Question

Let $M_n=[M_{ij}]$ be a $n \times n$ matrix such that $M_{ij} = \begin{cases} a, & i=j \\ 1, & |i-j|=1 \\ 0, & otherwise \end{cases}$

Let $D_n$ denotes the determinant value of matrix $M_n$. On the basis of above information, answer the following questions:

If a = 2 then unit's place digit of $D_{2017}$ is

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (D)

8

Answer 2

The matrix $M_n$ is a tridiagonal matrix with $a$ on the main diagonal and 1 on the sub-diagonals. $M_n = \begin{pmatrix} a & 1 & 0 & \dots & 0 \\ 1 & a & 1 & \dots & 0 \\ 0 & 1 & a & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & a \end{pmatrix}_{n \times n}$

Let $D_n = \det(M_n)$. We can find a recurrence relation for $D_n$ by expanding along the first row. $D_n = a \cdot \det(M_{n-1}) - 1 \cdot \det(\text{matrix obtained by removing row 1 and column 2 of } M_n)$. The matrix obtained by removing row 1 and column 2 of $M_n$ is: $\begin{pmatrix} 1 & 1 & 0 & \dots & 0 \\ 0 & a & 1 & \dots & 0 \\ 0 & 1 & a & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & a \end{pmatrix}_{(n-1) \times (n-1)}$

Expanding the determinant of this matrix along the first column, we get $1 \cdot \det(M_{n-2})$. So, the recurrence relation is $D_n = a D_{n-1} - D_{n-2}$ for $n \ge 3$.

We need the initial values $D_1$ and $D_2$. $M_1 = [a]$, so $D_1 = a$. $M_2 = \begin{pmatrix} a & 1 \\ 1 & a \end{pmatrix}$, so $D_2 = a^2 - 1$.

We are given $a=2$. The recurrence relation becomes $D_n = 2 D_{n-1} - D_{n-2}$ for $n \ge 3$. The initial conditions are $D_1 = 2$ and $D_2 = 2^2 - 1 = 3$.

Let's calculate the first few terms of the sequence $D_n$ for $a=2$: $D_1 = 2$ $D_2 = 3$ $D_3 = 2 D_2 - D_1 = 2(3) - 2 = 6 - 2 = 4$ $D_4 = 2 D_3 - D_2 = 2(4) - 3 = 8 - 3 = 5$ $D_5 = 2 D_4 - D_3 = 2(5) - 4 = 10 - 4 = 6$ $D_6 = 2 D_5 - D_4 = 2(6) - 5 = 12 - 5 = 7$ $D_7 = 2 D_6 - D_5 = 2(7) - 6 = 14 - 6 = 8$ $D_8 = 2 D_7 - D_6 = 2(8) - 7 = 16 - 7 = 9$ $D_9 = 2 D_8 - D_7 = 2(9) - 8 = 18 - 8 = 10$ $D_{10} = 2 D_9 - D_8 = 2(10) - 9 = 20 - 9 = 11$

We can observe a pattern in the sequence $D_n$: $2, 3, 4, 5, 6, 7, 8, 9, 10, 11, \dots$ It appears that $D_n = n+1$. Let's verify this formula with the recurrence relation $D_n = 2 D_{n-1} - D_{n-2}$. If $D_n = n+1$, then $D_{n-1} = (n-1)+1 = n$ and $D_{n-2} = (n-2)+1 = n-1$. Substituting these into the recurrence relation: $n+1 = 2(n) - (n-1)$ $n+1 = 2n - n + 1$ $n+1 = n+1$. The formula $D_n = n+1$ satisfies the recurrence relation. The initial conditions are also satisfied: $D_1 = 1+1 = 2$ (Correct) $D_2 = 2+1 = 3$ (Correct) So, for $a=2$, the determinant $D_n = n+1$.

We need to find the unit's place digit of $D_{2017}$. Using the formula $D_n = n+1$ with $n=2017$: $D_{2017} = 2017 + 1 = 2018$. The unit's place digit of 2018 is 8.