.
How does the inverse method handle AX = 0?
X = A⁻¹·0 = 0, so a nonsingular homogeneous system has only the trivial solution; nontrivial solutions force |A| = 0.
Why must the cofactor matrix be transposed?
Because adj A is defined so that A(adj A) = |A|I; the transpose is what makes the (i, j) entry of the product sum aᵢₖAⱼₖ collapse to the Kronecker pattern.