Solving a System of Equations by the Matrix Method
A system of linear equations can be written as a single matrix equation and solved in one clean step using the inverse of a matrix. This is the Class-12 "matrix method" for simultaneous equations.
If det(A)=0, then A−1 exists, and multiplying both sides on the left by A−1 gives
X=A−1B,where A−1=det(A)1adj(A).
So you compute det(A), then adj(A), form A−1, and multiply by B. The single column X=A−1B hands you x, y, z at once, and because A−1 is unique, the solution is unique.
Watch out
Multiply in the correct order: X=A−1B, not BA−1. Matrix multiplication is not commutative, and BA−1 is not even defined here.
When det(A)=0
If det(A)=0, A−1 does not exist and the inverse method fails. The system is then either inconsistent (no solution) or has infinitely many solutions. Decide which by computing (adjA)B:
(adjA)B=O → no solution (inconsistent).
(adjA)B=O → infinitely many solutions (consistent, dependent). …
Writing the system as AX=B, we get detA=51=0, so X=A−1B=(3,2,−2). Thus x=3,y=2,z=−2.
The idea
The matrix method packs the whole system into one equation AX=B, where A holds the coefficients, X the unknowns and B the constants. If detA=0 then A−1 exists and multiplying on the left gives X=A−1B — the three values at once.
Method: Solving a 3×3 System via the Inverse (Adjoint) when the Determinant is Non-zero
When the coefficient determinant of a 3-variable system is non-zero, the system is automatically consistent with a unique solution, and that solution can be obtained in one formula using the adjoint.
Steps
Step 1: Write AX=B and compute det(A)
Expand along the first row using the checkerboard sign pattern. A non-zero result confirms A−1 exists and the system is consistent with exactly one solution.
Step 2: Compute all nine cofactors of A
For entry aij, the cofactor Cij=(−1)i+jMij, where Mij is the 2×2 minor left after deleting row i and column j. Keep the sign pattern +−+−+−+−+ in front of you while doing this.
Step 3: Assemble the adjoint by transposing the cofactor matrix
Mistake 1: Using the cofactor matrix directly as the adjoint, forgetting to transpose it
Why it's wrong: the adjoint is the TRANSPOSE of the cofactor matrix, not the cofactor matrix itself. Here the cofactor matrix is 28−2−171310−17−19517, and only after transposing does it become adj(A)=2813−19−2105−17−1717. Skipping the transpose swaps off-diagonal entries and produces a completely wrong A−1, hence wrong x,y,z. Correct approach: always write out the cofactor matrix first, then explicitly transpose it (flip across the main diagonal) to get the adjoint.
Mistake 2: A checkerboard sign error on one of the nine cofactors …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KEAM 2026Set eng-2026-04194 marksMCQ
Q.If (x3−1)1−1110011−1231=0, then the values of x are
(A) -2
(B) 3−1
(C) -3
(D) 32
(E) 3−2
›Reveal solutionSolution
Do the matrix multiplication in stages (right pair first), reduce the row-vector × column-vector to a scalar, set it to 0, and solve for x.
Concept. The product of a 1×3 row, a 3×3 matrix, and a 3×1 column is a 1×1 scalar. Multiplication is associative, so evaluate the 3×3 times the column first.
Step 1 — multiply the matrix by the column (2,3,1)T.
Q.If A is a non-singular matrix of order n satisfying the matrix equation I+A+A2+A3+…+A10=O, where I and O are, respectively, unit and null matrices of order n, then A10=
(A) A−1
(B) I
(C) A
(D) I+A
(E) O
›Reveal solutionSolution
Multiply the series by A, subtract, to collapse it to A11=I.