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Determine the ranks of: a. \(\left[\begin{array}{rrrr}7 & 9 & -8 & 4 \\ 0 & 1 & 5 & 9 \\ 0 & 0 & 3 & 7 \\ 0 & 0 & 0 & -2\end{array}\right]\) b. \(\left[\begin{array}{rrrr}7 & 9 & -8 & 4 \\ 0 & 0 & 5 & 9 \\ 0 & 0 & 3 & 7 \\ 0 & 0 & 0 & -2\end{array}\right]\) c. \(\left[\begin{array}{rrrr}7 & 9 & -8 & 4 \\ 0 & 0 & 5 & 9 \\ 0 & 0 & 3 & 7 \\ 0 & 0 & 0 & 0\end{array}\right]\) d. Determine the rank of an arbitrary \(4 \times 4\) matrix of the form $$ \left[\begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \\ 0 & a_{22} & a_{23} & a_{24} \\ 0 & 0 & a_{33} & a_{34} \\ 0 & 0 & 0 & a_{44} \end{array}\right] $$ in terms of the entries \(a_{11}, a_{22}, a_{33}\), and \(a_{44}\). e. Generalize the result of part d to \(n \times n\) matrices in which the only nonzero entries are in the upper right triangular region of the matrix.

Short Answer

Expert verified
The ranks are: for matrix a is 4, for matrix b is 4, for matrix c is 3. For an arbitrary 4x4 matrix, the rank is 4 only if all elements on the main diagonal are non-zero. For a generalized n x n upper triangular matrix, the rank equals the number of non-zero elements on the main diagonal.

Step by step solution

01

Find the rank for matrix a

Examine the main diagonal of the matrix. Since none of the elements is zero, the rank of matrix a is 4.
02

Find the rank for matrix b

Inspect the diagonal elements of this matrix. Since no element is zero, the rank of the matrix b is also 4.
03

Find the rank for matrix c

In the case of matrix c, we have a zero in the main diagonal. So, when counting the number of non-zero elements on the diagonal, we find the rank of the matrix c to be 3.
04

Determine the rank of an arbitrary 4x4 upper triangular matrix

For an arbitrary 4x4 triangle matrix, the rank will be equal to the number of non-zero elements along the main diagonal. So, the matrix rank would be 4 if and only if none of \(a_{11},a_{22},a_{33},a_{44}\) is a zero.
05

Generalize for n x n upper triangular matrices

For a generalized case of an n x n upper triangular matrix, the number of non-zero elements on the main diagonal will determine the rank of the matrix. So if there are k non-zero elements on the main diagonal, the rank will be k.

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Key Concepts

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

Upper Triangular Matrix
An upper triangular matrix is a special type of square matrix. In this matrix, all elements below the main diagonal are zero. This means if you have a matrix represented by \(\begin{array}{cccc} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \ 0 & a_{22} & a_{23} & \cdots & a_{2n} \ 0 & 0 & a_{33} & \cdots & a_{3n} \ \vdots & \vdots & \vdots & \ddots & \vdots \ 0 & 0 & 0 & \cdots & a_{nn} \end{array} \)

This outer structure simplifies many computations. For example, when determining the rank of such a matrix, you focus mainly on the entries along the main diagonal. Upper triangular matrices are widely useful in linear algebra, particularly in solving systems of linear equations. With these matrices, back-substitution is a straightforward process, making calculations more efficient.
Diagonal Elements
The diagonal elements of a matrix are those that lie on the path from the upper-left to the lower-right of a square matrix. In terms of indices in a matrix, these are the elements where the row number equals the column number, often written as \(a_{ii}\).

For an upper triangular matrix, these elements are crucial because they determine the matrix's rank. If you're inspecting a matrix like:\(\begin{array}{cccc} a_{11} & a_{12} & a_{13} & a_{14} \ 0 & a_{22} & a_{23} & a_{24} \ 0 & 0 & a_{33} & a_{34} \ 0 & 0 & 0 & a_{44} \end{array}\)

The rank relies on how many of these diagonal entries are non-zero. In matrix computations, these diagonal elements serve not only as an indicator of rank but also affect properties like invertibility.
Non-Zero Elements
Non-zero elements in a matrix have a significant impact on determining the matrix's rank. Specifically for upper triangular matrices, where the rank directly correlates with the number of non-zero diagonal elements. This simplifies the calculation process considerably.

For instance, if a diagonal element of an upper triangular matrix is zero, it reduces the rank by 1. This is because the zero element does not contribute to the span of the matrix's rows. For us to find the rank of a matrix, especially an upper triangular one, simply tally the non-zero diagonal elements. This focus minimizes the need to evaluate every element individually, streamlining the process.
n x n Matrix
An \(n \times n\) matrix signifies a square matrix containing \(n\) rows and \(n\) columns. This symmetry allows many unique properties not found in non-square matrices. These matrices lay the groundwork for understanding complex matrix operations and transformations.

In the context of upper triangular matrices, an \(n \times n\) matrix maintains its properties efficiently due to having a systematic zeroing of both sides of the diagonal. With this understanding, generalized computations such as determining rank become straightforward. Exploiting the properties of the main diagonal streamlines calculations across any size matrix, simplifying linear algebra tasks down to counting non-zero diagonal elements. Understanding these contributes vastly to algebraic solutions in mathematical and real-world applications.

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Most popular questions from this chapter

Many characteristics of plants and animals are determined genetically. Suppose the height of a variety of corm is determined by a gene that comes in two forms, \(T\) and \(t\). Each plant has a pair of these genes and can thus be pure dominant type \(T T\), pure recessive type \(t\), or hybrid \(T\). Each seed obtains one of the two genes from one plant and the second from another plant. In each generation, a geneticist randomly selects a pair of plants from all possible offspring. These are used to breed the plants for the next generation. a. Set up a Markov chain to model this experiment. The states will be the pairs of genotypes of the parents. Since it does not matter which parent plant contributed which gene to the offspring, you need only consider six states: \((T T, T T),(T T, T t),(T T, t t),(T t, T t),(T t, t t)\), and \((t t, t t)\). b. Show that this is an absorbing Markov chain. c. Interpret the results of the Fundamental Theorem of Absorbing Markov Chains for this model.

Suppose \(A \in \mathbb{M}(m, n)\) and \(\mathbf{b} \in \mathbb{M}(m, 1)\). Use properties of matrix multiplication and the identification of \(\mathrm{R}^{n}\) with \(\mathbb{M}(n, 1)\) to prove the following facts about systems of linear equations. a. If \(v_{0} \in \mathbb{R}^{n}\) is any solution to the linear system \(A x=b\) and if \(v \in R^{n}\) is any solution to the homogeneous system \(A \mathbf{x}=\mathbf{0}\), then \(\mathbf{v}_{0}+\mathbf{v}\) is a solution to the system \(A x=b\). b. If \(v_{0} \in \mathbb{R}^{n}\) and \(v_{1} \in \mathbb{R}^{n}\) are any two solutions of the linear system \(A \mathrm{x}=\mathrm{b}\), then the difference \(\mathrm{v}_{0}-\mathrm{v}_{1}\) is a solution to the homogeneous system \(A \mathrm{x}=\mathbf{0}\). c. If \(\mathbf{v}_{0} \in \mathbb{R}^{n}\) is any solution to the linear system \(A \mathbf{x}=\mathbf{b}\) and \(S\) is the solution space of the homogeneous system \(A \mathbf{x}=\mathbf{0}\), then the set that is naturally enough denoted \(\mathbf{v}_{0}+S\) and defined by $$ \mathbf{v}_{0}+S=\left\\{\mathbf{v}_{0}+\mathbf{v} \mid \mathbf{v} \in S\right\\} $$ is the solution space of the original system.

Suppose a particle moves among the positions \(1,2,3,4,5\), and 6 along a line. Suppose states 1 and 6 are absorbing; otherwise, each step consists of the particle moving one position to the left with probability 4 , one step to the right with probability .5, or remaining fixed with probability . 1 . a. If the particle starts in position 3 , what are the expected numbers of times the particle will visit the four nonabsorbing positions? b. Show that starting position 3 results in the longest expected time for the particle to be absorbed. c. If the particle starts in position 3, what is the probability it will be absorbed in position 1? (Suggestion: Notice that the only way the particle can reach position 1 is through position 2. Each time the particle visits position 2, it has a \(40 \%\) chance of being absorbed into position I at the next step.)

a. Show that \(\left[\begin{array}{r}2 \\ -1\end{array}\right]\) is a right inverse of the matrix \(\left[\begin{array}{ll}1 & 1\end{array}\right]\). b. Show that \(\left[\begin{array}{r}2 \\ -1\end{array}\right]\) is not a left inverse of the matrix \(\left[\begin{array}{ll}1 & 1\end{array}\right]\). c. Show that \(\left[\begin{array}{ll}1 & 1\end{array}\right]\) does not have a left inverse.

a. Suppose \(A\) is a \(4 \times 3\) matrix. Find a \(3 \times 1\) matrix \(b\) such that \(A b\) is the first column of \(A\). b. Suppose \(A\) is an \(m \times n\) matrix. Find an \(n \times 1\) matrix b such that \(A b\) is the \(j\) th column of \(A\). c. How can the ith row of a matrix be obtained in terms of matrix multiplication?

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