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Chapter 8: Problem 44

(Linear algebra required) Let \(\mathbf{A}_{n}\) be the \(n \times n\) matrix with 2 \(\mathrm{s}\) on its main diagonal, 1 \(\mathrm{s}\) in all positions next to a diagonal element, and 0 \(\mathrm{s}\) everywhere else. Find a recurrence relation for \(d_{n},\) the determinant of \(\mathbf{A}_{n} .\) Solve this recurrence relation to find a formula for \(d_{n} .\)

Short Answer

Expert verified
The recurrence relation is \( d_n = 2d_{n-1} - d_{n-2} \) with the solution \( d_n = 1 + n \).

Step by step solution

01

Identify the matrix structure

Understand that the given matrix \( \textbf{A}_n \) is a tridiagonal matrix with 2s on its main diagonal, 1s on its off-diagonals, and 0s elsewhere. This structure is crucial for setting up the recurrence relation.
02

Express the determinant

Let \( d_n \) be the determinant of the \( n \)-th order matrix. We need to find a recurrence relation for \( d_n \). Note that the n x n determinant can be expressed in terms of smaller matrices.
03

Use Laplace Expansion

Expand the determinant \( d_n \) along the first row: \[ d_n = 2d_{n-1} - d_{n-2} \] This comes from the fact that expanding by the first row gives the term with the main diagonal's 2 times the \( (n-1) \times (n-1) \) determinant minus one for the 1s at the off-diagonals.
04

Set initial conditions

Set the base cases: \[ d_1 = 2 \] and \[ d_2 = 3 \]. These are found by directly calculating the determinants for \( n = 1 \) and \( n = 2 \).
05

Solve the recurrence relation

Use the characteristic equation method for second-order linear recurrence relations: \[ x^2 - 2x + 1 = 0 \] The roots are \( x = 1 \) with multiplicity 2. So the general solution is: \[ d_n = (A + Bn)1^n = A + Bn \]
06

Find constants using initial conditions

Use the initial conditions to solve for A and B: \[ d_1 = 2 = A + B(1) \] \[ d_2 = 3 = A + B(2) \] Solving, \[ B = 1 \] \[ A = 1 \]. Therefore, \[ d_n = 1 + n \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tridiagonal Matrix
A tridiagonal matrix is a type of square matrix that has non-zero elements only on the main diagonal, the first diagonal above it, and the first diagonal below it. This means that all other elements in the matrix are zeros.
For matrix \(\textbf{A}_n\), given in the problem, we see the following:
  • Main diagonal: All elements are 2.
  • First sub-diagonal: All elements are 1.
  • First super-diagonal: All elements are 1.

Understanding the structure of the matrix is crucial because it tells us that we can employ specific methods like Laplace expansion effectively. Tridiagonal matrices have a simplified determinant calculation, and they often lead to recurrence relations.
Characteristic Equation
The characteristic equation is used in many areas of linear algebra, especially when dealing with recurrence relations. It is derived from the determinant of a matrix and is essential for solving linear recurrence relations.
For the recurrence relation identified in the solution,
\[ d_n = 2d_{n-1} - d_{n-2} \],
we turn this into a characteristic equation:
\[ x^2 - 2x + 1 = 0 \],
Solving this quadratic equation gives us the roots, which are used to find the general solution to the recurrence relation. In this case, the root \(x = 1\) has multiplicity 2, leading to a specific form for the solution.
Laplace Expansion
The Laplace expansion is a method used to calculate the determinant of a matrix. It involves expanding the determinant along a row or column, thereby breaking it into smaller matrices that are easier to handle.
For our matrix \(\textbf{A}_n\), expanding along the first row allows us to express the determinant in terms of smaller determinants. Specifically:
\[ d_n = 2d_{n-1} - d_{n-2} \],
This makes use of the structure of the tridiagonal matrix and simplifies the computing process significantly. By doing this, we set up a recurrence relation that we can solve using other methods.
Initial Conditions
Initial conditions are specific values that help us start the process of solving a recurrence relation. They provide the baseline from which the behavior of the sequence is determined.
For our determinant sequence \(d_n\), we have:
  • \(d_1 = 2\)
  • \(d_2 = 3\)

These values are calculated directly from the definitions of 1x1 and 2x2 matrices. By plugging these initial conditions back into our general solution for the recurrence relation, we can solve for any constants involved.
Linear Recurrence Relations
Linear recurrence relations are equations that recursively define sequences. Each term of the sequence is a linear combination of previous terms.
In mathematics, these relations play a crucial role in various fields, such as the study of algorithms, sequences, and matrix theory.
For the matrix \(\textbf{A}_n\), the recurrence relation derived is:
\[ d_n = 2d_{n-1} - d_{n-2} \],
This is a second-order linear recurrence relation because it involves the two preceding terms \(d_{n-1}\) and \(d_{n-2}\). These relations help simplify complex problems by reducing them to a series of simpler calculations.

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Most popular questions from this chapter

Find the number of positive integers not exceeding 100 that are not divisible by 5 or by 7 .

Find the number of positive integers not exceeding 1000 that are not divisible by \(3,17,\) or \(35 .\)

a) Find a recurrence relation for the number of ways to completely cover a \(2 \times n\) checkerboard with \(1 \times 2\) dominoes. [Hint: Consider separately the coverings where the position in the top right corner of the checkerboard is covered by a domino positioned horizontally and where it is covered by a domino positioned vertically.] b) What are the initial conditions for the recurrence relation in part (a)? c) How many ways are there to completely cover a \(2 \times\) 17 checkerboard with \(1 \times 2\) dominoes?

Exercises \(38-45\) involve the Reve's puzzle, the variation of the Tower of Hanoi puzzle with four pegs and \(n\) disks. Before presenting these exercises, we describe the Frame-Stewart algorithm for moving the disks from peg 1 to peg 4 so that no disk is ever on top of a smaller one. This algorithm, given the number of disks \(n\) as input, depends on a choice of an integer \(k\) with \(1 \leq k \leq n .\) When there is only one disk, move it from peg 1 to peg 4 and stop. For \(n>1\) , the algorithm proceeds recursively, using these three steps. Recursively move the stack of the \(n-k\) smallest disks from peg 1 to peg \(2,\) using all four pegs. Next move the stack of the \(k\) largest disks from peg 1 to peg \(4,\) using the three-peg algorithm from the Tower of Hanoi puzzle without using the peg holding the \(n-k\) smallest disks. Finally, recursively move the smallest \(n-k\) disks to peg \(4,\) using all four pegs. Frame and Stewart showed that to produce the fewest moves using their algorithm, \(k\) should be chosen to be the smallest integer such that \(n\) does not exceed \(t_{k}=k(k+1) / 2,\) the \(k\) th triangular number, that is, \(t_{k-1}

a) Find a recurrence relation for the number of bit strings of length n that do not contain three consecutive 0s. b) What are the initial conditions? c) How many bit strings of length seven do not contain three consecutive 0s?
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