/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Choose \(a\) and \(b\) so that e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 2

Choose \(a\) and \(b\) so that each of the following matrices becomes symmetric: a) \(\left[\begin{array}{cc}1 & 3 a-1 \\ 2 a & 3\end{array}\right]\) b) \(\left[\begin{array}{ccc}2 & a & 3 \\ b-a & 0 & 4+a \\ 3 & b & 5\end{array}\right]\)

Short Answer

Expert verified
Answer: To make the matrices symmetric, the values are \(a=1\) and \(b=4+a\).

Step by step solution

01

Transpose the matrices

For matrix (a), we write its transpose: \(A^T = \left[\begin{array}{cc}1 & 2a \\\ 3a-1 & 3\end{array}\right]\) Similarly, for matrix (b), we find the transpose: \(B^T = \left[\begin{array}{ccc}2 & b-a & 3 \\\ a & 0 & b \\\ 3 & 4+a & 5\end{array}\right]\)
02

Compare matrices and find values of \(a\) and \(b\)

For a matrix to be symmetric, we need \(A = A^T\). So we compare the entries of \(A\) and \(A^T\): \(\left[\begin{array}{cc}1 & 3a-1 \\\ 2a & 3\end{array}\right] = \left[\begin{array}{cc}1 & 2a \\\ 3a-1 & 3\end{array}\right]\) To make these matrices equal, we need \(3a-1=2a\): \(3a-1=2a \Rightarrow a=1\) Similarly, for matrix (b), we need \(B = B^T\): \(\left[\begin{array}{ccc}2 & a & 3 \\\ b-a & 0 & 4+a \\\ 3 & b & 5\end{array}\right] = \left[\begin{array}{ccc}2 & b-a & 3 \\\ a & 0 & b \\\ 3 & 4+a & 5\end{array}\right]\) Here, we can see that the matrices are already equal in all the entries except for the \(a\) in the row 2, column 3 and the \(b\) in row 3, column 2. So, to make the matrices equal, we need \(4+a=b\). So, the values to make the matrices symmetric are \(a=1\) and \(4+a=b\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Matrix Transpose
In linear algebra, the transpose of a matrix is a fundamental concept. Transposing a matrix means flipping it over its diagonal. This transforms the rows into columns and the columns into rows.

For example, if you have a matrix \(A\) like this: \( A = \begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix}\), its transpose, denoted \(A^T\), would be: \(A^T = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}\).

Transposing a matrix is easy:
  • Identify the diagonal of the matrix (elements from top left to bottom right).
  • Swap elements over this diagonal. The element in row \(i\), column \(j\) becomes the element in row \(j\), column \(i\).
This operation is crucial when checking for symmetric matrices or when solving systems of linear equations.
Matrix Equality
Two matrices are considered equal if they have the same dimensions and their corresponding elements are equal. This means that for each element in matrix \(A\) located at position \((i, j)\), it must match the element in matrix \(B\) located at the same position for \(A = B\).

Suppose we have two matrices: \(A = \begin{bmatrix} 1 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 3 \end{bmatrix}\). These matrices are not equal because the second element in matrix \(A\) does not equal the second element in matrix \(B\).

Matrix equality is especially important when determining if a matrix is symmetric, as a symmetric matrix must be identical to its transpose. This requires each corresponding entry to be the same.
Matrices Comparison
Comparing matrices involves checking whether their dimensions align and if their corresponding elements are equal. This task often arises when determining if a matrix is symmetric.

To compare matrices:
  • Ensure both matrices have the same order (number of rows and columns).
  • Check each corresponding element for equality.

An essential application of matrix comparison is to determine symmetry, where the transpose \(A^T\) must equal the original matrix \(A\). If any corresponding elements do not match, the matrices are not equal, and thus, the matrix is not symmetric.

The matrix comparison process in linear algebra is used not only for symmetry but also in verifying solutions to matrix equations and in transformations.
Linear Algebra
Linear algebra is the branch of mathematics that deals with vectors, vector spaces (also called linear spaces), linear transformations, and systems of linear equations. Matrices play a crucial role in linear algebra because they can represent linear transformations and solve systems of equations efficiently.

Key concepts include:
  • Vectors: Objects with both magnitude and direction in space.
  • Matrix Operations: Including addition, subtraction, multiplication, and finding the transpose.
  • Systems of Linear Equations: Collections of equations to be solved jointly, often represented as matrices.
Understanding matrices and their properties, such as symmetry and equality, is fundamental in solving problems in linear algebra, as it provides tools for manipulating and interpreting mathematical models and systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(Diagonal matrices) a) Let \(B=\) diag \((\lambda, \mu)\). Show that, if \(\lambda \neq \mu\), then the eigenvectors associated with \(\lambda\) are all nonzero vectors \((1,0)\) and those associated with \(\mu\) are all nonzero vectors \(c(0,1)\); show that, if \(\lambda=\mu\), then the eigenvectors associated with \(\lambda\) are all nonzcro vectors \(\left(v_{1}, v_{2}\right)\). b) Let \(B=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)\) and let \(\mathbf{e}_{1}=(1,0,0)\). \(\mathbf{e}_{2}=(0,1,0), \mathbf{e}_{3}=(0,0.1)\) (column vectors). Show that if \(\lambda_{1} \cdot \lambda_{2}, \lambda_{3}\) are distinct, then for each \(\lambda_{k}\) the associated eigenvectors are the vectors \(c e_{k}\) for \(c \neq 0\); show that if \(\lambda_{1}=\lambda_{2} \neq \lambda_{3}\), then the cigenvectors associated with \(\lambda_{1}\) are all nonzero vectors \(c_{1} \mathbf{e}_{1}+c_{2} \mathbf{e}_{2}\) and those associated with \(\lambda_{3}\) are all nonzero vectors ce \(e_{3}\); show that if \(\lambda_{1}=\lambda_{2}=\lambda_{3}\), then the cigenvectors associated with \(\lambda_{1}\) are all nonzero vectors \(\mathbf{v}=\left(v_{1}, v_{2}, v_{3}\right)\). c) Let \(B=\operatorname{diag}\left(\lambda_{1}, \ldots, \lambda_{n}\right)\). Show that the eigenvectors associated with the eigenvalue \(\lambda_{k}\) are all nonzero vectors \(\mathbf{v}=\left(v_{1}, \ldots v_{n}\right)\) such that \(v_{i}=0\) for all \(i\) such that \(\lambda_{t} \neq \lambda_{k}\).

Prove the following: a) Every square matrix is similar to itself. b) If \(A\) is similar to \(B\) and \(B\) is similar to \(C\). then \(A\) is similar to \(C\).

In the following systems the elimination process shows that the determinant of coefficients is 0 . Carry out the process and find all solutions: a) \(2 x-y+z=3, x+2 y-z=1,5 x-5 y+4 z=8\) b) \(x+y-z=1,2 x-y-z=2, x+4 y-2 z=2\) c) \(x-y+z+w=0, x+2 y-z-w=0\). \(3 x-y-z+2 w=0, x+3 y+z-2 w=0\) d) \(x+y-2 z+3 w=0,2 x-y+z-w=0\). \(3 x-z+2 w=0,5 x+2 y-5 z+8 w=0\)

We consider complex matrices, that is, matrices with complex entries. If \(A=\left(a_{i j}\right)\) is such a matrix, we denote by \(\bar{A}\) the conjugate of \(A\), that is, the matrix \(\left(\bar{a}_{i j}\right)\), where \(\bar{z}\) is the conjugate of the complex number \(z\) (for example, if \(z=3+5 i\), then \(\bar{z}=3-5 i\) ). Prove the following: a) If \(z_{1}\) and \(z_{2}\) are complex numbers, then \(\overline{\left(z_{1}+z_{2}\right)}=\bar{z}_{1}+\bar{z}_{2}\) and \(\overline{z_{1} z_{2}}=\bar{z}_{1} \bar{z}_{2}\). b) If \(\frac{A}{A} \frac{\text { and } B}{B}\) are matrices and \(c\) is a complex scalar, then \(\overline{(A+B)}=\bar{A}+\bar{B}, \overline{A B}=\) \(\bar{A} \bar{B}, \overline{(c A)}=\bar{c} \bar{A}\). c) Every complex matrix \(A\) can be written uniquely as \(A_{1}+i A_{2}\), where \(A_{1}\) and \(A_{2}\) are real matrices, and \(A_{1}=\frac{1}{2}(A+\bar{A}), A_{2}=(2 i)^{-1}(A-\bar{A})\). We call \(A_{1}\) the real part of \(A, A_{2}\) the imaginary part of \(A\). d) If \(A\) is a square matrix, then \((\bar{A})^{\prime}=\overline{\left(A^{\prime}\right)}\) and, if \(A\) is nonsingular, then \(\overline{\left(A^{-1}\right)}=(\bar{A})^{-1}\). e) \(A\) is a real matrix if and only if \(A=\bar{A}\).

Verify that each of the following matrices is nonsingular and find the inverse of each: a) \(A=\left[\begin{array}{ll}3 & 5 \\ 2 & 4\end{array}\right]\). b) \(B=\left[\begin{array}{ll}4 & 7 \\ 1 & 6\end{array}\right]\). c) \(C=\left[\begin{array}{rrr}1 & 0 & 1 \\ 2 & 2 & 1 \\ 0 & 1 & -1\end{array}\right]\). d) \(D=\left[\begin{array}{lll}2 & 0 & 1 \\ 3 & 1 & 2 \\ 4 & 0 & 3\end{array}\right]\).
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.