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Chapter 5: Problem 12

Given the matrices $$ \boldsymbol{A}=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right], \quad \boldsymbol{B}=\left[\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right] $$ and $$ \boldsymbol{C}=\left[\begin{array}{rr} 0 & 2 \\ 1 & 1 \\ -1 & -1 \end{array}\right] $$ evaluate \(A B, A C, B C, C A\) and \(B A^{\mathrm{T}}\). Which if any of these are diagonal, unit or symmetric?

Short Answer

Expert verified
None of the matrices (AB, AC, BC, CA, BA^T) are diagonal, unit, or symmetric.

Step by step solution

01

Compute A B

To find the product of matrices \( A \) and \( B \), we multiply each element of row \( i \) in \( A \) by each element of column \( j \) in \( B \) and sum these products:\[AB = \left[\begin{array}{lll} 1 & 1 & 1 \ 1 & 1 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & 1 & 0 \ 1 & 1 & 0 \ 0 & 0 & -1 \end{array}\right] = \left[\begin{array}{ccc} (1)(1)+(1)(1)+(1)(0) & (1)(1)+(1)(1)+(1)(0) & (1)(0)+(1)(0)+(1)(-1) \ (1)(1)+(1)(1)+(1)(0) & (1)(1)+(1)(1)+(1)(0) & (1)(0)+(1)(0)+(1)(-1) \end{array}\right]\]\[= \left[\begin{array}{ccc} 2 & 2 & -1 \ 2 & 2 & -1 \end{array}\right]\].
02

Compute A C

Multiply matrix \( A \) by matrix \( C \):\[AC = \left[\begin{array}{lll} 1 & 1 & 1 \ 1 & 1 & 1 \end{array}\right] \left[\begin{array}{rr} 0 & 2 \ 1 & 1 \ -1 & -1 \end{array}\right] = \left[\begin{array}{rr} (1)(0)+(1)(1)+(1)(-1) & (1)(2)+(1)(1)+(1)(-1) \ (1)(0)+(1)(1)+(1)(-1) & (1)(2)+(1)(1)+(1)(-1) \end{array}\right]\]\[= \left[\begin{array}{rr} 0 & 2 \ 0 & 2 \end{array}\right]\].
03

Compute B C

Multiply matrix \( B \) by matrix \( C \):\[BC = \left[\begin{array}{ccc} 1 & 1 & 0 \ 1 & 1 & 0 \ 0 & 0 & -1 \end{array}\right] \left[\begin{array}{rr} 0 & 2 \ 1 & 1 \ -1 & -1 \end{array}\right] = \left[\begin{array}{rr} (1)(0)+(1)(1)+(0)(-1) & (1)(2)+(1)(1)+(0)(-1) \ (1)(0)+(1)(1)+(0)(-1) & (1)(2)+(1)(1)+(0)(-1) \ (0)(0)+(0)(1)+(-1)(-1) & (0)(2)+(0)(1)+(-1)(-1) \end{array}\right]\]\[= \left[\begin{array}{rr} 1 & 3 \ 1 & 3 \ 1 & 1 \end{array}\right]\].
04

Compute C A

Multiply matrix \( C \) by matrix \( A \):\[CA = \left[\begin{array}{rr} 0 & 2 \ 1 & 1 \ -1 & -1 \end{array}\right] \left[\begin{array}{lll} 1 & 1 & 1 \ 1 & 1 & 1 \end{array}\right] = \left[\begin{array}{lll} (0)(1)+(2)(1) & (0)(1)+(2)(1) & (0)(1)+(2)(1) \ (1)(1)+(1)(1) & (1)(1)+(1)(1) & (1)(1)+(1)(1) \ (-1)(1)+(-1)(1) & (-1)(1)+(-1)(1) & (-1)(1)+(-1)(1) \end{array}\right]\]\[= \left[\begin{array}{lll} 2 & 2 & 2 \ 2 & 2 & 2 \ -2 & -2 & -2 \end{array}\right]\].
05

Compute B A^T

First, compute \( A^{\mathrm{T}} \), the transpose of \( A \), by swapping rows and columns of \( A \):\[A^{\mathrm{T}} = \left[\begin{array}{ll} 1 & 1 \ 1 & 1 \ 1 & 1 \end{array}\right]\].Now multiply \( B \) by \( A^{\mathrm{T}} \):\[BA^{\mathrm{T}} = \left[\begin{array}{ccc} 1 & 1 & 0 \ 1 & 1 & 0 \ 0 & 0 & -1 \end{array}\right] \left[\begin{array}{ll} 1 & 1 \ 1 & 1 \ 1 & 1 \end{array}\right] = \left[\begin{array}{ll} (1)(1)+(1)(1)+(0)(1) & (1)(1)+(1)(1)+(0)(1) \ (1)(1)+(1)(1)+(0)(1) & (1)(1)+(1)(1)+(0)(1) \ (0)(1)+(0)(1)+(-1)(1) & (0)(1)+(0)(1)+(-1)(1) \end{array}\right]\]\[= \left[\begin{array}{ll} 2 & 2 \ 2 & 2 \ -1 & -1 \end{array}\right]\].
06

Identify Matrix Types

Examine each resultant matrix to determine if it is diagonal, unit, or symmetric:- **Diagonal Matrix**: Only diagonal elements are non-zero.- **Unit Matrix**: Diagonal matrix with all diagonal elements equal to 1.- **Symmetric Matrix**: Matrix equal to its transpose.Answer:- None of the resultant matrices (\( AB \), \( AC \), \( BC \), \( CA \), \( BA^T \)) are diagonal or unit.- None are symmetric because they do not satisfy \( M = M^{\mathrm{T}} \).

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Key Concepts

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

Matrix Types
Matrices come in various types, each defined by specific characteristics of their elements and structure. Understanding these types can help simplify many mathematical problems.
  • Square Matrix: Has the same number of rows and columns, like a 3x3 matrix.
  • Diagonal Matrix: Only the diagonal elements (from top left to bottom right) are non-zero. All other elements are zero.
  • Symmetric Matrix: A square matrix that is equal to its transpose, meaning it looks the same if flipped along its main diagonal.
  • Identity Matrix: A diagonal matrix where all diagonal elements are 1, often symbolized as "I". It acts like the number 1 when multiplying matrices, leaving other matrices unchanged: such as "AI = A".
  • Zero Matrix: Every element is zero, denoted as "0".
Each type has unique properties and applications. Understanding these will help you identify and utilize matrices more effectively in calculations.
Symmetric Matrix
A symmetric matrix is a special type of square matrix where the matrix is equal to its transpose. In simple terms, this means that the matrix "mirrors" itself along its main diagonal.
  • Definition: A matrix \( M \) is symmetric if \( M = M^{ ext{T}} \).
  • Properties:
    • Symmetric matrices can only be square.
    • The entries across the diagonal are matched: \( a_{ij} = a_{ji} \).
    • Commonly found in quadratic forms and real-valued representation spaces.
For example, when examining the matrices from our problem, none of the resulting matrices from the provided matrix products are symmetric because they do not satisfy the condition \( M = M^{ ext{T}} \). This means each matrix differs from its transpose.
Transpose of a Matrix
The transpose of a matrix is an operation that flips a matrix over its diagonal, switching the matrix's row and column indices.
  • Notation: Denoted as \( A^{ ext{T}} \) for a matrix \( A \).
  • Operation:
    • If \( A = [a_{ij}] \), then \( A^{ ext{T}} = [a_{ji}] \).
  • Properties:
    • The transpose of a transpose is the original matrix: \((A^{ ext{T}})^{ ext{T}} = A \).
    • \((A + B)^{ ext{T}} = A^{ ext{T}} + B^{ ext{T}} \).
    • \((AB)^{ ext{T}} = B^{ ext{T}} A^{ ext{T}} \), demonstrating that transposing a product reverses the order of multiplication.
In this exercise, computing the transpose was crucial when evaluating \( B A^T \). It means swapping the rows and columns of matrix \( A \) before multiplying it with \( B \). This rearrangement leads to the result \( \text{BA}^{ ext{T}} \).
Diagonal Matrix
A diagonal matrix is a unique type of square matrix where all elements outside the main diagonal are zero. Simply put, any non-zero elements are constrained to the diagonal.
  • Definition: A matrix is diagonal if \( a_{ij} = 0 \) for all \( i eq j \).
  • Key Features:
    • The main diagonal may have non-zero elements, but all other elements are zero.
    • Easy to compute determinants and inverses (when applicable) due to the simplicity of their structure.
    • Multiplying any matrix by a diagonal matrix scales the rows or columns of that matrix by the diagonal values.
None of the matrices from the original matrix products in this problem result in a diagonal matrix. This is determined by observing that not all off-diagonal components are zero in any of the resultant matrices.

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Most popular questions from this chapter

For the matrices $$ \boldsymbol{A}=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] \text { and } \quad \boldsymbol{B}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] $$ (a) evaluate \((\boldsymbol{A}+\boldsymbol{B})^{2}\) and \(\boldsymbol{A}^{2}+2 \boldsymbol{A B}+\boldsymbol{B}^{2}\) (b) evaluate \((\boldsymbol{A}+\boldsymbol{B})(\boldsymbol{A}-\boldsymbol{B})\) and \(\boldsymbol{A}^{2}-\boldsymbol{B}^{2}\) Repeat the calculations with the matrices $$ \boldsymbol{A}=\left[\begin{array}{ll} 1 & 2 \\ 5 & 2 \end{array}\right] \text { and } \quad \boldsymbol{B}=\left[\begin{array}{rr} 2 & -2 \\ -5 & 1 \end{array}\right] $$ and explain the differences between the results for the two sets.

Find the rank of \(A\) and of the augmented matrix \((\boldsymbol{A}: \boldsymbol{b})\). Solve \(\boldsymbol{A} \boldsymbol{X}=\boldsymbol{b}\) where possible and check that there are \((n-\operatorname{rank}(\boldsymbol{A}))\) free parameters. (a) \(\boldsymbol{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \quad \boldsymbol{b}=\left[\begin{array}{l}0 \\ 1\end{array}\right]\) (b) \(\boldsymbol{A}=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right] \quad \boldsymbol{b}=\left[\begin{array}{l}0 \\ 1\end{array}\right]\) (c) \(\boldsymbol{A}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{array}\right] \quad \boldsymbol{b}=\left[\begin{array}{l}1 \\ 0 \\\ 0\end{array}\right]\) (d) \(\boldsymbol{A}=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right] \quad \boldsymbol{b}=\left[\begin{array}{l}2 \\\ 1\end{array}\right]\) (e) \(\boldsymbol{A}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1 \\ 1 & 0\end{array}\right] \quad \boldsymbol{b}=\left[\begin{array}{l}1 \\ 0 \\\ 0\end{array}\right]\) (f) \(\boldsymbol{A}=\left[\begin{array}{llll}0 & 0 & 0 & 1 \\ 0 & 0 & 2 & 0 \\\ 0 & 3 & 0 & 0 \\ 4 & 0 & 0 & 0\end{array}\right] \quad \boldsymbol{b}=\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 0\end{array}\right]\)

Given the matrices $$ \boldsymbol{A}=\left[\begin{array}{lll} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 1 & 2 \end{array}\right] \text { and } \quad \boldsymbol{B}=\left[\begin{array}{rrr} -1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 1 \end{array}\right] $$ evaluate \(\boldsymbol{C}\) in the three cases. (a) \(\boldsymbol{C}=\boldsymbol{A}+\boldsymbol{B}\) (b) \(2 A+3 C=4 B\) (c) \(A-C=B+C\)

In finite-element calculations the bilinear function $$ u(x, y)=a+b x+c y+d x y $$ is commonly used for interpolation over a quadrilateral and data is always stored in matrix form. If the function fits the data \(u(0,0)=u_{1}\), \(u(p, 0)=u_{2}, u(0, q)=u_{3}\) and \(u(p, q)=u_{4}\) at the four corners of a rectangle, use matrices to find the coefficients \(a, b, c\) and \(d\).

Show that for any \(x\) the matrix $$ \mathbf{A}=\left[\begin{array}{lc} \cos (2 x) & \sin (2 x) \\ \sin (2 x) & -\cos (2 x) \end{array}\right] $$ satisfies the relation \(\boldsymbol{A}^{2}=\boldsymbol{I}\).
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