Linear Algebra and Its Applications, Exercise 2.2.7

Exercise 2.2.7. Consider the following system of linear equations:

$\begin{bmatrix} 1&0 \\ 0&1 \\ 2&3 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$

Find the values of $b$ for which the system has a solution. What is the rank? How many basic variables and free variables are there?

Answer: We perform elimination by first subtracting 2 times the first row from the third row:

$\begin{bmatrix} 1&0 \\ 0&1 \\ 2&3 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0 \\ 0&1 \\ 0&3 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 - 2b_1 \end{bmatrix}$

and then subtracting 3 times the second row from the third:

$\begin{bmatrix} 1&0 \\ 0&1 \\ 0&3 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 - 2b_1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0 \\ 0&1 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 - 2b_1 -3b_2 \end{bmatrix}$

This completes elimination and produces the system $Ux = c$ .

From the third equation we must have $b_3 -2b_1 -3b_2 = 0$ in order for the system to have a solution. The matrix $U$ has pivots in columns 1 and 2, so that the basic variables are $u$ and $v$ ; there are no free variables. The rank is 2 (the number of basic variables).

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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2 Responses to Linear Algebra and Its Applications, Exercise 2.2.7

mungaramanoj says:

August 9, 2012 at 4:56 pm

could you please give how to get particular solutions

- hecker says:
  
  August 12, 2012 at 4:51 pm
  
  The previous exercise, 2.2.6, shows an example of finding a particular solution to a system of linear equations. If you search for the word “particular” in the search box on this blog (in the upper right-hand corner) you may find more examples.