Linear Algebra and Its Applications, Exercise 2.2.5

Exercise 2.2.5. Consider the system of linear equations represented by the following matrix:

$A = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix}$

(This is the transpose of the matrix from exercise 2.2.4.) Find the echelon matrix $U$ , a set of basic variables, a set of free variables, and the general solution to $Ax = 0$ . Then use elimination to find when the system $Ax = b$ has a solution, and express that solution as the sum of a particular solution and the general solution to $Ax = 0$ . Finally, find the rank of $A$ .

Answer: We must do a row exchange to exchange the first and second rows. This is equivalent to multiplying by a permutation matrix $P$ as follows:

$PA = \begin{bmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 1&2 \\ 0&0 \\ 4&8 \\ 0&0 \end{bmatrix}$

We then perform elimination on $PA$ by subtracting 4 times the first row from the third row (i.e., using the multiplier $l_{31} = 4$ ):

$\begin{bmatrix} 1&2 \\ 0&0 \\ 4&8 \\ 0&0 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

This completes elimination, and leaves us with the echelon matrix

$U = \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

and the factorization $PA = LU$ :

$\begin{bmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 4&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

We now solve for

$Ax = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = 0$

which is equivalent to

$Ux = \begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = 0$

The pivot in $U$ is in column 1, so the (only) basic variable is $u$ and the free variable is $v$ .

From the first row of $U$ we have $u + 2v = 0$ and thus $u = -2v$ . We then have

$x = \begin{bmatrix} -2v \\ v \end{bmatrix} = v \begin{bmatrix} -2 \\ 1 \end{bmatrix}$

In considering the inhomogeneous system $Ax = b$ or

$\begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix}$

the elimination sequence above (including the initial row exchange) would produce the system $Ux = c$ or

$\begin{bmatrix} 1&2 \\ 0&0 \\ 0&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} b_2 \\ b_1 \\ b_3 - 4b_2 \\ b_4 \end{bmatrix} \Rightarrow \begin{bmatrix} u+2v \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} b_2 \\ b_1 \\ b_3 - 4b_2 \\ b_4 \end{bmatrix}$

From the second and fourth equations we must have $b_1 = b_4 = 0$ . From the third equation we must have $b_3 - 4b_2 = 0$ or $b_3 = 4b_2$ . Thus for $Ax = b$ to have a solution the vector $b$ must have the form $(0, b_2, 4b_2, 0)$ .

Taking $b = (0, b_2, 4b_2, 0)$ produces the system

$\begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ 4b_2 \\ 0 \end{bmatrix}$

which after row exchange and elimination becomes

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

The first equation gives $u + 2v = b_2$ or $u = b_2 - 2v$ . Setting the free variable $v$ to zero produces the particular solution $x = (u, v) = (b_2, 0)$ to the system

$\begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ 4b_2 \\ 0 \end{bmatrix}$

We can combine the particular solution to this system with the solution to $Ax = 0$ to produce the general solution $x$ for the system $Ax = b$

$x = \begin{bmatrix} b_2 \\ 0 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \end{bmatrix}$

We can check this solution as follows:

$Ax = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \left( \begin{bmatrix} b_2 \\ 0 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \end{bmatrix} \right) = \begin{bmatrix} 0&0 \\ 1&2 \\ 4&8 \\ 0&0 \end{bmatrix} \begin{bmatrix} b_2 - 2v \\ v \end{bmatrix}$

$= \begin{bmatrix} 0 \\ (b_2 - 2v) + 2v \\ 4(b_2 - 2v) + 8v \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 - 2v + 2v \\ 4b_2 - 8v + 8v \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ b_2 \\ 4b_2 \\ 0 \end{bmatrix}$

The rank of $A$ is 1, the number of basic variables (or pivots).

UPDATE: Corrected a typo found by Argiris.

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.

2 Responses to Linear Algebra and Its Applications, Exercise 2.2.5

Argiris says:

September 17, 2017 at 3:30 am

In the last b side 3rd row u need 4b2 and not 4b2 -2

- hecker says:
  
  September 17, 2017 at 1:14 pm
  
  Thank you for finding this error! I have updated the post to correct the problem.