Value of a Determinant

4.1.1 Value of a Determinant

A determinant of order two is the square arrangement

Here $a, b, c, d$ are called the elements of the determinant. There are two rows ($R_1 = [a\ \ b]$, $R_2=[c\ \ d]$) and two columns ($C_1=\begin{bmatrix}a\\c\end{bmatrix}$, $C_2=\begin{bmatrix}b\\d\end{bmatrix}$).

Definition (value of an order-2 determinant).

that is, the product of the elements on the leading (top-left to bottom-right) diagonal, minus the product of the elements on the other diagonal.

Worked Examples

Example (i). Evaluate $\begin{vmatrix} 7 & -4 \\ 9 & 3 \end{vmatrix}$.

Step 1: Identify $a=7, b=-4, c=9, d=3$.

Step 2: Apply $ad-bc$: $7 \times 3 - (-4)\times 9 = 21 - (-36) = 21+36$.

Example (ii). Evaluate $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$.

Step 1: $a=\cos\theta, b=-\sin\theta, c=\sin\theta, d=\cos\theta$.

Step 2: $ad - bc = \cos\theta\cdot\cos\theta - (-\sin\theta)(\sin\theta) = \cos^2\theta + \sin^2\theta$.

Step 3: By the Pythagorean identity this equals $1$.

Example (iii). Evaluate $\begin{vmatrix} 4 & -2i \\ i & 7 \end{vmatrix}$, where $i^2=-1$.

Step 1: $ad-bc = 4\times 7 - (-2i)(i) = 28 - (-2i^2) = 28 + 2i^2$.

Step 2: Substitute $i^2=-1$: $28 + 2(-1) = 28-2$.

Example (iv). Evaluate $\begin{vmatrix} \log_4 2 & 2 \\ \log_4 2 & 4 \end{vmatrix}$. …