Matrices: Lecture Notes

A Few Matrix Applications

Step 1: Compute the determinant

$\det(A) = (1)(3) - (1)(2) = 3 - 2 = 1$

Step 2: Swap and negate entries

$\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ -2 & 1 \end{bmatrix}$

Step 3: Multiply by $\frac{1}{\det(A)}$

Since $\det(A) = 1$, we get:

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

Verification (optional but insightful). Check that $AA^{-1} = I$:

$\begin{bmatrix}1 & 1 \\ 2 & 3\end{bmatrix} \begin{bmatrix} 3 & -1 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} (3 - 2) & (-1 + 1) \\ (6 - 6) & (-2 + 3) \end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$

So the inverse is correct.