## Linear Algebra and Its Applications, Exercise 2.2.1

Exercise 2.2.1. Consider the set of 2 by 3 echelon matrices. How many patterns are there for such matrices?

Answer: Suppose the first row of a 2 by 3 echelon matrix has a pivot in column 1. The second row could then have a pivot in column 2 or in column 3, or could have no pivot at all. This produces three possible patterns: $\begin{bmatrix} *&*&* \\ 0&*&* \end{bmatrix} \qquad \begin{bmatrix} *&*&* \\ 0&0&* \end{bmatrix} \qquad \begin{bmatrix} *&*&* \\ 0&0&0 \end{bmatrix}$

If the first row has no pivot in column 1 but a pivot in column 2, then the second row could have a pivot in column 3 or no pivots at all: $\begin{bmatrix} 0&*&* \\ 0&0&* \end{bmatrix} \qquad \begin{bmatrix} 0&*&* \\ 0&0&0 \end{bmatrix}$

If the pivot for the first row is in column 3, then the second row can have no pivots: $\begin{bmatrix} 0&0&* \\ 0&0&0 \end{bmatrix}$

Finally, there could be no pivots in the first row, in which case the second row would not have pivots either; in other words, this corresponds to the zero matrix: $\begin{bmatrix} 0&0&0 \\ 0&0&0 \end{bmatrix}$

There are thus seven possible patterns in all.

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 .

This entry was posted in linear algebra. Bookmark the permalink.