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Chapter 5: Problem 4

If $$ \boldsymbol{A}=\left[\begin{array}{rrr} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right], \quad \boldsymbol{B}=\left[\begin{array}{lll} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 5 \end{array}\right] $$ and \(\quad \boldsymbol{C}=\left[\begin{array}{rrr}4 & 0 & -2 \\ 5 & 3 & 1 \\\ 2 & 5 & 4\end{array}\right]\) (a) show that $$ \operatorname{trace}(\boldsymbol{A}+\boldsymbol{B})=\operatorname{trace} \boldsymbol{A}+\operatorname{trace} \boldsymbol{B} $$ (b) find \(\boldsymbol{D}\) so that \(\boldsymbol{A}+\boldsymbol{D}=\boldsymbol{C}\); (c) verify the associative law $$ (A+B)+C=A+(B+C) $$

Short Answer

Expert verified
(a) Verified. (b) \(\boldsymbol{D} = \begin{bmatrix}3 & -2 & 1\\0 & 3 & -1\\1 & 6 & 3\end{bmatrix}\). (c) Verified.

Step by step solution

01

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

To find \(\boldsymbol{A} + \boldsymbol{B}\), add corresponding elements of matrices \(\boldsymbol{A}\) and \(\boldsymbol{B}\).\[\begin{array}{ccc}1+3 & 2-1 & -3+2 \5+4 & 0+2 & 2+5 \1+2 & -1+0 & 1+5 \\end{array} =\begin{array}{ccc}4 & 1 & -1 \9 & 2 & 7 \3 & -1 & 6 \\end{array}\]
02

Compute the Trace of \(\boldsymbol{A} + \boldsymbol{B}\)

Compute the trace by adding the diagonal elements of \(\boldsymbol{A} + \boldsymbol{B}\).\[\operatorname{trace}(\boldsymbol{A} + \boldsymbol{B}) = 4 + 2 + 6 = 12\]
03

Compute the Trace of \(\boldsymbol{A}\) and \(\boldsymbol{B}\) Separately

Find the trace of \(\boldsymbol{A}\) by adding its diagonal elements: \[\operatorname{trace}(\boldsymbol{A}) = 1 + 0 + 1 = 2\]Then find the trace of \(\boldsymbol{B}\):\[\operatorname{trace}(\boldsymbol{B}) = 3 + 2 + 5 = 10\]
04

Verify the Trace Equality

Check if \[\operatorname{trace}(\boldsymbol{A} + \boldsymbol{B}) = \operatorname{trace}(\boldsymbol{A}) + \operatorname{trace}(\boldsymbol{B})\]\[12 = 2 + 10 = 12\] which holds true.
05

Solve for \(\boldsymbol{D}\) in \(\boldsymbol{A} + \boldsymbol{D} = \boldsymbol{C}\)

Rearrange the equation to find \(\boldsymbol{D}\): \(\boldsymbol{D} = \boldsymbol{C} - \boldsymbol{A}\).Subtract corresponding elements of \(\boldsymbol{C}\) and \(\boldsymbol{A}\):\[\begin{array}{ccc}4-1 & 0-2 & -2-(-3) \5-5 & 3-0 & 1-2 \2-1 & 5-(-1) & 4-1 \\end{array} =\begin{array}{ccc}3 & -2 & 1 \0 & 3 & -1 \1 & 6 & 3 \\end{array}\]
06

Establish \((\boldsymbol{A} + \boldsymbol{B}) + \boldsymbol{C}\)

Calculate\(\boldsymbol{A} + \boldsymbol{B}\) using Step 1:\[\begin{array}{ccc}4 & 1 & -1 \9 & 2 & 7 \3 & -1 & 6 \\end{array}\]Then add \(\boldsymbol{C}\):\[\begin{array}{ccc}4+4 & 1+0 & -1+(-2) \9+5 & 2+3 & 7+1 \3+2 & -1+5 & 6+4 \\end{array} =\begin{array}{ccc}8 & 1 & -3 \14 & 5 & 8 \5 & 4 & 10 \\end{array}\]
07

Establish \(\boldsymbol{A} + (\boldsymbol{B} + \boldsymbol{C})\)

First compute \(\boldsymbol{B} + \boldsymbol{C}\) using matrices \(\boldsymbol{B}\) and \(\boldsymbol{C}\):\[\begin{array}{ccc}3+4 & -1+0 & 2+(-2) \4+5 & 2+3 & 5+1 \2+2 & 0+5 & 5+4 \\end{array} =\begin{array}{ccc}7 & -1 & 0 \9 & 5 & 6 \4 & 5 & 9 \\end{array}\]Then add \(\boldsymbol{A}\):\[\begin{array}{ccc}1+7 & 2+(-1) & -3+0 \5+9 & 0+5 & 2+6 \1+4 & -1+5 & 1+9 \\end{array} =\begin{array}{ccc}8 & 1 & -3 \14 & 5 & 8 \5 & 4 & 10 \\end{array}\]
08

Confirm Associative Property Equality

Compare results from Steps 6 and 7. Both are equal:\[\begin{array}{ccc}8 & 1 & -3 \14 & 5 & 8 \5 & 4 & 10 \\end{array}\]Thus, \((\boldsymbol{A} + \boldsymbol{B}) + \boldsymbol{C} = \boldsymbol{A} + (\boldsymbol{B} + \boldsymbol{C})\).

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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 concept in matrix algebra where two matrices of the same dimensions are added together by summing their corresponding elements. For example, in the matrices \( \boldsymbol{A} \) and \( \boldsymbol{B} \) given in the problem, each element from \( \boldsymbol{A} \) is added to the corresponding element of \( \boldsymbol{B} \). The process looks like this:
  • First row: \( (1 + 3), (2 + (-1)), (-3 + 2) \)
  • Second row: \( (5 + 4), (0 + 2), (2 + 5) \)
  • Third row: \( (1 + 2), ((-1) + 0), (1 + 5) \)
This results in a new matrix:\[ \begin{bmatrix} 4 & 1 & -1 \ 9 & 2 & 7 \ 3 & -1 & 6 \end{bmatrix} \]
Understanding matrix addition is crucial for solving more complex matrix equations and applying the associative property in matrix operations.
Matrix Trace
The trace of a matrix is a useful concept that involves summing all the diagonal elements of a square matrix. This operation is denoted as \( \operatorname{trace}(\boldsymbol{M}) \). For the matrices \( \boldsymbol{A} \) and \( \boldsymbol{B} \), the trace is calculated as follows:
  • For \( \boldsymbol{A} \), the diagonal elements are 1, 0, and 1, so its trace is \( 1 + 0 + 1 = 2 \).
  • For \( \boldsymbol{B} \), the diagonal elements are 3, 2, and 5, giving a trace of \( 3 + 2 + 5 = 10 \).
  • The trace of \( \boldsymbol{A} + \boldsymbol{B} \) follows the same rule, calculated as \( 4 + 2 + 6 = 12 \).
Thus, we observe the property that \( \operatorname{trace}(\boldsymbol{A} + \boldsymbol{B}) = \operatorname{trace}(\boldsymbol{A}) + \operatorname{trace}(\boldsymbol{B}) \), highlighting an essential rule of matrix algebra.
Associative Property
In matrix algebra, the associative property is a fundamental property that states, for any three matrices \( \boldsymbol{A} \), \( \boldsymbol{B} \), and \( \boldsymbol{C} \), the equation \( (\boldsymbol{A} + \boldsymbol{B}) + \boldsymbol{C} = \boldsymbol{A} + (\boldsymbol{B} + \boldsymbol{C}) \) holds. This property implies that how matrices are grouped during addition does not affect the result.
To verify this property for matrices \( \boldsymbol{A} \), \( \boldsymbol{B} \), and \( \boldsymbol{C} \) in the problem, we calculate:
  • Step 1: Compute \( (\boldsymbol{A} + \boldsymbol{B}) + \boldsymbol{C} \).
  • Step 2: Compute \( \boldsymbol{A} + (\boldsymbol{B} + \boldsymbol{C}) \).
  • Confirm that both results are equal, which they indeed are: \( \begin{bmatrix} 8 & 1 & -3 \ 14 & 5 & 8 \ 5 & 4 & 10 \end{bmatrix} \).
This establishes that the associative property holds true for these matrices, reaffirming its nature as a fundamental rule in matrix operations.
Matrix Equations
Finding unknown matrices in equations is a common task in matrix algebra. To solve for \( \boldsymbol{D} \) in the equation \( \boldsymbol{A} + \boldsymbol{D} = \boldsymbol{C} \), you rearrange the equation to isolate \( \boldsymbol{D} \), which yields \( \boldsymbol{D} = \boldsymbol{C} - \boldsymbol{A} \).
This subtraction is element-wise, meaning you subtract each element of \( \boldsymbol{A} \) from the corresponding element of \( \boldsymbol{C} \):
  • First row: \( (4 - 1), (0 - 2), ((-2) - (-3)) \)
  • Second row: \( (5 - 5), (3 - 0), (1 - 2) \)
  • Third row: \( (2 - 1), (5 - (-1)), (4 - 1) \)
The result is:\[ \begin{bmatrix} 3 & -2 & 1 \ 0 & 3 & -1 \ 1 & 6 & 3 \end{bmatrix} \]
Understanding how to manipulate such matrix equations is essential for solving many real-world problems that use matrices.

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Most popular questions from this chapter

Find the inverse of the matrix $$ \boldsymbol{A}=\left[\begin{array}{rrr} -1 & 2 & 1 \\ 0 & 1 & -2 \\ 1 & 4 & -1 \end{array}\right] $$ and hence solve the equations $$ \begin{array}{r} -x+2 y+z=2 \\ y-2 z=-3 \\ x+4 y-z=4 \end{array} $$

Given the matrix $$ \mathbf{A}=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 1 & 0 & 1 \\ 2 & 2 & 2 \end{array}\right] $$ determine \(|\boldsymbol{A}|,\left|\boldsymbol{A} \boldsymbol{A}^{\mathrm{T}}\right|,\left|\boldsymbol{A}^{2}\right|\) and \(|\boldsymbol{A}+\boldsymbol{A}|\)

Solve, using Gaussian elimination with partial pivoting, the following equations: (a) \(\left[\begin{array}{lll}1.17 & 2.64 & 7.41 \\ 3.37 & 1.22 & 9.64 \\\ 4.10 & 2.89 & 3.37\end{array}\right]\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=\left[\begin{array}{c}1.27 \\ 3.91 \\\ 4.63\end{array}\right]\) (b) \(\left[\begin{array}{rrr}3.21 & 4.18 & -2.31 \\ -4.17 & 3.63 & 4.20 \\\ 1.88 & -8.14 & 0.01\end{array}\right]\left[\begin{array}{c}x \\ y \\\ z\end{array}\right]=\left[\begin{array}{c}3.27 \\ -1.21 \\\ 4.88\end{array}\right]\) (c) \(\left[\begin{array}{rrrr}1 & 7 & 2 & -1 \\ 11 & 4 & -3 & 9 \\ 7 & 6 & 4 & -2 \\ 5 & 8 & -5 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z \\\ t\end{array}\right]=\left[\begin{array}{r}12 \\ -12 \\ 7 \\\ -7\end{array}\right]\)

Show that the matrix $$ \boldsymbol{B}=\left[\begin{array}{lll} 1 & 0 & 2 \\ 3 & 4 & 0 \\ 6 & -2 & 1 \end{array}\right] $$ is non-singular and verify Cauchy's theorem, namely \(|\operatorname{adj} \boldsymbol{B}|=|\boldsymbol{B}|^{2}\).

A popular method of numerical integration - see the work in Chapter 8 - involves Gaussian integration; it is used in finite-element calculations which are well used in most of engineering. As a simple example, the numerical integral over the interval \(-1 \leqslant x \leqslant 1\) is written $$ \int_{-1}^{1} f(x) \mathrm{d} x=C_{1} f\left(x_{1}\right)+C_{2} f\left(x_{2}\right) $$ and the formula is made exact for the four functions \(f=1, f=x, f=x^{2}\) and \(f=x^{3}\), so it must be accurate for all cubics. This leads to the four equations $$ \begin{gathered} C_{1}+C_{2}=2 \\ C_{1} x_{1}+C_{2} x_{2}=0 \\ C_{1} x_{1}^{2}+C_{2} x_{2}^{2}=\frac{2}{3} \\ C_{1} x_{1}^{3}+C_{2} x_{2}^{3}=0 \end{gathered} $$ Use Gaussian elimination to reduce the equations and hence deduce that the equations are only consistent if \(x_{1}\) and \(x_{2}\) are chosen at the 'Gauss' points \(\pm \frac{1}{\sqrt{3}}\)
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