Question

Let $D_n = [a_{ij}]_{n \times n}$ be a $(n \times n)$ determinant with the following conditions

$a_{ij} = \begin{cases} 4 & i=j \\ 2 & |i-j| = 1 \\ 0 & \text{otherwise} \end{cases}$

Then,

• A. $D_3 = 32$
• B. $D_6 + D_4 = 2D_5$
• C. $D_6 + 4D_4 = 4D_5$
• D. $D_{12} + D_{10} = D_{11}$

Answer 1

Answer: (A), (C)

A, C

Answer 2

Here's how to solve this problem:

1. Calculate initial determinants:

   • $D_1 = 4$
   • $D_2 = \begin{vmatrix} 4 & 2 \\ 2 & 4 \end{vmatrix} = 16 - 4 = 12$
   • $D_3 = \begin{vmatrix} 4 & 2 & 0 \\ 2 & 4 & 2 \\ 0 & 2 & 4 \end{vmatrix} = 4(16-4) - 2(8) = 48 - 16 = 32$

   This directly verifies option (A).

2. Derive the recurrence relation:

   Expand $D_n$ along the first row. The cofactor of $a_{11}$ is $D_{n-1}$. The cofactor of $a_{12}$ is $-2D_{n-2}$. This leads to the recurrence:

   $D_n = 4D_{n-1} - 4D_{n-2}$

3. Verify options using the recurrence relation:

   • Option (B): $D_6 + D_4 = 2D_5$. Substitute $D_6 = 4D_5 - 4D_4$. This simplifies to $4D_5 - 4D_4 + D_4 = 2D_5$, which further simplifies to $2D_5 = 3D_4$. This is false.

   • Option (C): $D_6 + 4D_4 = 4D_5$. Substitute $D_6 = 4D_5 - 4D_4$. This simplifies to $4D_5 - 4D_4 + 4D_4 = 4D_5$, which simplifies to $4D_5 = 4D_5$. This is true.

   • Option (D): $D_{12} + D_{10} = D_{11}$. Substitute $D_{12} = 4D_{11} - 4D_{10}$. This simplifies to $4D_{11} - 4D_{10} + D_{10} = D_{11}$, which further simplifies to $3D_{11} = 3D_{10}$ or $D_{11} = D_{10}$. This is false.

Therefore, the correct options are (A) and (C).