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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

Use generating functions to solve the recurrence relation \(a_{k}=3 a_{k-1}+2\) with the initial condition \(a_{0}=1\)

Use generating functions to solve the recurrence rela- tion \(a_{k}=2 a_{k-1}+3 a_{k-2}+4^{k}+6\) with initial conditions \(a_{0}=20, a_{1}=60\)

How many onto functions are there from a set with seven elements to one with five elements?

Let \(S(m, n)\) denote the number of onto functions from a set with \(m\) elements to a set with \(n\) elements. Show that \(S(m, n)\) satisfies the recurrence relation $$ S(m, n)=n^{m}-\sum_{k=1}^{n-1} C(n, k) S(m, k) $$ whenever \(m \geq n\) and \(n>1,\) with the initial condition \(S(m, 1)=1\).

In this exercise we will develop a dynamic programming algorithm for finding the maximum sum of consecutive terms of a sequence of real numbers. That is, given a sequence of real numbers \(a_{1}, a_{2}, \ldots, a_{n}\) the algorithm computes the maximum sum \(\sum_{i=j}^{k} a_{i}\) where \(1 \leq j \leq k \leq n\). a) Show that if all terms of the sequence are nonnegative, this problem is solved by taking the sum of all terms. Then, give an example where the maximum sum of consecutive terms is not the sum of all terms. b) Let \(M(k)\) be the maximum of the sums of consecutive terms of the sequence ending at \(a_{k} .\) That is, \(M(k)=\) \(\max _{1 \leq j \leq k} \sum_{i=j}^{k} a_{i}\) Explain why the recurrence relation \(M(k)=\max \left(M(k-1)+a_{k}, a_{k}\right)\) holds for \(k=2, \ldots, n\). c) Use part (b) to develop a dynamic programming algorithm for solving this problem. d) Show each step your algorithm from part (c) uses to find the maximum sum of consecutive terms of the sequence \(2,-3,4,1,-2,3\) e) Show that the worst-case complexity in terms of the number of additions and comparisons of your algorithm from part (c) is linear.
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