Many problems reduce to a system of linear equations, for example
2x+3yx−y=8=−1.
The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.
Writing the system as AX=B
Collect the coefficients, the unknowns and the constants:
A=(213−1),X=(xy),B=(8−1),
so the whole system becomes AX=B.
The idea
For numbers, ax=b gives x=a−1b provided a=0. The same works for matrices: if A is invertible, multiply AX=B on the left by A−1:
A−1(AX)=A−1B⇒IX=A−1B⇒X=A−1B.
X=A−1B
Multiplying on the left matters — matrix products do not commute, so BA−1 would be wrong.
When it works
The inverse A−1 exists only when detA=0, so:
detA=0: the system is consistent with the unique solution X=A−1B.
detA=0: no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.
Worked steps
For the system above, detA=(2)(−1)−(3)(1)=−5=0, and
This method finds the inverse of any square matrix whose determinant is non-zero, using cofactors and the adjoint — the standard NCERT route for a 3×3 inverse.
Steps
Step 1: Compute detA and confirm it is non-zero
The inverse of a matrix exists only when detA=0 (a singular matrix has no inverse). Expand the determinant along any convenient row or column — pick one with the most zeros to minimise the arithmetic.
A−1 exists⟺detA=0
Step 2: Compute all nine cofactors
For each entry aij, delete its row and column to get the minor Mij, then apply the checkerboard sign:
Cij=(−1)i+jMij
Work systematically row by row (C11,C12,C13, then C21,…) so no entry is skipped, keeping the alternating +,−,+ sign pattern in front of you rather than re-deriving it each time.
Step 3: Form the adjoint by transposing the cofactor matrix …
Mistake 1: Forgetting to transpose the cofactor matrix into the adjoint
Why it's wrong: the adjoint is defined as the transpose of the cofactor matrix, adj(A)=[Cij]T — using the cofactor matrix unchanged silently swaps every off-diagonal pair (e.g. C12 and C21), giving a wrong inverse that still "looks" plausible. Correct approach: write the cofactor matrix out fully first, then explicitly transpose it as its own step before dividing by detA.
Mistake 2: A sign slip in the checkerboard pattern when computing cofactors …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL6 marks
Q.Using matrix method solve the following system of linear equations: x−y+z=4, 2x+y−3z=0, x+y+z=2.
OR
Using elementary transformation find the inverse of the following matrix: A=1−32305−2−50.
›Reveal solutionSolution
Solve AX=B via X=A−1B using the adjoint method; for the OR, row-reduce [A∣I] to [I∣A−1].
Matrix method: x−y+z=4,2x+y−3z=0,x+y+z=2
A=121−1111−31,X=xyz,B=402
detA=1(1+3)−(−1)(2+3)+1(2−1)=4+5+1=10=0, so a unique solution exists.
Since ∣A∣=−17e0, A−1 exists and X=A−1B gives a unique solution. Solving (by X=A−1B, or by elimination): from the second equation y=1−2x+z; substituting into the first gives 7x+z=10, and into the third gives 17x=17.