Matrix Inverses and Linear Systems

In matrix algebra, the inverse of a matrix allows us to perform an operation similar to division. The inverse of a matrix $A$, denoted as $A^{-1}$, is a matrix such that when multiplied by $A$, it yields the Identity matrix $I$: $AA^{-1} = A^{-1}A = I$

Finding the Inverse of a $2 \times 2$ Matrix

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse exists if and only if its determinant is not zero.

The determinant is calculated as: $\det(A) = ad - bc$

If $\det(A) \neq 0$, the inverse is given by the formula: $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ Notice that we swap the positions of $a$ and $d$, change the signs of $b$ and $c$, and multiply everything by $1$ over the determinant.

Example 1: Finding an Inverse

Find the inverse of $A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$.

1. Find the determinant: $\det(A) = (3)(2) - (1)(5) = 6 - 5 = 1$.
2. Apply the inverse formula: $A^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}$

Solving Linear Systems Using Matrices

We can write a system of linear equations as a single matrix equation $AX = B$, where:

• $A$ is the coefficient matrix.
• $X$ is the variable matrix.
• $B$ is the constant matrix.

To solve for $X$, we multiply both sides by $A^{-1}$ (on the left): $X = A^{-1}B$

Example 2: Solving a System

Solve the system $2x + 3y = 7$ and $x - y = 1$ using matrices.

1. Set up the matrix equation $AX = B$: $\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$

2. Find the inverse of $A$: The determinant is $(2)(-1) - (3)(1) = -2 - 3 = -5$. $A^{-1} = \frac{1}{-5} \begin{bmatrix} -1 & -3 \\ -1 & 2 \end{bmatrix}$

3. Multiply $A^{-1}B$ to find $X$: $X = \frac{1}{-5} \begin{bmatrix} -1 & -3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 7 \\ 1 \end{bmatrix}$ $X = -\frac{1}{5} \begin{bmatrix} (-1)(7) + (-3)(1) \\ (-1)(7) + (2)(1) \end{bmatrix} = -\frac{1}{5} \begin{bmatrix} -10 \\ -5 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$

The solution is $x = 2$ and $y = 1$.