## 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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