/*! 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 20 For the matrices $$ \boldsym... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
Non-commutativity results in different expressions producing different matrices when A and B are changed.

Step by step solution

01

Compute \( \boldsymbol{A} + \boldsymbol{B} \)

To add the matrices \( \boldsymbol{A} = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \) and \( \boldsymbol{B} = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), sum the corresponding elements:\[ \boldsymbol{A} + \boldsymbol{B} = \begin{bmatrix} 1+0 & 1+1 \ 0+1 & 1+0 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 1 & 1 \end{bmatrix} \]
02

Calculate \((\boldsymbol{A} + \boldsymbol{B})^2\)

To find \((\boldsymbol{A} + \boldsymbol{B})^2\), multiply \(\boldsymbol{A} + \boldsymbol{B}\) by itself:\[ \begin{bmatrix} 1 & 2 \ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1\cdot1 + 2\cdot1 & 1\cdot2 + 2\cdot1 \ 1\cdot1 + 1\cdot1 & 1\cdot2 + 1\cdot1 \end{bmatrix} = \begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix} \]
03

Determine \( \boldsymbol{A}^2 \)

Compute \(\boldsymbol{A}^2\) by multiplying \(\boldsymbol{A}\) by itself:\[ \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1\cdot1 + 1\cdot0 & 1\cdot1 + 1\cdot1 \ 0\cdot1 + 1\cdot0 & 0\cdot1 + 1\cdot1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \]
04

Compute \( 2\boldsymbol{A}\boldsymbol{B} \)

First find \(\boldsymbol{A}\boldsymbol{B}\), and then multiply by 2:\[ \boldsymbol{A}\boldsymbol{B} = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1\cdot0 + 1\cdot1 & 1\cdot1 + 1\cdot0 \ 0\cdot0 + 1\cdot1 & 0\cdot1 + 1\cdot0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix} \]Then,\[ 2\boldsymbol{A}\boldsymbol{B} = 2 \times \begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 2 \ 2 & 0 \end{bmatrix} \]
05

Compute \( \boldsymbol{B}^2 \)

Compute \(\boldsymbol{B}^2\) by multiplying \(\boldsymbol{B}\) by itself:\[ \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0\cdot0 + 1\cdot1 & 0\cdot1 + 1\cdot0 \ 1\cdot0 + 0\cdot1 & 1\cdot1 + 0\cdot0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]
06

Evaluate \( \boldsymbol{A}^2 + 2\boldsymbol{A}\boldsymbol{B} + \boldsymbol{B}^2 \)

Add the results from Steps 3, 4, and 5:\[ \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 2 \ 2 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1+2+1 & 2+2+0 \ 0+2+0 & 1+0+1 \end{bmatrix} = \begin{bmatrix} 4 & 4 \ 2 & 2 \end{bmatrix} \]
07

Evaluate \( (\boldsymbol{A} + \boldsymbol{B})(\boldsymbol{A} - \boldsymbol{B}) \)

Compute \(\boldsymbol{A} - \boldsymbol{B}\):\[ \boldsymbol{A} - \boldsymbol{B} = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ -1 & 1 \end{bmatrix} \]Now multiply:\[ \begin{bmatrix} 1 & 2 \ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1\cdot1 + 2\cdot-1 & 1\cdot0 + 2\cdot1 \ 1\cdot1 + 1\cdot-1 & 1\cdot0 + 1\cdot1 \end{bmatrix} = \begin{bmatrix} -1 & 2 \ 0 & 1 \end{bmatrix} \]
08

Evaluate \( \boldsymbol{A}^2 - \boldsymbol{B}^2 \)

Subtract the result of \(\boldsymbol{B}^2\) from \(\boldsymbol{A}^2\):\[ \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1-1 & 2-0 \ 0-0 & 1-1 \end{bmatrix} = \begin{bmatrix} 0 & 2 \ 0 & 0 \end{bmatrix} \]
09

Evaluate with the second set of matrices \( \boldsymbol{A}' \) and \( \boldsymbol{B}' \)

For matrices \(\boldsymbol{A}' = \begin{bmatrix} 1 & 2 \ 5 & 2 \end{bmatrix}\) and \(\boldsymbol{B}' = \begin{bmatrix} 2 & -2 \ -5 & 1 \end{bmatrix}\), repeat Steps 1-8:* Compute \(\boldsymbol{A}' + \boldsymbol{B}'\), \((\boldsymbol{A}' + \boldsymbol{B}')^2\), \(\boldsymbol{A}'^2\), \(2\boldsymbol{A}'\boldsymbol{B}'\), and \(\boldsymbol{B}'^2\).* Compare \((\boldsymbol{A}' + \boldsymbol{B}')^2\) with \(\boldsymbol{A}'^2 + 2\boldsymbol{A}'\boldsymbol{B}' + \boldsymbol{B}'^2\).* Compute \((\boldsymbol{A}' + \boldsymbol{B}')(\boldsymbol{A}' - \boldsymbol{B}')\) and \(\boldsymbol{A}'^2 - \boldsymbol{B}'^2\).Note the non-commuting property of matrix multiplication that leads to different results.

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Key Concepts

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

Matrix Addition
Matrix addition is a fundamental operation where you combine two matrices by adding their corresponding elements. To successfully add matrices, they must be of the same dimensions. For example, if you have matrices \( \boldsymbol{A} \) and \( \boldsymbol{B} \) both with the dimension \(2\times2\), matrix addition involves:
  • Taking the element from row 1, column 1 of \( \boldsymbol{A} \) and adding it to the element from row 1, column 1 of \( \boldsymbol{B} \),
  • Doing the same for each corresponding position in the two matrices.
In our exercise, after performing the addition of matrices \( \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \) and \( \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), we get \( \begin{bmatrix} 1 & 2 \ 1 & 1 \end{bmatrix} \). Remember, matrix addition is commutative, which means \( \boldsymbol{A} + \boldsymbol{B} = \boldsymbol{B} + \boldsymbol{A} \).
Matrix Multiplication
Matrix multiplication is a bit more complex than addition. To multiply two matrices, the number of columns in the first matrix must match the number of rows in the second matrix. The result will have dimensions equivalent to the number of rows in the first matrix and the number of columns in the second matrix. For our matrices \( \boldsymbol{A} \) and \( \boldsymbol{B} \), the multiplication works by:
  • Taking the dot product of the rows of \( \boldsymbol{A} \) with the columns of \( \boldsymbol{B} \).
  • This involves matching corresponding elements from each row and column, multiplying them, and summing these products for each position in the resulting matrix.
With our given matrices, when multiplying \( \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \) and \( \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), the resulting matrix is \( \begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix} \). This operation is not commutative; generally, \( \boldsymbol{A}\boldsymbol{B} eq \boldsymbol{B}\boldsymbol{A} \).
Matrix Powers
Matrix powers refer to multiplying a matrix by itself one or more times, much like raising numbers to a power. When you square a matrix \( \boldsymbol{A} \), you multiply it by itself: \( \boldsymbol{A}^{2} = \boldsymbol{A} \times \boldsymbol{A} \). For the matrix \( \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \), squaring it involves taking the dot product of rows and columns within the matrix, resulting in \( \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \). Matrix powers can reveal insights into the properties of matrices, like stability and transformations in systems represented by the matrices. Interestingly, due to matrix multiplication's nature, the ordering in power computations must remain consistent.
Non-Commutative Property of Matrices
Perhaps one of the most intriguing aspects of matrices is their non-commutative property in multiplication. This means that for matrices \( \boldsymbol{A} \) and \( \boldsymbol{B} \), generally, \( \boldsymbol{A}\boldsymbol{B} eq \boldsymbol{B}\boldsymbol{A} \). This property is crucial, especially in fields such as quantum mechanics or 3D graphics, where the order of transformations (represented by matrices) affects the outcome. In our exercise, the observed differences in expressions, like \((\boldsymbol{A} + \boldsymbol{B})^2\) versus \(\boldsymbol{A}^2 + 2 \boldsymbol{A} \boldsymbol{B} + \boldsymbol{B}^2\), highlight this non-commutative nature. Even if some results align, the paths to these results can vary widely when order changes, showcasing the importance of the matrix order in computations.

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Most popular questions from this chapter

Solve the matrix equation \(\boldsymbol{A} \boldsymbol{X}=\boldsymbol{b}\) for the vector \(\boldsymbol{X}\) in the following: (a) \(\boldsymbol{A}=\left[\begin{array}{lr}2 & 3 \\ 5 & -2\end{array}\right] \quad \boldsymbol{b}=\left[\begin{array}{l}8 \\ 1\end{array}\right]\) (b) \(\boldsymbol{A}=\left[\begin{array}{rrr}1 & 0 & 0 \\ 2 & -1 & 0 \\ 2 & 2 & 2\end{array}\right] \quad \boldsymbol{b}=\left[\begin{array}{r}1 \\ 6 \\\ -6\end{array}\right]\) (c) \(\boldsymbol{A}=\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right] \quad \boldsymbol{b}=-\left[\begin{array}{l}3 \\ 1\end{array}\right]\) (d) \(\boldsymbol{A}=\left[\begin{array}{llll}1 & 0 & 0 & 0 \\ 0 & 3 & 1 & 0 \\\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1\end{array}\right] \quad \boldsymbol{b}=\left[\begin{array}{c}4 \\ 11 \\ 7 \\ 1\end{array}\right]\)

(a) If \(\boldsymbol{P}=\frac{1}{3}\left[\begin{array}{rrr}2 & 1 & 2 \\ -2 & 2 & 1 \\ -1 & -2 & 2\end{array}\right]\) write down the transpose matrix \(\boldsymbol{P}^{\mathrm{T}}\). Calculate \(\boldsymbol{P P}^{\mathrm{T}}\) and hence show that \(\boldsymbol{P}^{\mathrm{T}}=\boldsymbol{P}^{-1}\). What does this mean about the solution of the matrix equation \(\boldsymbol{P} \boldsymbol{x}=\boldsymbol{b}\) ? (b) The matrix \(\boldsymbol{F}=\left[\begin{array}{ccc}I_{x} & I_{x y} & Q_{x} \\\ I_{x y} & I_{y} & Q_{y} \\ Q_{x} & Q_{y} & A\end{array}\right]\) occurs in the structural analysis of an arch. If $$ \boldsymbol{B}=\left[\begin{array}{ccc} 1 & 0 & -Q_{x} / A \\ 0 & 1 & -Q_{y} / A \\ 0 & 0 & 1 \end{array}\right] $$ find \(\boldsymbol{E}=\boldsymbol{B F B}^{\mathrm{T}}\) and show that it is a symmetric matrix.

A computer screen has dimensions \(20 \mathrm{~cm} \times 30 \mathrm{~cm}\). Axes are set up at the centre of the screen, as illustrated in Figure 5.5. A box containing an arrow, has dimensions \(2 \mathrm{~cm} \times 2 \mathrm{~cm}\) and is situated with its centre at the point \((-16,10)\). It is first to be rotated through \(45^{\circ}\) in an anticlockwise direction. Find this transformation in the form $$ \left[\begin{array}{c} x^{\prime}+16 \\ y^{\prime}-10 \end{array}\right]=\boldsymbol{A}\left[\begin{array}{c} x+16 \\ y-10 \end{array}\right] $$

Find the values of \(x, y, z\) and \(t\) from the equation $$ \left[\begin{array}{cc} x & y-x+t \\ t-z & z-1 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] $$

Given \(\boldsymbol{A}=\left[\begin{array}{rrr}2 & -1 & 0 \\ -4 & 3 & -1 \\ 1 & -1 & 1\end{array}\right]\) and $$ \boldsymbol{B}=\left[\begin{array}{lrl} 1 & 0 & 2 \\ 3 & 4 & 0 \\ 6 & -2 & 1 \end{array}\right] $$ verify that \(\operatorname{adj}(\boldsymbol{A B})=(\operatorname{adj} \boldsymbol{B})(\operatorname{adj} \boldsymbol{A})\).

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