Precalculus / Unit 8: Systems and Matrices
8.4 Inverse Matrices
Notes
The inverse $A^{-1}$ satisfies $AA^{-1} = A^{-1}A = I$, where $I$ is the identity matrix.
- For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$: $\displaystyle A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
- Swap the diagonal entries, negate the other two, then divide by the determinant.
- Example: $\begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$ has determinant $1$, so its inverse is $\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$.
Practice
What is the inverse of $\begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$?
Determinant is $6 - 5 = 1$; swap 2 and 3 and negate 1 and 5.
What is the inverse of $\begin{bmatrix} 4 & 3 \\ 2 & 2 \end{bmatrix}$?
The determinant is $8 - 6 = 2$. Then $\displaystyle \frac{1}{2}\begin{bmatrix} 2 & -3 \\ -2 & 4 \end{bmatrix}$.
For which value of $k$ does $\begin{bmatrix} 1 & 2 \\ 3 & k \end{bmatrix}$ have no inverse?
The determinant $k - 6$ must be 0.
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