Q.Evaluate the determinant $\begin{vmatrix} 6 & -3 \\ 4 & 2 \end{vmatrix}$.

A straightforward second-order determinant, found by cross-multiplying and subtracting.

The subtracted term contains a negative entry, so subtracting it adds to the total.

Using $\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - b_1a_2$ with $a_1=6,\ b_1=-3,\ a_2=4,\ b_2=2$: $\begin{vmatrix} 6 & -3 \\ 4 & 2 \end{vmatrix} = (6)(2) - (-3)(4) = 12 - (-12) = 12 + 12 = 24.$

Verification. Interchanging the rows should reverse the sign (Property 2): $\begin{vmatrix} 4 & 2 \\ 6 & -3 \end{vmatrix} = (4)(-3)-(2)(6) = -12-12 = -24$, which is $-1$ times $24$ — confirming the value.

As with any determinant containing a negative entry, the risk is mishandling the subtraction of a negative product. Write 12 - (-12) out fully as 12 + 12 rather than doing the sign change mentally.