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
Method: Solving a Linear System with the Inverse Matrix Method — Matching the Coefficient Matrix Carefully
Use this method whenever you're given a matrix A (and asked to find A−1), then a separate system of equations to solve "using" it — the key extra step beyond a routine inverse-matrix solve is confirming which matrix actually equals the system's coefficient matrix.
Steps
Step 1: Find A−1 using the adjoint
Compute ∣A∣, the cofactor matrix, and adj(A), then assemble A−1=∣A∣1adj(A). Do this once, before touching the system.
Step 2: Write the system with every variable shown explicitly
Rewrite each equation so every variable appears with an explicit coefficient, including any that are "missing" (write them with coefficient 0). This exposes the true coefficient matrix — skipping this step is the most common source of error in this question type.
Step 3: Compare the coefficient matrix against A …
Mistake 1: Assuming the given matrix A is automatically the system's coefficient matrix
Why it's wrong: writing the three equations in full (x−2y+0z=10, 2x−y−z=8, 0x−2y+z=7) gives a coefficient matrix that is actually AT, not A — the rows and columns of A have been swapped relative to the equations. Using X=A−1B directly here solves the wrong system entirely. Correct approach: always write out the coefficient matrix explicitly from the equations (filling in 0s for missing variables) and compare it to A before deciding whether to use A−1 or (A−1)T.
Mistake 2: Forgetting to include zero coefficients for missing variables …
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.