## Linear Algebra and Its Applications, Exercise 2.2.8

Exercise 2.2.8. Consider the following system of linear equations: $\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ 2u&+&3v&-&w&=&5 \\ 3u&+&4v&+&w&=&c \end{array}$

For what value of $c$ does this system have a solution?

Answer: We use elimination to attempt to solve the system, starting by multiplying the first equation by 2 and subtracting it from the second, and multiplying the first equation by 3 and subtracting it from the third: $\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ 2u&+&3v&-&w&=&5 \\ 3u&+&4v&+&w&=&c \end{array} \Rightarrow \begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ &&v&-&5w&=&1 \\ &&v&-&5w&=&c-6 \end{array}$

and then substract 1 times the second equation from the third: $\begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ &&v&-&5w&=&1 \\ &&v&-&5w&=&c-6 \end{array} \Rightarrow \begin{array}{rcrcrcl}u&+&v&+&2w&=&2 \\ &&v&-&5w&=&1 \\ &&&&0&=&c-7 \end{array}$

In order for this system to have a solution we must have $c - 7 = 0$ or $c = 7$.

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.