Question

An even number is the determinant of $\text{(A)} \ \begin{bmatrix} 1 & -1 \\\ -1 & 5 \end{bmatrix} \quad \text{(B)} \ \begin{bmatrix} 13 & -1 \\\ -1 & 15 \end{bmatrix} \quad \text{(C)} \ \begin{bmatrix} 16 & -1 \\\ -11 & 15 \end{bmatrix} \quad \text{(D)} \ \begin{bmatrix} 6 & -12 \\\ 11 & 15 \end{bmatrix}$
Choose the $\textbf{correct}$ answer from the options given below:

• A. (A), (B) and (D) only
• B. (A), (B) and (C) only
• C. (A), (B), (C) and (D)
• D. (B), (C) and (D) only

Answer 1

Answer: (A)

(A), (B) and (D) only

Answer 2

The determinant of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\\ c & d \end{bmatrix}$ is:
$\text{Determinant} = ad - bc.$

Calculate the determinants:

Step 1: For (A): $\text{Determinant} = (1)(5) - (-1)(-1) = 5 - 1 = 4 \text{ (even)}.$

Step 2: For (B): $\text{Determinant} = (13)(15) - (-1)(-1) = 195 - 1 = 194 \text{ (even)}.$

Step 3: For (C): $\text{Determinant} = (16)(15) - (-1)(-11) = 240 + 11 = 251 \text{ (odd)}.$

Step 4: For (D): $\text{Determinant} = (6)(15) - (-12)(11) = 90 + 132 = 222 \text{ (even)}.$

Thus, matrices (A), (B), and (D) have even determinants.