Matrices: Edge Cases and Subtle Traps (7)

Question

If $A$ and $B$ are $2 \times 2$ matrices with $AB = O$ (the zero matrix), does it follow that $A = O$ or $B = O$? Many students answer “yes”. What’s the actual situation?

Solution — Step by Step

The conclusion: $AB = O$ does not imply $A = O$ or $B = O$. Matrices have zero divisors.

Why This Works

For real numbers, $ab = 0$ means $a = 0$ or $b = 0$. This works because reals form a field — no zero divisors. Matrices form a ring but not a field, and rings can have zero divisors. The example above is the standard one.

If $A$ is invertible, then $AB = O \implies B = A^{-1}O = O$. So the implication does hold when one of the matrices is invertible. The failure happens specifically for singular (non-invertible) matrices.

Alternative Method

Use determinants. If $AB = O$, then $\det(A)\det(B) = 0$, so at least one of $A$, $B$ has determinant zero — i.e. is singular. But singular doesn’t mean zero. Same conclusion.

Common Mistake

Carrying scalar intuition into matrix algebra. The fix: whenever you “divide” by a matrix, ask whether it’s invertible. If not, you can’t divide. JEE Advanced loves this — singular-matrix traps appear almost every year.