Q.Find the inverse (if it exists) of the matrix $A = \begin{bmatrix}1 & 2 & -2\\-1 & 3 & 0\\0 & -2 & 1\end{bmatrix}$.

What is Matrix Inversion?

Imagine you have a number, say 5. Its multiplicative inverse is $\frac{1}{5}$, because $5 \times \frac{1}{5} = 1$. That "1" is the identity for multiplication — the number that does nothing when you multiply by it.

Matrix inversion is the exact same idea, but for matrices. For a square matrix $A$, its inverse $A^{-1}$ is the matrix that, when multiplied with $A$, gives the identity matrix $I$:

The identity matrix $I$ is the matrix equivalent of the number 1 — it has 1s on the main diagonal and 0s everywhere else. For a $2 \times 2$ matrix, $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Why Does This Matter?

In algebra, if you have $5x = 20$, you solve for $x$ by multiplying both sides by $\frac{1}{5}$: $x = \frac{1}{5} \times 20 = 4$.

In matrix algebra, if you have $A\mathbf{x} = \mathbf{b}$, you solve for $\mathbf{x}$ by multiplying both sides by $A^{-1}$:

This is how you solve systems of linear equations — the heart of everything from engineering to economics.

The Precise Definition

Let $A$ be an $n \times n$ square matrix. If there exists an $n \times n$ matrix $B$ such that:

then $B$ is called the inverse of $A$, written $A^{-1}$. If such a $B$ exists, $A$ is called invertible or non-singular. If no such $B$ exists, $A$ is singular.

How to Find the Inverse (for $2 \times 2$)

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$:

1. Compute the determinant: $\det(A) = ad - bc$
2. If $\det(A) = 0$, stop — no inverse exists.
3. If $\det(A) \neq 0$, the inverse is:

$A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ …