Question
If A=(2314), find A−1 using the cofactor method, and verify by computing A⋅A−1. CBSE 2024 boards 4-mark.
Solution — Step by Step
det(A)=(2)(4)−(1)(3)=8−3=5
Since det(A)=0, A is invertible.
For a 2×2 matrix (acbd), the cofactor matrix is (d−b−ca).
cof(A)=(4−1−32)
adj(A)=cof(A)T=(4−3−12)
A−1=det(A)1⋅adj(A)=51(4−3−12)=(4/5−3/5−1/52/5)
Verification: A⋅A−1=(2314)⋅51(4−3−12)=51(5005)=I. ✓
Why This Works
The formula A−1=det(A)1adj(A) comes from the matrix identity A⋅adj(A)=det(A)⋅I. So dividing the adjugate by the determinant gives the inverse.
For 2×2 matrices, there’s a memorable shortcut: swap the diagonal entries, negate the off-diagonal entries, divide by the determinant. That’s exactly what cofactor + transpose does for the 2×2 case.
Alternative Method
For 2×2, the direct formula is unbeatable:
(acbd)−1=ad−bc1(d−c−ba)
Memorise this. It saves 60 seconds on JEE Main MCQs and gives the answer in 5 seconds.
The biggest matrix-inverse mistake: forgetting to transpose the cofactor matrix. The cofactor matrix and the adjugate differ by a transpose. For 2×2 matrices, the transpose effect is just swapping the off-diagonal entries — easy to miss.
Always verify your inverse by multiplying A⋅A−1. If you don’t get the identity matrix, you have an arithmetic error somewhere. Takes 30 seconds, catches 90% of mistakes.