Q.If $A$, $B$ and $C$ are square matrices of same order, then $AB = AC$ always implies that $B = C$.

The key idea is that matrix multiplication is not cancellative unless the matrix $A$ is invertible. The statement is false in general.

Step 1: If $A$ is invertible, we can multiply both sides of $AB = AC$ on the left by $A^{-1}$:

Step 2: But if $A$ is not invertible (singular), $A^{-1}$ does not exist. For example, take

$A = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \quad B = \begin{pmatrix}1 & 2 \\ 3 & 4\end{pmatrix}, \quad C = \begin{pmatrix}1 & 2 \\ 5 & 6\end{pmatrix}.$ …

The statement looks tempting because it mimics ordinary algebra, where if $a \neq 0$ and $ab = ac$, you can cancel $a$ to get $b = c$. But matrices are not numbers. The key difference: matrix multiplication is not commutative, and more importantly, a matrix can be singular (determinant zero), meaning it has no inverse. Without an inverse, cancellation fails.

Think of it this way: if $A$ has a non-trivial nullspace, then $A$ can send two different vectors to the same result. When you multiply $A$ by a matrix $B$, each column of $B$ is transformed by $A$. If $A$ collapses some directions to zero, then two different columns in $B$ and $C$ could produce the same column in $AB$ and $AC$.

1. What would make cancellation valid?

   If $A$ is invertible (i.e., $\det A \neq 0$), then multiplying both sides of $AB = AC$ on the left by $A^{-1}$ gives $B = C$. That works. But the problem statement says "always implies" — it must hold for every square matrix $A$, $B$, $C$ of the same order. That's a much stronger claim.

2. Find a counterexample.

   We need a singular $A$ (non-invertible) and two different matrices $B$ and $C$ such that $AB = AC$. The simplest choice: take $A$ as the zero matrix. Then $AB = 0$ and $AC = 0$ for any $B, C$, so $AB = AC$ holds, but $B$ and $C$ can be completely different. That already disproves the statement.

   But maybe you think "that's cheating — $A$ is zero". Fine, take a non-zero singular $A$. For instance, let

This matrix kills the second coordinate. Now pick

Compute:

$AC = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$ …

Method: Testing Cancellation Laws in Matrix Algebra

Use this whenever a claim "cancels" a common factor as if matrices were non-zero numbers. Cancellation is only valid when the cancelled matrix is invertible.

Steps

Step 1: Recall when cancellation is legitimate.

From $AB=AC$ you may left-multiply by $A^{-1}$ to get $B=C$ — but only if $A^{-1}$ exists (i.e. $A$ is non-singular, $\det A\neq 0$).

Step 2: Reduce the equation to expose the obstruction. …

Common Mistakes

Mistake 1: Cancelling $A$ from both sides like a non-zero number.

Why it's wrong: cancellation needs $A^{-1}$ to exist; a singular $A$ cannot be cancelled. Correct approach: verify $A$ is invertible before writing $B=C$.

Mistake 2: Thinking $A\neq O$ is enough for cancellation. …