Linear Algebra and Its Applications, Exercise 2.2.13

Exercise 2.2.13. What is a 3 by 3 system of equations Ax = b that has the following general solution (the same as in exercise 2.2.12)

x = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + w \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}

and that has no solution if b_1 + b_2 \ne b_3?

Answer: As in exercise 2.2.12, the general solution above is the sum of a particular solution and a homogeneous solution, where

x_{particular} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}

and

x_{homogeneous} = w \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}

Also as in exercise 2.2.12, since w is the only variable referenced in the homogeneous solution it must be the only free variable, with u and v being basic. Since u is basic we must have a pivot in column 1, and since v is basic we must have a second pivot in column 2. After performing elimination on A the resulting echelon matrix U must therefore have the form

U = \begin{bmatrix} *&*&* \\ 0&*&* \\ 0&0&0 \end{bmatrix}

Unlike in exercise 2.2.12 we cannot simply assume that A = U because we need to account for the particular condition b_1 + b_2 = b_3 that is required for the system Ax = b to have a solution. This condition arises from the particular sequence of elimination steps that takes the system Ax = b and produces the equivalent system Ux = c with c resulting from the elimination operations applied to b. In particular, we assume that the system Ux = c will look as follows:

\begin{bmatrix} *&*&* \\ 0&*&* \\ 0&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 - b_1 - b_2 \end{bmatrix}

The third row produces the equation 0 = b_3 - b_1 - b_2 or b_1 + b_2 = b_3. This corresponds to the assumption that there is no solution if b_1 + b_2 \ne b_3.

We can’t assume that A = U but we can assume that the first two rows of A are the same as the first two rows of U. That removes the need to do elimination for the second row of A and corresponds to our assumption that the second element of c is the same as the second element of b (namely b_2). We then have to do elimination operations only for the third row, and those operations will produce the third element of c as listed above.

As stated above the general solution to Ax = b for this exercise is the same as the general solution in exercise 2.2.12, so the homogeneous solution is the same as well. Since we are assuming that the first two rows of A are the same as the first two rows of U we have

Ux_{homogeneous} = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ 0&a_{22}&a_{23} \\ 0&0&0 \end{bmatrix} w \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = 0

Note that the first two rows produce the same equations as in exercise 2.2.12. The third row is simply equivalent to the equation 0 = 0 and so adds no new information. We can therefore follow the same process as in exercise 2.2.12 to find values for the first two rows of U and A:

U = \begin{bmatrix} 1&-1&1 \\ 0&1&-2 \\ 0&0&0 \end{bmatrix} \qquad A = \begin{bmatrix} 1&-1&1 \\ 0&1&-2 \\ *&*&* \end{bmatrix}

so that we have

Ux_{homogeneous} = \begin{bmatrix} 1&-1&1 \\ 0&1&-2 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

We now turn to determining the last row of A:

A = \begin{bmatrix} 1&-1&1 \\ 0&1&-2 \\ a_{31}&a_{32}&a_{33} \end{bmatrix}

In order to produce b_3 - b_1 - b_2 as the last element of c on the right-hand side (i.e., of Ux = c) we must subtract the first row from the third row on the left-hand side to produce a zero as the first element of the third row, and then subsequently subtract the second row from the third row to produce a zero as the second element of the third row. That first operation means we must have a_{31} - 1 = 0 or a_{31} = 1. The second operation means we must have a_{32} - (-1) - 1 = 0 or a_{32} = 0 as well as a_{33} - 1 - (-2) = 0 or a_{33} = -1. The proposed value for A is then

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

We next turn to the general system Ax = b. We now have a value for A, and we were given the value of the particular solution. We can multiply the two to calculate the value of b:

b = Ax_{particular} = \begin{bmatrix} 1&-1&1 \\ 0&1&-2 \\ 1&0&-1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}

This gives us the following as an example 3 by 3 system that has the general solution specified above:

\begin{bmatrix} 1&-1&1 \\ 0&1&-2 \\ 1&0&-1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}

or

\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}u&-&v&+&w&=&0 \\ &&v&-&2w&=&1 \\ u&&&-&w&=&1 \end{array}

Note that we have b_3 = b_1 + b_2. If this were not the case then the system would have no solution.

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 2.2.13

  1. Kostas says:

    I quess that’s not the only 3 by 3 system that has the above general solution,right?

    • hecker says:

      That’s a good question. I did make some assumptions as to the form of A relative to U, and it’s possible that if you change those assumptions then it’s possible to find different values for A and b. However I have not tried to do this myself.

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