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Chapter 2: Problem 31

Find \(A^{H}\) where (a) \(A=\left[\begin{array}{ll}3-5 i & 2+4 i \\ 6+7 i & 1+8 i\end{array}\right]\) (b) \(A=\left[\begin{array}{cc}2-3 i & 5+8 i \\ -4 & 3-7 i \\ -6-i & 5 i\end{array}\right]\)

Short Answer

Expert verified
(a) The Hermitian conjugate of the given matrix A is: \(A^H=\left[\begin{array}{ll}3+5 i & 6-7 i \\\ 2-4 i & 1-8 i\end{array}\right]\). (b) The Hermitian conjugate of the given matrix A is: \(A^H=\left[\begin{array}{cc}2+3 i & -4 & -6+i \\\ 5-8 i & 3+7 i & -5 i\end{array}\right]\).

Step by step solution

01

Find the transpose of matrix A.

To find the transpose of matrix A, we exchange its rows with columns. The transpose of the given matrix is: \(A^T=\left[\begin{array}{ll}3-5 i & 6+7 i \\\ 2+4 i & 1+8 i\end{array}\right]\)
02

Take the complex conjugate of each entry in \(A^T\).

We now take the complex conjugate of each entry of the matrix \(A^T\): \(A^H=\left[\begin{array}{ll}3+5 i & 6-7 i \\\ 2-4 i & 1-8 i\end{array}\right]\) (a) So the Hermitian conjugate of the given matrix A is: \(A^H=\left[\begin{array}{ll}3+5 i & 6-7 i \\\ 2-4 i & 1-8 i\end{array}\right]\). (b) \(A=\left[\begin{array}{cc}2-3 i & 5+8 i \\\ -4 & 3-7 i \\\ -6-i & 5 i\end{array}\right]\)
03

Find the transpose of matrix A.

To find the transpose of matrix A, we exchange its rows with columns. The transpose of the given matrix is: \(A^T=\left[\begin{array}{cc}2-3 i & -4 & -6-i \\\ 5+8 i & 3-7 i & 5 i\end{array}\right]\)
04

Take the complex conjugate of each entry in \(A^T\).

We now take the complex conjugate of each entry of the matrix \(A^T\): \(A^H=\left[\begin{array}{cc}2+3 i & -4 & -6+i \\\ 5-8 i & 3+7 i & -5 i\end{array}\right]\) (b) So the Hermitian conjugate of the given matrix A is: \(A^H=\left[\begin{array}{cc}2+3 i & -4 & -6+i \\\ 5-8 i & 3+7 i & -5 i\end{array}\right]\).

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Key Concepts

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

Complex Conjugate
In linear algebra, the concept of a complex conjugate is fundamental when dealing with complex numbers. A complex number is composed of a real part and an imaginary part, expressed as a + bi, where a is the real part, b is the imaginary part, and i is the imaginary unit with the property that i^2 = -1. The complex conjugate of this number is a - bi. It is obtained by changing the sign of the imaginary part.

The importance of the complex conjugate comes into play for various operations in mathematics, including finding the magnitude of a complex number and division of complex numbers. In matrix operations, taking the complex conjugate of each entry is a crucial step towards computing the Hermitian conjugate of a matrix.
Matrix Transpose
The matrix transpose is another operation vital to the field of linear algebra. When we transpose a matrix, denoted as A^T, we essentially flip the matrix over its diagonal. This means that the rows of the original matrix A become the columns of its transpose A^T, and vice versa.

This operation is used in various matrix computations and has interesting properties such as (AB)^T = B^T A^T, where A and B are matrices. Transposing is also integral to finding the Hermitian conjugate of a matrix, which combines both the transpose and the concept of complex conjugates.
Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It is a foundational topic in most scientific fields, including engineering, physics, computer science, and even economics. Linear algebra allows the manipulation of large sets of equations or transformations in a systematic way. Operations such as matrix multiplication, finding determinants, and calculating eigenvectors and eigenvalues are all part of linear algebra. Mastering these operations enables one to solve more complex problems like systems of differential equations and optimization problems.
Matrices Operations
In matrices operations, we deal with various computations involving matrices. These operations include addition, subtraction, multiplication, finding the determinant, and the inverse of a matrix. Each operation has specific rules and applications. For instance, matrix multiplication is not commutative, meaning AB is not necessarily equal to BA.

Understanding matrix operations is crucial for solving linear systems and performing transformations in space. These operations are not only essential in pure mathematics but also practical in various applications like graphics, statistics, and solving simultaneous equations.

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Most popular questions from this chapter

Find the inverses of \(A=\left[\begin{array}{lll}1 & 1 & 2 \\ 1 & 2 & 5 \\ 1 & 3 & 7\end{array}\right]\) and \(B=\left[\begin{array}{rrr}1 & -1 & 1 \\ 0 & 1 & -1 \\ 1 & 3 & -2\end{array}\right] .\) [Hint: See Problem 2.19.]

Prove Theorem \(2.2(\text { iii })\) and (iv): (iii) \((B+C) A=B A+C A, \quad\) (iv) \(k(A B)=(k A) B=A(k B)\)

Prove Theorem 2.1 (i) and (v): (i) \((A+B)+C=A+(B+C),\) (v) \(k(A+B)=k A+k B\)

Prove Theorem \(2.2(\mathrm{i}):(A B) C=A(B C)\) Let \(A=\left[a_{i j}\right], \quad B=\left[b_{j k}\right], \quad C=\left[c_{k i}\right], \quad\) and let \(A B=S=\left[s_{i k}\right], \quad B C=T=\left[t_{j l}\right] .\) Then $$ s_{i k}=\sum_{j=1}^{m} a_{i j} b_{j k} \quad \text { and } \quad t_{j l}=\sum_{k=1}^{n} b_{j k} c_{k l} $$ Multiplying \(S=A B\) by \(C\), the il-entry of \((A B) C\) is $$ s_{i 1} c_{1 l}+s_{i 2} c_{2 l}+\cdots+s_{i n} c_{n l}=\sum_{k=1}^{n} s_{i k} c_{k l}=\sum_{k=1}^{n} \sum_{j=1}^{m}\left(a_{i j} b_{j k}\right) c_{k l} $$ On the other hand, multiplying \(A\) by \(T=B C\), the il-entry of \(A(B C)\) is $$ a_{i 1} t_{1 l}+a_{i 2} t_{2 l}+\cdots+a_{i m} t_{m l}=\sum_{j=1}^{m} a_{i j} t_{j l}=\sum_{j=1}^{m} \sum_{k=1}^{n} a_{i j}\left(b_{j k} c_{k l}\right) $$ The above sums are equal; that is, corresponding elements in \((A B) C\) and \(A(B C)\) are equal. Thus, \((A B) C=A(B C)\)

Given \(A=\left[\begin{array}{rrr}1 & -2 & 3 \\ 4 & 5 & -6\end{array}\right]\) and \(B=\left[\begin{array}{rrr}3 & 0 & 2 \\ -7 & 1 & 8\end{array}\right]\), find: (a) \(A+B\), (b) \(2 A-3 B\). (a) Add the corresponding elements: $$ A+B=\left[\begin{array}{rrr} 1+3 & -2+0 & 3+2 \\ 4-7 & 5+1 & -6+8 \end{array}\right]=\left[\begin{array}{rrr} 4 & -2 & 5 \\ -3 & 6 & 2 \end{array}\right] $$ (b) First perform the scalar multiplication and then a matrix addition: $$ 2 A-3 B=\left[\begin{array}{rrr} 2 & -4 & 6 \\ 8 & 10 & -12 \end{array}\right]+\left[\begin{array}{rrr} -9 & 0 & -6 \\ 21 & -3 & -24 \end{array}\right]=\left[\begin{array}{rrr} -7 & -4 & 0 \\ 29 & 7 & -36 \end{array}\right] $$ (Note that we multiply \(B\) by \(-3\) and then add, rather than multiplying \(B\) by 3 and subtracting. This usually prevents errors.)
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