Q.Two students claim that the matrix $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$ has two different inverses, $B=\begin{bmatrix}-2&1\\1.5&-0.5\end{bmatrix}$ and $C=\begin{bmatrix}-2&1\\3/2&-1/2\end{bmatrix}$. Using the uniqueness-of-inverse theorem, explain why this cannot actually happen.

A square matrix $A$ of order $n$ is invertible (non-singular) if some square matrix $B$ of the same order satisfies $AB=BA=I$; such a $B$ is called the inverse of $A$, denoted $A^{-1}$. A complete proof, resting only on the associativity of matrix multiplication, shows that this inverse -- when it exists -- is always unique: if $B$ and $C$ both satisfy the defining condition, then $B=BI=B(AC)=(BA)C=IC=C$, so $B=C$. This uniqueness is what justifies the single notation $A^{-1}$ without ambiguity. For a $2\times2$ matrix $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the practical shortcut $A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d …