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### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Question:A linear system of the form ${\mathbit{A}}\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{=}}{\mathbf{0}}$ is called homogeneous. Justify the following facts: a. All homogeneous systems are consistent.b. A homogeneous system with fewer equations than unknowns has infinitely many solutions.c. If $\stackrel{\mathbf{\to }}{{\mathbf{x}}_{\mathbf{1}}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}\stackrel{\mathbf{\to }}{{\mathbf{x}}_{\mathbf{2}}}$ are solutions of the homogeneous system ${\mathbit{A}}\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{=}}{\mathbf{0}}$, then $\stackrel{\mathbf{\to }}{{\mathbf{x}}_{\mathbf{1}}}{\mathbf{+}}\stackrel{\mathbf{\to }}{{\mathbf{x}}_{\mathbf{2}}}$ is a solution as well.d. If $\stackrel{\mathbf{\to }}{\mathbf{x}}$ is a solution of the homogeneous system ${\mathbit{A}}\stackrel{\mathbf{\to }}{\mathbf{x}}{\mathbf{=}}{\mathbf{0}}$ and kis an arbitrary constant, then $k\stackrel{\to }{x}$ is a solution as well.

a. $\stackrel{\to }{x}=0$ is a solution for a linear system of the form and hence all the homogeneous systems are consistent.

b. Using theorem 1.3.3., homogeneous system with lesser equations than unknowns have infinitely many solutions.

c. Yes, $\stackrel{\to }{{\mathrm{x}}_{1}}+\stackrel{\to }{{\mathrm{x}}_{2}}$ is also a solution, if $\stackrel{\to }{{\mathrm{x}}_{1}}and\stackrel{\to }{{\mathrm{x}}_{2}}$ are solutions of the homogeneous system $A\stackrel{\to }{x}=0$.

d. Yes, $k\stackrel{\to }{x}$ is also a solution, if $\stackrel{\to }{x}$ is a solution of the homogeneous system $A\stackrel{\to }{x}=0$.

See the step by step solution

## Step 1: (a) Consider the system.

If A is an $n×m$matrix with row vectors $\stackrel{\to }{\omega 1},...,\stackrel{\to }{\omega n}$ and $\stackrel{\to }{x}$ is a vector in ${\mathrm{ℝ}}^{m}$ then,

$\stackrel{\to }{Ax}=\left[\begin{array}{c}-{\stackrel{\to }{\omega }}_{1}-\\ ⋮\\ {\stackrel{\to }{\omega }}_{n}-\end{array}\right]\stackrel{\to }{x}=\left[\begin{array}{c}-{\stackrel{\to }{\omega }}_{1}\cdot \stackrel{\to }{x}-\\ ⋮\\ -{\stackrel{\to }{\omega }}_{n}\cdot \stackrel{\to }{x}-\end{array}\right]$

Consider a linear system,

$A\stackrel{\to }{x}=\stackrel{\to }{b}$

## Step 2: Determine the solution for the system.

When $\stackrel{\to }{b}=0,A\stackrel{\to }{x}=0$

Thus, $\stackrel{\to }{x}=0$ will also be a solution for homogeneous system, $A\stackrel{\to }{x}=0$.

## Step 3: (b) Consider a theorem

Theorem 1.3.3

a. For a linear system with at least one solution, then, the number of unknowns must be equal to the number of equations.

b. A linear system will have no solution or infinitely many solutions for number of equations less than the number of unknowns.

Considering the theorem 1.3.3, it can be stated that, a homogeneous system with fewer equations than unknowns has infinitely many solutions.

## Step 4: (c) Consider the system.

Given:

$A\stackrel{\to }{x}=0$ is a homogeneous linear system.

$\stackrel{\to }{{x}_{1}}$and $\stackrel{\to }{{x}_{2}}$are homogeneous solutions.

Consider the system,

$\begin{array}{r}A\stackrel{\to }{x}=0\\ ⇒A\left({\stackrel{\to }{x}}_{1}+\stackrel{\to }{{x}_{2}}\right)=0\\ ⇒A{\stackrel{\to }{x}}_{1}+A{\stackrel{\to }{x}}_{2}=0\\ \therefore {\stackrel{\to }{x}}_{1}+{\stackrel{\to }{x}}_{2}=0\end{array}$

## Step 5:(d) Consider the system.

Given:

$A\stackrel{\to }{x}=0$ is a homogeneous linear system.

$\stackrel{\to }{x}$ is a homogeneous solution.

k is a arbitrary constant.

Consider the system,

$\begin{array}{r}A\stackrel{\to }{x}=0\\ ⇒A\left(k\stackrel{\to }{x}\right)=0\\ ⇒k\left(\stackrel{\to }{Ax}\right)=0\\ ⇒k\left(0\right)\stackrel{\to }{x}=0\\ \therefore \stackrel{\to }{kx}=0\end{array}$

a. $\stackrel{\to }{x}=0$ is a solution.

b. Using theorem 1.3.3., homogeneous system with lesser equations than unknowns have infinitely many solutions.

c. Yes, $\stackrel{\to }{{x}_{1}}+\stackrel{\to }{{x}_{2}}$ is also a solution, if $\stackrel{\to }{{\mathrm{x}}_{1}}\mathrm{and}\stackrel{\to }{{x}_{2}}$ are solutions of the homogeneous system $A\stackrel{\to }{x}=0$.

d. Yes, $k\stackrel{\to }{x}$ is also a solution, if $\stackrel{\to }{x}$ is a solution of the homogeneous system $A\stackrel{\to }{x}=0$.