## 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

1. mungaramanoj says:

could you please give how to get particular solutions

• hecker says:

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.