Linear Algebra and Its Applications, Exercise 2.2.4

Exercise 2.2.4. Consider the system of linear equations represented by the following matrix:

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

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 perform elimination on A by subtracting 2 times the first row from the second row (i.e., using the multiplier l_{21} = 2):

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

This completes elimination, and leaves us with the echelon matrix

U = \begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix}

and the factorization A = LU:

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

We now solve for

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

which is equivalent to

Ux = \begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = 0

The pivot in U is in column 2, so the (only) basic variable is v and the free variables are u, w and y.

From the first row of U we have v + 4w = 0 and thus v = -4w. We then have

x = \begin{bmatrix} u \\ -4w \\ w \\ y \end{bmatrix} = u \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ -4 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}

In considering the inhomogeneous system Ax = b or

\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}

the elimination sequence above would produce the system Ux = c or

\begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}

From the second equation we must have b_2 - 2b_1 = 0 or b_2 = 2b_1. Thus for Ax = b to have a solution the vector b must lie on the line passing through the origin and the point (1, 2) so that b = (b_1, 2b_1).

Taking b = (b_1, 2b_2) produces the system

\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ 2b_1 \end{bmatrix}

which after elimination becomes

\begin{bmatrix} 0&1&4&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ 0 \end{bmatrix}

The first equation gives v + 4w = b_1 or v = b_1 - 4w. Setting the free variables u, w, and y all to zero produces the particular solution x = (0, b_1, 0, 0) to the system

\begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ 2b_1 \end{bmatrix}

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

x = \begin{bmatrix} 0 \\ b_1 \\ 0 \\ 0 \end{bmatrix} + u \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ -4 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}

We can check this solution as follows:

Ax = \begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \left( \begin{bmatrix} 0 \\ b_1 \\ 0 \\ 0 \end{bmatrix} + u \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ -4 \\ 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right)

= \begin{bmatrix} 0&1&4&0 \\ 0&2&8&0 \end{bmatrix} \begin{bmatrix} u \\ b_1 - 4w \\ w \\ y \end{bmatrix} = \begin{bmatrix} (b_1-4w) + 4w \\ 2(b_1-4w) + 8w \end{bmatrix}

= \begin{bmatrix} b_1 - 4w + 4w \\ 2b_1 - 8w + 8w \end{bmatrix} = \begin{bmatrix} b_1 \\ 2b_1 \end{bmatrix}

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

UPDATE: I expanded the answer to conform to the presentation in the answer to exercise 2.2.5.

UPDATE 2: I corrected the value for U used in various places. Thanks go to Zoi for catching this error!

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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4 Responses to Linear Algebra and Its Applications, Exercise 2.2.4

  1. Zoi says:

    Hi! You have a mistake – some times you add an extra 1 in U.

    Thanks for all your work!

    • hecker says:

      Thank you for finding this error; I have updated the post to correct it. And thanks for reading the blog! I’m glad you find it useful.

  2. Jeff Berhow says:

    I came across this blog post because when checking my solution in the back of the Third Edition, there were discrepancies. Thank you so much for this post. I was afraid I had the concepts wrong.

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