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). …
X must be 2×2; equating the entries of XA with the right-hand side gives X=[12−20].
We need X with X[142536]=[−72−84−96]. Since A is 2×3, it has no ordinary inverse, so we can't just multiply by A−1. Instead we fix the shape of X and solve for its entries.
Step 1 — shape of X
For XA to be defined and to come out 2×3 (to match the right side), X must be 2×2. Write
Mistake 1: Trying to write X=BA−1 when A is not square.
Why it's wrong: only square matrices can have an inverse; a 2×3 matrix has none. Correct approach: set up X with unknown entries and compare, don't invert.
Mistake 2: Getting the order of X wrong.
Why it's wrong: an incorrect shape makes XA either undefined or the wrong size. Correct approach: fix the rows of X from the rows of B and the columns of X from the rows of A.