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.

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