Lesson 32: Solving Linear Systems with Matrix Inverses (Exploration: Exploring Matrix Inverses)

New Concept

Two matrices $A$ and $B$ are inverses of each other if $A \cdot B = B \cdot A = I$.

What’s next

Next, you'll calculate the inverse of a matrix and use it as a powerful new tool to solve systems of linear equations.

The inverse of a matrix $A$, notated $A^{-1}$, is a special matrix where multiplying it by $A$ results in the identity matrix, $I$. For two square matrices $A$ and $B$ to be inverses of each other, their product must be the identity matrix, regardless of the order of multiplication: $A \cdot B = B \cdot A = I$.

Are $P = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $Q = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$ inverses? Yes, because $P \cdot Q = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Are $R = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}$ and $S = \begin{bmatrix} 3 & -1 \\ -2 & 0 \end{bmatrix}$ inverses? No, because $R \cdot S = \begin{bmatrix} 1 & -1 \\ 0 & -2 \end{bmatrix} \neq I$.

Think of it like a secret decoder ring! Multiplying a matrix by its inverse is like undoing a secret code, getting you back to the simple identity matrix, which is the matrix equivalent of the number 1. This 'undo' feature is super powerful for solving complex matrix equations, just like division undoes multiplication with regular numbers.

If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and its determinant $ad - cb \neq 0$, then the inverse of $A$ is given by the formula:

For $T = \begin{bmatrix} 5 & 2 \\ -4 & -2 \end{bmatrix}$, the determinant is $(5)(-2) - (2)(-4) = -2$. So, $T^{-1} = \frac{1}{-2} \begin{bmatrix} -2 & -2 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -2 & -2.5 \end{bmatrix}$.
For $M = \begin{bmatrix} 1 & 8 \\ 1 & 9 \end{bmatrix}$, the determinant is $(1)(9) - (8)(1) = 1$. So, $M^{-1} = \frac{1}{1} \begin{bmatrix} 9 & -8 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 9 & -8 \\ -1 & 1 \end{bmatrix}$.

Finding the inverse of a 2x2 matrix is like a magic trick with four simple steps! First, calculate the determinant—it can't be zero. Then, swap the top-left and bottom-right numbers. After that, change the signs of the other two numbers. Finally, divide the entire new matrix by the determinant you found earlier. Voila, you have the inverse!

An $n \times n$ matrix $A$ is called a singular matrix if it does not have an inverse. This situation occurs if and only if the determinant of the matrix is equal to zero. So, for a singular matrix $A$, we have $\det A = 0$. A matrix must have a non-zero determinant for its inverse to exist.

Matrix $Y = \begin{bmatrix} 3 & 6 \\ 4 & 8 \end{bmatrix}$ is singular because its determinant is $(3)(8) - (6)(4) = 24 - 24 = 0$. It has no inverse.
Matrix $Z = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ is singular because its determinant is $(1)(1) - (1)(1) = 0$. Therefore, $Z^{-1}$ does not exist.

A singular matrix is like a lock with no key! It's a special type of square matrix that has a determinant of zero. Because its determinant is zero, you can't complete the final step of the inverse formula—dividing by zero is a big no-no in math. So, a singular matrix is stuck without an inverse buddy forever.

If a matrix $A$ has an inverse and the matrix equation $AX = B$ has a solution, then the solution is $X = A^{-1}B$. To solve, you must multiply both sides of the equation by $A^{-1}$ from the left side: $(A^{-1})AX = (A^{-1})B$, which simplifies to $X = A^{-1}B$.

To solve $\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} X = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, first find the inverse: $A^{-1} = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$.
Then, solve for $X$ by multiplying: $X = A^{-1}B = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$.
The system $2x+y=5$, $3x+2y=8$ becomes $AX=B$, where $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$.

Solving $AX=B$ is like solving the simple equation $5x=10$. To get $x$, you multiply by the inverse of 5, which is $1/5$. Similarly, to get the matrix $X$, you multiply matrix $B$ by the inverse of matrix $A$. Just remember, matrix multiplication order matters, so you must multiply from the left side: $X = A^{-1}B$.