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Chapter 1: Problem 13

Given a homogeneous system of linear equations, if the system is overdetermined, what are the possibilities as to the number of solutions? Explain.

Short Answer

Expert verified
An overdetermined homogeneous system of linear equations can have either no solution or a unique trivial solution (\(x_1 = x_2 = \dots = x_n = 0\)). The outcome depends on whether the equations in the system are consistent with each other or not. If at least one equation is inconsistent, there will be no solution; otherwise, the unique trivial solution exists.

Step by step solution

01

Definition of Homogeneous System

A homogeneous system of linear equations is a system where all the constant terms are equal to zero. In a matrix form, it can be represented as \(Ax = 0\), where A is the coefficient matrix, x is the variable vector, and 0 is the zero vector.
02

Definition of Overdetermined System

An overdetermined system of linear equations is a system in which there are more equations than unknowns (variables). In other words, the system has more restrictions than variables to satisfy them.
03

Possible Number of Solutions

For an overdetermined homogeneous system, there are two possibilities for the number of solutions: no solution or a unique trivial solution. 1. No Solution: This case occurs when at least one equation in the system is inconsistent with others, i.e., there exists no solution that would satisfy all the equations simultaneously. Since the system is homogeneous (all constant terms are zero), any inconsistent equation would need to be of the form \(c_1x_1 + c_2x_2 + \dots + c_nx_n \ne 0\), where \(c_i\) are coefficients and \(x_i\) are variables. 2. Unique Trivial Solution: A trivial solution refers to the solution where all the variables are set to zero, i.e., \(x_1 = x_2 = \dots = x_n = 0\). If all equations in the overdetermined homogeneous system are consistent with each other, the trivial solution is the only solution (since the system has more restrictions than variables to adjust). In conclusion, for an overdetermined homogeneous system, there can only be no solution or a unique trivial solution, depending on the consistency of the equations in the system.

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Most popular questions from this chapter

Let \\[ A=\left(\begin{array}{rr} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{array}\right) \\] Compute \(A^{2}\) and \(A^{3} .\) What will \(A^{n}\) turn out to be?

The augmented matrices that follow are in reduced row echelon form. In each case, find the solution set of the corresponding linear system. (a) \(\left(\begin{array}{rrr|r}1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 3\end{array}\right)\) (b) \(\left(\begin{array}{lll|l}1 & 4 & 0 & 2 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1\end{array}\right)\) (c) \(\left(\begin{array}{rrr|r}1 & -3 & 0 & 2 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0\end{array}\right)\) (d) \(\left[\begin{array}{cccc|c}1 & 2 & 0 & 1 & 5 \\ 0 & 0 & 1 & 3 & 4\end{array}\right]\) (e) \(\left(\begin{array}{cccc|c}1 & 5 & -2 & 0 & 3 \\ 0 & 0 & 0 & 1 & 6 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right)\) (f) \(\left(\begin{array}{lll|r}0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0\end{array}\right)\)

Consider the matrix \\[ A=\left(\begin{array}{lllll} 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \end{array}\right) \\] (a) Draw a graph that has \(A\) as its adjacency matrix. Be sure to label the vertices of the graph. (b) By inspecting the graph, determine the number of walks of length 2 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\) (c) Compute the second row of \(A^{3},\) and use it to determine the number of walks of length 3 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\).

The matrix \\[ A=\left(\begin{array}{ll} 1 & -1 \\ 1 & -1 \end{array}\right) \\] has the property that \(A^{2}=O .\) Is it possible for a nonzero symmetric \(2 \times 2\) matrix to have this property? Prove your answer.

Let \\[ A=\left(\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right) \\] Show that if \(d=a_{11} a_{22}-a_{21} a_{12} \neq 0,\) then \\[ A^{-1}=\frac{1}{d}\left(\begin{array}{rr} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{array}\right) \\]
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