/*! 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 9 Given the matrix $$ \boldsym... [FREE SOLUTION] | 91Ó°ÊÓ

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Given the matrix $$ \boldsymbol{A}=\lambda\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\mu\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]+v\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] $$ (a) find the value of \(\lambda, \mu, \nu\) so that \(\boldsymbol{A}=\left[\begin{array}{ll}0 & -1 \\ 0 & 3\end{array}\right]\) (b) show that no solution is possible if $$ \mathbf{A}=\left[\begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array}\right] $$

Short Answer

Expert verified
(a) \( \lambda = 1, \mu = -1, \nu = 3 \); (b) No solution possible.

Step by step solution

01

Express Matrix A

Matrix \( \boldsymbol{A} \) is given as the sum of three matrices, weighted by \( \lambda, \mu, u \):\[\boldsymbol{A} = \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \mu \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} + u \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}.\]
02

Simplify Expression for A

Substitute and simplify to obtain:\[\begin{aligned}\boldsymbol{A} &= \begin{bmatrix}\lambda & 0 \ 0 & \lambda \end{bmatrix} + \begin{bmatrix}\mu & \mu \ 0 & \mu \end{bmatrix} + \begin{bmatrix} 0 & 0 \ 0 & u \end{bmatrix} \&= \begin{bmatrix}\lambda + \mu & \mu \ 0 & \lambda + \mu + u \end{bmatrix}.\end{aligned}\]
03

Set Up Equations for Part (a)

Equate the expression we have for \( \boldsymbol{A} \) to the given matrix:\[\begin{bmatrix} \lambda + \mu & \mu \ 0 & \lambda + \mu + u \end{bmatrix} = \begin{bmatrix} 0 & -1 \ 0 & 3 \end{bmatrix}\]This provides the system of equations:\[\begin{aligned}\lambda + \mu &= 0, \\mu &= -1, \\lambda + \mu + u &= 3.\end{aligned}\]
04

Solve the System for Part (a)

Substitute \( \mu = -1 \) into the other equations:1. \( \lambda + (-1) = 0 \Rightarrow \lambda = 1 \)2. \( 1 + (-1) + u = 3 \Rightarrow u = 3 \)Thus, \( \lambda = 1, \mu = -1, u = 3 \).
05

Set Up Equations for Part (b)

Equate the expression for \( \boldsymbol{A} \) to the second given matrix:\[\begin{bmatrix} \lambda + \mu & \mu \ 0 & \lambda + \mu + u \end{bmatrix} = \begin{bmatrix} 1 & -1 \ 1 & 0 \end{bmatrix}\]This gives us the system of equations:\[\begin{aligned}\lambda + \mu &= 1, \\mu &= -1, \\lambda + \mu + u &= 0, \0 &= 1.\end{aligned}\]
06

Analyze Inconsistency in Part (b)

The equation \( 0 = 1 \) is a contradiction, indicating that no values of \( \lambda, \mu, u \) can satisfy the equations simultaneously. Thus, no solution is possible for this part.

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Key Concepts

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

Matrix Algebra
Matrix algebra is a fundamental area in mathematics, often used to solve linear equations. In matrix algebra, matrices are structures that allow you to perform algebraic calculations, which can be crucial in various applications.
\[ \mathbf{A} = \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \mu \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} + v \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix} \]
The given problem uses matrix algebra to express a matrix \( \mathbf{A} \) as a combination of three smaller matrices, each multiplied by a scalar \( \lambda \), \( \mu \), or \( v \). This illustrates how matrices can be scalable through scalar multiplication and additive transformations.
Matrix algebra involves operations such as matrix addition, scalar multiplication, and in many complex cases, matrix multiplication. In the exercise, these operations are applied to find the specific values for \( \lambda \), \( \mu \), and \( v \). Understanding these operations helps in dealing with systems of linear equations represented in matrix form.
Linear Equations
Linear equations can be effectively solved using matrices. The given problem translates a set of linear equations into matrix form, showcasing a real-world application of matrix algebra.
A key part of understanding linear equations within matrices is the concept of matrix equality. If two matrices are equal, each corresponding element must be equal, leading to a system of linear equations.
  • The matrix equation \( \mathbf{A} = \begin{bmatrix} 0 & -1 \ 0 & 3 \end{bmatrix} \) translates to:
    • \( \lambda + \mu = 0 \)
    • \( \mu = -1 \)
    • \( \lambda + \mu + v = 3 \)
This set of linear equations stems from the requirement that the matrices must be identical in every element. Solving these equations involves substitution or elimination methods, yielding the values of \( \lambda \), \( \mu \), and \( v \).
These techniques are essential for solving more complex linear systems where direct solutions are cumbersome or impossible using standard algebraic methods.
Inconsistent System
An inconsistent system of linear equations occurs when no solution satisfies all equations simultaneously. This concept is crucial, especially in linear algebra and matrix studies.
In part (b) of the exercise, the aim was to equate matrix \( \mathbf{A} \) to another given matrix:\[ \mathbf{A} = \begin{bmatrix} 1 & -1 \ 1 & 0 \end{bmatrix} \]
When examining the corresponding equations derived from matrix equality:
  • \( \lambda + \mu = 1 \)
  • \( \mu = -1 \)
  • \( \lambda + \mu + v = 0 \)
  • \( 0 = 1 \)
The last equation \( 0 = 1 \) is a contradiction, revealing that the system is inconsistent. It shows that no possible values for \( \lambda \), \( \mu \), or \( v \) can satisfy all these equations.
This inconsistency indicates that although the system is fully defined, the conditions set by the given matrices cannot be met. Understanding how inconsistency arises helps to identify and interpret systems of equations in various mathematical applications.

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Most popular questions from this chapter

Given the matrices $$ \boldsymbol{a}=\left[\begin{array}{l} 1 \\ 2 \\ 0 \end{array}\right], \quad \boldsymbol{b}=\left[\begin{array}{lll} 0 & 1 & 1 \end{array}\right], $$ $$ \boldsymbol{C}=\left[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 2 & -1 \end{array}\right], \quad \boldsymbol{D}=\left[\begin{array}{cc} 5 & 6 \\ 7 & 8 \\ 9 & 10 \end{array}\right] $$ evaluate, where possible, (a) \(\boldsymbol{a}+\boldsymbol{b}\), (b) \(\boldsymbol{b}^{\mathrm{T}}+\boldsymbol{a}\), (c) \(\boldsymbol{b}+\boldsymbol{C}^{\mathrm{T}},(\mathrm{d}) \boldsymbol{C}+\boldsymbol{D},(\mathrm{e}) \boldsymbol{D}^{\mathrm{T}}+\boldsymbol{C}\).

Find the rank of the coefficient matrix and of the augmented matrix in the matrix equation $$ \left[\begin{array}{cc} 1 & 1-\alpha \\ \alpha & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \alpha^{2} \\ \alpha \end{array}\right] $$ For each value of \(\alpha\), find, where possible, the solution of the equation.

Show that $$ \begin{aligned} &\left|\begin{array}{ccc} x & a & b \\ x^{2} & a^{2} & b^{2} \\ a+b & x+b & x+a \end{array}\right| \\ &=(b-a)(x-a)(x-b)(x+a+b) \end{aligned} $$ Such an exercise can be solved in two lines of code of a symbolic manipulation package such as MAPLE or MATLAB's Symbolic Math Toolbox.

Rearrange the equations $$ \begin{gathered} x_{1}-x_{2}+3 x_{3}=8 \\ 4 x_{1}+x_{2}-x_{3}=3 \\ x_{1}+2 x_{2}+x_{3}=8 \end{gathered} $$ so that they are diagonally dominant to ensure convergence of the Gauss-Seidel method. Write a MATLAB program to obtain the solution of these equations using this method, starting from \((0,0,0)\). Compare your solution with that from a program when the equations are not rearranged. Use SOR, with \(\omega=1.3\), to solve the equations. Is there any improvement?

A builder's yard organizes its stock in the form of a vector Bricks - type A Bricks - type B Bricks - type C Bags of cement Tons of sand The current stock, \(\mathbf{S}\), and the minimum stock, \(\boldsymbol{M}\), required to avoid running out of materials, are given as $$ \mathbf{S}=\left[\begin{array}{r} 45 \\ 23 & 750 \\ 17 & 170 \\ 462 \\ 27 \end{array}\right] \text { and } \quad \boldsymbol{M}=\left[\begin{array}{r} 5000 \\ 4000 \\ 3500 \\ 100 \\ 10 \end{array}\right] $$ The firm has five lorries which take materials from stock for deliveries: Lorry 1 makes three deliveries in the day with the same load each time; Lorry2 makes two deliveries in the day with the same load each time; the other lorries make one delivery. The loads are $$ \begin{aligned} &\boldsymbol{L}_{1}=\left[\begin{array}{c} 5500 \\ 0 \\ 3800 \\ 75 \\ 3 \end{array}\right] \quad L_{2}=\left[\begin{array}{c} 2500 \\ 1500 \\ 0 \\ 40 \\ 2 \end{array}\right] \quad \boldsymbol{L}_{3}=\left[\begin{array}{c} 7500 \\ 2000 \\ 1500 \\ 0 \\ 3 \end{array}\right]\\\ &L_{4}=\left[\begin{array}{c} 0 \\ 4000 \\ 2500 \\ 20 \\ 2 \end{array}\right] \quad L_{5}=\left[\begin{array}{c} 2000 \\ 0 \\ 1500 \\ 15 \\ 0 \end{array}\right] \end{aligned} $$ How much material has gone from stock, what is the current stock position and has any element gone below the minimum?

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