Q.Evaluate the determinant $\begin{vmatrix} 5 & 3 \\ 2 & 4 \end{vmatrix}$.

A second-order determinant is evaluated by cross-multiplying the two diagonals and subtracting.

The leading-diagonal product is $5\times 4 = 20$ and the secondary-diagonal product is $3\times 2 = 6$; subtracting gives $20-6=14$.

By definition, $\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1 b_2 - b_1 a_2$.

Here $a_1=5,\ b_1=3,\ a_2=2,\ b_2=4$, so $\begin{vmatrix} 5 & 3 \\ 2 & 4 \end{vmatrix} = (5)(4) - (3)(2) = 20 - 6 = 14.$

Verification. Interchanging the two rows should reverse the sign (Property 2): $\begin{vmatrix} 2 & 4 \\ 5 & 3 \end{vmatrix} = (2)(3)-(4)(5) = 6-20 = -14$, which is exactly $-1$ times the original $14$ — confirming the value.

A very common slip is to add instead of subtract, or to mishandle the sign when an entry is negative. Always write the two diagonal products out fully with their signs before subtracting the second from the first.