Lesson 11.7: Solving Systems with Inverses

New Concept

We'll explore the inverse of a matrix, $A^{-1}$. Similar to how $a \cdot a^{-1} = 1$, a matrix and its inverse multiply to equal the identity matrix, $I$. This powerful tool allows us to solve systems of linear equations.

What’s next

Now that you have the big picture, you'll master finding inverses and applying them through a series of practice cards and interactive examples.

Property

The identity matrix, $I_n$, is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by $I_n$ where $n$ represents the dimension of the matrix.

If $A$ is an $n \times n$ matrix and $B$ is an $n \times n$ matrix such that $AB = BA = I_n$, then $B = A^{-1}$, the multiplicative inverse of a matrix $A$. A matrix that has a multiplicative inverse has the properties $AA^{-1} = I$ and $A^{-1}A = I$.

Examples

• Given matrix $A = \begin{bmatrix} 5 & 2 \\ -1 & 3 \end{bmatrix}$, we can show that $AI = IA = A$. $AI = \begin{bmatrix} 5 & 2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5(1)+2(0) & 5(0)+2(1) \\ -1(1)+3(0) & -1(0)+3(1) \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ -1 & 3 \end{bmatrix}$.

• To show that $A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$ are inverses, we multiply them: $AB = \begin{bmatrix} 2(3)+5(-1) & 2(-5)+5(2) \\ 1(3)+3(-1) & 1(-5)+3(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_2$.

Property

If $A$ is a $2 \times 2$ matrix, such as

the multiplicative inverse of $A$ is given by the formula

where $ad - bc \neq 0$. If $ad - bc = 0$, then $A$ has no inverse.

Examples

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

• Find the inverse of $A = \begin{bmatrix} 4 & 2 \\ 5 & 3 \end{bmatrix}$. The determinant is $4(3) - 2(5) = 2$. So, $A^{-1} = \frac{1}{2} \begin{bmatrix} 3 & -2 \\ -5 & 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & -1 \\ -\frac{5}{2} & 2 \end{bmatrix}$.

Property

To find the multiplicative inverse, augment the matrix with the identity. When matrix $A$ is transformed into $I$ using row operations, the augmented identity matrix $I$ transforms into $A^{-1}$.

1. Write the original matrix augmented with the identity matrix on the right: $[A|I]$.
2. Use elementary row operations so that the identity appears on the left.
3. The matrix on the right is the inverse: $[I|A^{-1}]$
4. You can check your answer by showing $AA^{-1} = I$.

Examples

• To find the inverse of $A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$, we start with $\left[ \begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 5 & 2 & 0 & 1 \end{array} \right]$. Row operations lead to $\left[ \begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 0 & 1 & -5 & 3 \end{array} \right]$. Thus, $A^{-1} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}$.

• Find the inverse for $A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$. Augmenting and using row operations gives $\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & 1 & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{array} \right]$. So $A^{-1} = \frac{1}{2}\begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix}$.

Property

Given a system of equations, write the coefficient matrix $A$, the variable matrix $X$, and the constant matrix $B$. The system can be expressed as:

Multiply both sides by the inverse of $A$ to obtain the solution.

Examples

• Solve the system $2x+5y=1$ and $x+3y=0$. Here, $A=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$, $X=\begin{bmatrix} x \\ y \end{bmatrix}$, and $B=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. The inverse is $A^{-1}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$. So, $X = A^{-1}B = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$. The solution is $(3, -1)$.

• Solve the system $3x+8y=5$ and $4x+11y=7$. The matrices are $A=\begin{bmatrix} 3 & 8 \\ 4 & 11 \end{bmatrix}$ and $B=\begin{bmatrix} 5 \\ 7 \end{bmatrix}$. The inverse is $A^{-1}=\begin{bmatrix} 11 & -8 \\ -4 & 3 \end{bmatrix}$. The solution is $X = A^{-1}B = \begin{bmatrix} 11 & -8 \\ -4 & 3 \end{bmatrix} \begin{bmatrix} 5 \\ 7 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$. The solution is $(-1, 1)$.