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Chapter 9: Problem 1

\(\mathbf{A}=\left[ \begin{array}{rr}{3} & {-2} \\ {0} & {3}\end{array}\right]\)

Short Answer

Expert verified
\mathbf{A} is a 2x2 matrix with elements \(a_{11}=3, a_{12}=-2, a_{21}=0, a_{22}=3\).

Step by step solution

01

Identify the Structure

A matrix is a rectangular array of numbers arranged in rows and columns. The given matrix is a square matrix because it has the same number of rows and columns (2). The dimension of this matrix is 2x2.
02

Locate the Elements

In this 2x2 matrix, we have four elements. Each element in a matrix is denoted by a pair of indices, which represents its position in the matrix. The first index indicates the row number and the second index indicates the column number. The elements of the matrix \mathbf{A} can be represented as \(a_{ij}\), where \(i\) refers to the row number and \(j\) refers to the column number. Therefore, for matrix \mathbf{A}, \(a_{11}=3, a_{12}=-2, a_{21}=0, a_{22}=3\).
03

Representation of Matrix

We can represent this matrix \mathbf{A} as \(\mathbf{A}=\left[ \begin{array}{rr}{a_{11}} & {a_{12}} \ {a_{21}} & {a_{22}}\end{array}\right] = \left[ \begin{array}{rr}{3} & {-2} \ {0} & {3}\end{array}\right]\)

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Key Concepts

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

Square Matrix
A square matrix is a special type of matrix commonly encountered in matrix algebra. It is defined as a matrix that has an equal number of rows and columns. This particular property makes them easy to spot and unique in matrix operations. In the example provided, the matrix \(\mathbf{A}\) is a square matrix because it has 2 rows and 2 columns, forming a 2x2 configuration.
  • Important in various mathematical computations, such as calculating determinants and eigenvalues.
  • Symmetrical operations are possible, enhancing its role in advanced algebra.
  • Essential in understanding identity matrices and inverse matrices.
Square matrices serve as fundamental building blocks in linear algebra, providing a basis for more complex matrix operations and theories.
Matrix Elements
Matrix elements are the individual numbers or values that make up a matrix. In any matrix, each element is identified by its position in terms of row and column numbers. For the given matrix \(\mathbf{A}\), we denote the elements by \(a_{ij}\), where \(i\) is the row, and \(j\) is the column.

In the specific 2x2 matrix \(\mathbf{A}\), we have four elements:
  • \(a_{11} = 3\) 鈥� First row, first column
  • \(a_{12} = -2\) 鈥� First row, second column
  • \(a_{21} = 0\) 鈥� Second row, first column
  • \(a_{22} = 3\) 鈥� Second row, second column
These elements define the matrix structure and are crucial for performing any matrix manipulation or operation.
Matrix Notation
Matrix notation is a standardized format used to represent matrices in mathematical expressions and transformations. Matrices are typically denoted by bold uppercase letters, such as \(\mathbf{A}\), while their elements are represented using lowercase letters with subscript indices.

This notation helps in quickly identifying the matrix and its components without ambiguity:
  • \(\mathbf{A} = \left[ \begin{array}{rr} {a_{11}} & {a_{12}} \ {a_{21}} & {a_{22}} \end{array}\right]\)
Using this form of representation:
  • Makes comprehension of complex matrix equations easier.
  • Simplifies the process of writing and reading mathematical problems.
  • Supports visualizing bigger matrices due to organized arrangement in rows and columns.
Understanding matrix notation is essential for delving into deeper subjects involving matrices, such as transformations in geometry or systems of linear equations.

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Most popular questions from this chapter

\(\left| \begin{array}{rrr}{1} & {0} & {0} \\ {3} & {1} & {2} \\ {1} & {5} & {-2}\end{array}\right|\)

RLC-Network. The currents in the \(R L C\) network given by the schematic diagram in Figure 9.6 are governed by the following equations: $$ \begin{array}{l}{4 I_{2}^{\prime}(t)+52 q_{1}(t)=10} \\ {13 I_{3}(t)+52 q_{1}(t)=10} \\ {I_{1}(t)=I_{2}(t)+I_{3}(t)}\end{array} $$ where \(q_{1}(t)\) is the charge on the capacitor, \(I_{1}(t)=q_{1}^{\prime}(t)\) and initially \(q_{1}(0)=0\) coulombs and \(I_{1}(0)=0\) amps. Solve for the currents \(I_{1}, I_{2},\) and \(I_{3} .\) [Hint: Differentiate the first two equations, eliminate \(I_{1},\) and form a normal system with \(x_{1}=I_{2}, x_{2}=I_{2}^{\prime},\) and \(x_{3}=I_{3} . ]\)

In Problems \(31-34,\) solve the given initial value problem. \(\mathbf{x}^{\prime}(t)=\left[ \begin{array}{ll}{1} & {3} \\ {3} & {1}\end{array}\right] \mathbf{x}(t), \quad \mathbf{x}(0)=\left[ \begin{array}{l}{3} \\ {1}\end{array}\right]\)

37\. To illustrate the connection between a higher-order equation and the equivalent first-order system, consider the equation $$\quad y^{\prime \prime \prime}(t)-6 y^{\prime \prime}(t)+11 y^{\prime}(t)-6 y(t)=0.$$ (a) Show that \(\left\\{e^{t}, e^{2 t}, e^{3 t}\right\\}\) is a fundamental solution set for \((11)\) . (b) Using the definition in Section 6.1 , compute the Wronskian of \(\left(e^{t}, e^{2 t}, e^{3 t}\right)\) . (c) Setting \(x_{1}=y, x_{2}=y^{\prime}, x_{3}=y^{\prime \prime},\) show that equa- tion \((11)\) is equivalent to the first-order system $$(12) \quad \mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}$$ where $$\mathbf{A} :=\left[ \begin{array}{rrr}{0} & {1} & {0} \\ {0} & {0} & {1} \\\ {6} & {-11} & {6}\end{array}\right].$$ (d) The substitution used in part (c) suggests that $$S :=\left\\{\left[ \begin{array}{c}{e^{t}} \\ {e^{t}} \\\ {e^{t}}\end{array}\right], \left[ \begin{array}{c}{e^{2 t}} \\ {2 e^{2 t}} \\\ {4 e^{2 t}}\end{array}\right], \left[ \begin{array}{c}{e^{3 t}} \\ {3 e^{3 t}} \\ {9 e^{3 t}}\end{array}\right]\right\\}$$ is a fundamental solution set for system \((12) .\) Verify that this is the case. (e) Compute the Wronskian of \(S .\) How does it compare with the Wronskian computed in part (b)?

$$ \mathbf{x}^{\prime}(t)=\left[ \begin{array}{rrr}{1} & {0} & {-1} \\ {0} & {2} & {0} \\ {1} & {0} & {1}\end{array}\right] \mathbf{x}(t) $$ $$ \quad \text { (a) }\ \mathbf{x}(0)=\left[ \begin{array}{r}{-2} \\ {2} \\ {-1}\end{array}\right] \quad \text { (b) } \quad \mathbf{x}(-\pi)=\left[ \begin{array}{l}{0} \\ {1} \\\ {1}\end{array}\right] $$
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