11.7 Solving Systems with Inverses

Matrix Inverses and Solving Systems

Calculation of Matrix Inverses

A matrix inverse lets you "undo" matrix multiplication, similar to how dividing undoes multiplying with regular numbers. If you can find the inverse of a coefficient matrix, solving a system of equations becomes straightforward.

Inverse of a 2×2 Matrix

Given a 2×2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, its inverse is:

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

The expression $ad - bc$ is the determinant of the matrix, written $\det(A)$. The inverse only exists when $\det(A) \neq 0$. If the determinant equals zero, the matrix is called singular and has no inverse.

For example, if $A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$, then $\det(A) = (3)(2) - (1)(5) = 1$, and:

$A^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}$

Inverse by Augmentation (for any size matrix)

For matrices larger than 2×2, use row reduction on an augmented matrix:

1. Set up the augmented matrix $[A \mid I]$, placing the identity matrix $I$ next to $A$
2. Perform row operations (row swapping, multiplying a row by a nonzero scalar, adding a multiple of one row to another) to transform the left side into the identity matrix
3. Once the left side is $I$, the right side is $A^{-1}$

If at any point you get a row of all zeros on the left side, the matrix is singular and has no inverse.

Matrix Inverses for Linear Systems

Any system of linear equations can be written as a single matrix equation:

$A\vec{x} = \vec{b}$

• $A$ is the coefficient matrix (the numbers in front of each variable)
• $\vec{x}$ is the variable vector (the unknowns you're solving for)
• $\vec{b}$ is the constant vector (the numbers on the right side of each equation)

Solving with the inverse:

If $A$ is invertible, multiply both sides on the left by $A^{-1}$:

$A^{-1}A\vec{x} = A^{-1}\vec{b}$

Since $A^{-1}A = I$ and $I\vec{x} = \vec{x}$, this simplifies to:

$\vec{x} = A^{-1}\vec{b}$

That's the entire solution. You find the inverse once, then multiply it by the constant vector.

When does this work?

• If $\det(A) \neq 0$, the system has exactly one unique solution
• If $\det(A) = 0$, you can't use this method. The system is either inconsistent (no solution) or dependent (infinitely many solutions). You'd need a different approach, like row reduction, to figure out which case you're in.

Interpretation of Matrix Solutions

Translating a word problem into a matrix equation follows a consistent process:

1. Identify the unknowns and assign them as variables ($x, y, z, \ldots$)
2. Write a system of linear equations from the relationships described in the problem
3. Build the coefficient matrix $A$, variable vector $\vec{x}$, and constant vector $\vec{b}$
4. Find $A^{-1}$ and compute $\vec{x} = A^{-1}\vec{b}$
5. Interpret each component of the solution vector in context (e.g., $x = 5$ might mean "5 units of Product A")

The inverse method is especially useful when you need to solve the same system structure with different constant vectors. Since $A^{-1}$ stays the same, you just multiply by each new $\vec{b}$. This comes up in applications like analyzing circuits with different voltage sources or modeling economic scenarios with changing inputs.

Matrix Operations and Linear Algebra

The inverse method builds on core matrix operations you've already learned: addition, scalar multiplication, and matrix multiplication. A few things to keep straight:

• Matrix multiplication is not commutative: $AB \neq BA$ in general. That's why you must multiply by $A^{-1}$ on the left side of both sides of the equation.
• The identity matrix $I$ acts like the number 1: $AI = IA = A$ for any square matrix $A$.
• Only square matrices (same number of rows and columns) can have inverses, and only when their determinant is nonzero.

These properties are the foundation for everything in this section, so make sure they feel solid before moving on.