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

Hint:

When calculating the determinant of a \(2 \times 2\) matrix, use the formula \( \text{Determinant} = ad - bc \). The sign of the determinant determines whether the matrix is invertible (non-zero determinant) or singular (zero determinant). Keep in mind that the determinant can also help determine whether a system of linear equations has a unique solution. In this case, calculating and checking the parity (even or odd) of the determinant can be useful in some problems.

Answer 1

The Correct Option is A

To determine whether the determinants of the given matrices are even numbers, we need to calculate the determinant of each. The determinant of a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is computed using the formula:  
\(ad - bc\).

Let's compute the determinant for each matrix:

For matrix \(\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}\): 
\(1 \cdot 5 - (-1) \cdot (-1) = 5 - 1 = 4\). 
The determinant is 4, which is even.

For matrix \(\begin{bmatrix} 13 & -1 \\ -1 & 15 \end{bmatrix}\): 
\(13 \cdot 15 - (-1) \cdot (-1) = 195 - 1 = 194\). 
The determinant is 194, which is even.

For matrix \(\begin{bmatrix} 16 & -1 \\ -11 & 15 \end{bmatrix}\): 
\(16 \cdot 15 - (-1) \cdot (-11) = 240 + 11 = 251\). 
The determinant is 251, which is odd.

For matrix \(\begin{bmatrix} 6 & -12 \\ 11 & 15 \end{bmatrix}\): 
\(6 \cdot 15 - (-12) \cdot 11 = 90 + 132 = 222\). 
The determinant is 222, which is even.

Thus, the matrices for which the determinant is an even number are: options (A), (B), and (D). Therefore, the correct answer is: (A), (B) and (D) only

Answer 2

The determinant of a \(2 \times 2\) matrix \[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] is given by: \[ \text{Determinant} = ad - bc. \]

Step 1: Calculate the determinant for matrix (A):

For matrix (A): \[ \text{Determinant} = (1)(5) - (-1)(-1) = 5 - 1 = 4 \text{ (even)}. \]

Step 2: Calculate the determinant for matrix (B):

For matrix (B): \[ \text{Determinant} = (13)(15) - (-1)(-1) = 195 - 1 = 194 \text{ (even)}. \]

Step 3: Calculate the determinant for matrix (C):

For matrix (C): \[ \text{Determinant} = (16)(15) - (-1)(-11) = 240 + 11 = 251 \text{ (odd)}. \]

Step 4: Calculate the determinant for matrix (D):

For matrix (D): \[ \text{Determinant} = (6)(15) - (-12)(11) = 90 + 132 = 222 \text{ (even)}. \]

Conclusion: Thus, matrices (A), (B), and (D) have even determinants, while matrix (C) has an odd determinant.