Linear Algebra and Its Applications, Exercise 2.2.14

Exercise 2.2.14. Create a 2 by 2 system of equations that has many homogeneous solutions but no particular solution.

Answer: A trivial example of such a system is Ax = b where A = 0 and b \ne 0; for example

\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

The corresponding homogeneous system

\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

has any vector in \mathbf{R}^2 as a solution, but the general system has no particular solution since 0x = 0 for any x.

For a less trivial example we can have both rows of A equal to each other, but have the corresponding elements of b not be equal to each other. For example, consider the system

\begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}

The corresponding homogeneous system

\begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

has infinitely many solutions of the form

x_{homogeneous} = v \begin{bmatrix} -1 \\ 1 \end{bmatrix}

but there is no particular solution to the general system: Elimination produces the following system

Ux = \begin{bmatrix} 1&1 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = c

and the second row produces the contradictory equation 0 = 1.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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