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 and solving X=A−1B via the adjoint of A gives x=2, y=1, z=3.
The stem asks specifically for the matrix method, so we package the three equations into AX=B and invert the coefficient matrix rather than eliminate variables directly with row operations.
The equations are:
x−y+2z=7(1)
3x+4y−5z=−5(2)
2x−y+3z=12(3)
Step 1: Write as AX=B.
A=132−14−12−53,X=xyz,B=7−512.
Step 2: Find det(A) by expanding along the first row.
det(A)=14−1−53−(−1)32−53+2324−1
=1(12−5)+1(9+10)+2(−3−8)=7+19−22=4.
Since det(A)=4=0, A is invertible and the system has a unique solution.
Method: Solving a Linear System by Row Reduction (Augmented Matrix)
This method solves AX=B by transforming the augmented matrix [A∣B] into row-echelon (upper-triangular) form using row operations, then reading off the unknowns by back-substitution — useful when the numbers make Cramer's rule or adjoint computation messy.
Steps
Step 1: Write the augmented matrix
Stack the coefficient matrix A and the constants B side by side, separated by a vertical bar: [A∣B]. Every row operation from here on must act on the entire row, bar included.
Step 2: Eliminate the first variable from every row below the first
Using the first row's leading entry as the pivot, replace each lower row with (that row) − (a suitable multiple of the first row) so the first column becomes 0 below the pivot: R2→R2−kR1, R3→R3−mR1.
Step 3: Eliminate the second variable from the row below the second …
Mistake 1: Forgetting to carry the constant column through a row operation
Why it's wrong: a row operation acts on the WHOLE row, including the number after the bar — updating only the coefficient part and leaving the old constant behind silently corrupts the system. Correct approach: always compute and rewrite all four entries of a row together, never treat the constant column as separate bookkeeping.
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.