Matrix Equations and Inverse Matrices

By the end of this lesson, you will be able to find the determinant of a 2×2 matrix, find the inverse of a 2×2 matrix, write systems as matrix equations, and solve systems using $X = A^{-1}B$.

The Big Idea

A system of linear equations can be written as a single matrix equation $AX = B$. Solving for $X$ means multiplying both sides by the inverse matrix $A^{-1}$: $X = A^{-1}B$. This is the matrix analog of dividing both sides of $ax = b$ by $a$ — but for multiple equations simultaneously.

The Determinant of a 2×2 Matrix

$$\det\begin{pmatrix} a & b \ c & d \end{pmatrix} = ad - bc$$

If $\det(A) = 0$, the matrix has no inverse (it's singular).

The Inverse of a 2×2 Matrix

$$A = \begin{pmatrix} a & b \ c & d \end{pmatrix}, \quad A^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \ -c & a \end{pmatrix}$$

Recipe: swap the main diagonal, negate the other diagonal, divide by the determinant.

Verification: $AA^{-1} = A^{-1}A = I$

Example 1 — Finding the Inverse

$A = \begin{pmatrix} 3 & 1 \ 5 & 2 \end{pmatrix}$

$\det(A) = 6 - 5 = 1$

$A^{-1} = \dfrac{1}{1}\begin{pmatrix} 2 & -1 \ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \ -5 & 3 \end{pmatrix}$

Verify: $\begin{pmatrix} 3 & 1 \ 5 & 2 \end{pmatrix}\begin{pmatrix} 2 & -1 \ -5 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$ ✓

Writing a System as a Matrix Equation

The system:

$$3x + y = 7$$ $$5x + 2y = 13$$

becomes $AX = B$:

$$\underbrace{\begin{pmatrix} 3 & 1 \ 5 & 2 \end{pmatrix}}{A} \underbrace{\begin{pmatrix} x \ y \end{pmatrix}}{X} = \underbrace{\begin{pmatrix} 7 \ 13 \end{pmatrix}}_{B}$$

Solving with the Inverse

$$AX = B \implies X = A^{-1}B$$

Using $A^{-1} = \begin{pmatrix} 2 & -1 \ -5 & 3 \end{pmatrix}$:

$$X = \begin{pmatrix} 2 & -1 \ -5 & 3 \end{pmatrix}\begin{pmatrix} 7 \ 13 \end{pmatrix} = \begin{pmatrix} 14-13 \ -35+39 \end{pmatrix} = \begin{pmatrix} 1 \ 4 \end{pmatrix}$$

Solution: $x = 1$, $y = 4$

Check: $3(1) + 4 = 7$ ✓; $5(1) + 2(4) = 13$ ✓

Example 2 — Singular Matrix

$A = \begin{pmatrix} 2 & 4 \ 1 & 2 \end{pmatrix}$: $\det(A) = 4 - 4 = 0$ → no inverse.