AP DSC School Assistant Math Model Test 2

Concept:The determinant of a 3×3 matrix can be evaluated by direct expansion or by applying row/column operations. If any two rows or columns become identical, the determinant is zero.Explanation:Let the given determinant be $A = \begin{vmatrix} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{vmatrix}$.Method 1: Direct ExpansionUsing the formula $|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$:$|A| = 11(13 \times 15 - 14 \times 14) - 12(12 \times 15 - 13 \times 14) + 13(12 \times 14 - 13 \times 13)$$= 11(195 - 196) - 12(180 - 182) + 13(168 - 169)$$= 11(-1) - 12(-2) + 13(-1)$$= -11 + 24 - 13 = 0$.Method 2: Column OperationsPerform $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_2$ (these operations do not change the value):$A = \begin{vmatrix} 11 & 1 & 1 \\ 12 & 1 & 1 \\ 13 & 1 & 1 \end{vmatrix}$.Now columns 2 and 3 are identical. A determinant with two identical columns is always zero.Thus $|A| = 0$.Answer:Option B: 0.