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# Example Problems

## Calculations

We can either use the properties listed above (mainly derived property 1 and 3) or these two equation to solve for determinants in specific cases.

For 2 x 2 matrices: $$\text{det}\left(\begin{bmatrix}a & b\ c & d\end{bmatrix}\right) = ad - bc$$

For 3 x 3 matrices: $$\text{det}\left(\begin{bmatrix}a & b & c\ d & e & f\ g & h & i\end{bmatrix}\right) = a \cdot \text{det}\left(\begin{bmatrix}e & f \ h & i\end{bmatrix}\right) - b \cdot \text{det}\left(\begin{bmatrix}d & f \ g & i\end{bmatrix}\right) + c \cdot \text{det}\left(\begin{bmatrix}d & e\ g & h\end{bmatrix}\right)$$

## Example

Let $$A = \left(\begin{bmatrix}5 & 6 \ 2 & 5\end{bmatrix}\right)$$. Using the equation for 2 x 2 matrices: $$\text{det}(\textbf{A}) = 5 \cdot 5 - 2 \cdot 6 = 13$$

Using the properties listed above:

By derived property 2: $$\text{det}(\textbf{A}) = \text{det}\left(\begin{bmatrix}5 & 6 \ 2 - 2 & 5 - \frac{12}{5}\end{bmatrix}\right) =\text{det}\left(\begin{bmatrix}5 & 6 \ 0 & \frac{13}{5}\end{bmatrix}\right)$$

By derived property 3: $$\text{det}\left(\begin{bmatrix}5 & 6 \ 0 & \frac{13}{5}\end{bmatrix}\right) = 5 \cdot \frac{13}{5} = 13$$