## Linear Algebra and Its Applications, exercise 1.2.8

Exercise 1.2.8. We have the following system of three of equations in three unknowns: $\begin{array}{r} u + v + w = 2 \\ u + 2v + 3w = 1 \\ v + 2w = 0 \end{array}$

Show that the system is singular by showing that there is a combination of the three equations that produces the contradiction 0 = 1. Find a right-hand side for which the system is nonsingular, and find one of its solutions.

Answer: We start by subtracting the second equation from the first in order to eliminate u: $\begin{array}{r} u + v + w = 2 \\ u + 2v + 3w = 1 \\ v + 2w = 0 \end{array} \Rightarrow \begin{array}{r} u + v + w - (u + 2v +3w) = 2 - 1 \\ v + 2w = 0 \end{array} \Rightarrow \begin{array}{r} -v -2w = 1 \\ v + 2w = 0 \end{array}$

We then add the remaining two equations: $\begin{array}{r} -v -2w = 1 \\ v + 2w = 0 \end{array} \Rightarrow -v -2w + v + 2w = 1 + 0 \Rightarrow 0 = 1$

Since we’ve derived the contradiction 0 = 1 there is no solution to the system, i.e., it is singular.

However, if the zero on the right-hand side of the third equation were changed to -1 then there would be no contradiction: $\begin{array}{r} u + v + w = 2 \\ u + 2v + 3w = 1 \\ v + 2w = -1 \end{array}$

(The sequence of operations described above would instead generate the identity 1 = 1.)

In the above system we can discard the second equation as redundant, since it is equal to the sum of the first and third equations. We can solve the remaining system of equations by subtracting the third equation from the first: $\begin{array}{r} u + v + w = 2 \\ v + 2w = -1 \end{array} \Rightarrow \begin{array}{r} u + v + w - (v + 2w) = 2 - (-1) \\ v + 2w = 1 \end{array} \Rightarrow \begin{array}{r} u - w = 3 \\ v + 2w = -1 \end{array}$

We can then solve for u and v in terms of w: $\begin{array}{r} u - w = 3 \\ v + 2w = -1 \end{array} \Rightarrow \begin{array}{l} u = w + 3 \\ v = -2w -1 \end{array}$

For w = 0 we have u = 3 and v = -1, so (3, -1, 0) is a solution to this new system of equations.

UPDATE: Corrected the third equation in the original statement of the problem; thanks to joohyun for finding the error. Also corrected another error I found myself at the same time.

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 1.2.8

1. joohyun says:

hi thanks for making these post for learners. i appreciate it. by the way i found the error on this page. on top of the page,
u+v+w=2
u+2v+3w=1
u+ 2w=0
is actually
u+v+w=2
u+2v+3w=1
v+2w=0

• hecker says:

I’m glad you find these posts useful. Thank you for finding the error; I have corrected it.