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).
Note
A homogeneous system AX=O always has the trivial solution X=O; it has non-trivial solutions exactly when det(A)=0.
The takeaway
Package the equations as AX=B; if det(A)=0 the answer is the single formula X=A−1B. The determinant is your first check — it tells you whether a unique solution exists before you do any heavy computation.
Solving a system of linear equations using the matrix method (X = A⁻¹B) is a major application covered in the CBSE Class 12 Determinants chapter, and "solve system of equations using matrix method class 12" is one of the most searched topics in this unit given its near-guaranteed appearance in board exams. This same inverse-based technique is also tested in JEE Main questions on the consistency of linear systems.
Concept: Matrix Equation Solving – Represent the system as Ax=b and solve by finding A−1.
We solve the linear system by the matrix method: write it as AX=B, find A−1, and compute X=A−1B. The solution is x=3, y=−1.
Why the matrix approach?
A system like 2x+5y=1, 3x+2y=7 is just a compact way of asking: what pair (x,y) makes both equations true at the same time? Instead of elimination or substitution, we can think of it as a single matrix equation:
(2352)(xy)=(17)
If we call the coefficient matrix A, the variable column X, and the constant column B, then AX=B. The neat idea: if A has an inverse A−1, multiply both sides on the left by A−1 to get X=A−1B. That gives the solution directly — no guessing, no back-substitution.
Write the system in matrix form
A=(2352),X=(xy),B=(17)
So AX=B.
Check if A is invertible — compute its determinant
det(A)=(2)(2)−(5)(3)=4−15=−11
Since det(A)=0, A−1 exists.
Find A−1 using the formula for a 2×2 matrix
For A=(acbd), the inverse is det(A)1(d−c−ba).
A−1=−111(2−3−52)=(−112113115−112)
Tip
A quick check: multiply A−1A — you should get the identity matrix. If not, a sign or fraction is off.
Multiply A−1 by B to get X
X=A−1B=(−112113115−112)(17)
Compute each entry:
For x: (−112)(1)+(115)(7)=−112+1135=1133=3
For y: (113)(1)+(−112)(7)=113−1114=−1111=−1
So x=3, y=−1.
Verify by plugging back into the original equations
2(3)+5(−1)=6−5=1✓
3(3)+2(−1)=9−2=7✓
Watch out
A common mistake: forgetting that matrix multiplication is not commutative. When solving AX=B, always multiply on the left: A−1(AX)=(A−1A)X=IX=X. If you multiply on the right, you get XAA−1=X, which is not the same — and wrong.
✓Final answer
The solution is x=3, y=−1.
Method: Solving a 2-Variable Linear System by the Matrix (Inverse) Method
This method solves any system of two linear equations in two unknowns by packaging it as a single matrix equation AX=B and solving via X=A−1B.
Steps
Step 1: Write the system as AX=B
Collect the coefficients into a 2×2 matrix A, the unknowns into a column X, and the constants into a column B:
A=(a1a2b1b2),X=(xy),B=(d1d2)
Step 2: Compute det(A) and check it's nonzero
det(A)=a1b2−a2b1
If this is zero, A−1 doesn't exist and the matrix method can't be used directly — the system needs a different treatment (inconsistent or infinitely many solutions).
Step 3: Find A−1
A−1=det(A)1(b2−a2−b1a1)
Step 4: Multiply on the left: X=A−1B
(xy)=A−1(d1d2)
Carry out the 2×2-by-2×1 matrix multiplication carefully, term by term.
Step 5: Verify by substituting back
Plug the found x,y into both original equations to confirm — this is quick and catches an arithmetic slip before it's submitted as the final answer.
This is the base case (two unknowns) of the general matrix method — the same X=A−1B idea scales directly to three or more unknowns, just with a 3×3 (or larger) inverse via the adjoint instead of the 2×2 shortcut.