Linear Algebra and Its Applications, Exercise 2.2.9

Exercise 2.2.9. Consider the following A and b:

A = \begin{bmatrix} 1&2&0&3 \\ 2&4&0&7 \end{bmatrix} \qquad b = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}

For what values of b_1 and b_2 does the system Ax = b have a solution? Also, determine the nullspace of A and provide two examples of vectors within it, and find the general solution to Ax = b.

Answer: We perform elimination by subtracting 2 times the first row from the third row:

\begin{bmatrix} 1&2&0&3 \\ 2&4&0&7 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}

This completes elimination and produces the system Ux = c:

\begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \end{bmatrix}

Since the pivots of U are in columns 1 and 4, the basic variables are u and y and the free variables are v and w. There are no constraints on the values of b_1 and b_2.

To find the general solution to Ax = b we start with the system Ax = 0, which is equivalent to Ux = 0:

\begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

From the second equation we must have y = 0. Substituting the value of y into the first equation we have u + 2v = 0 or u = -2v. The general solution to Ax = 0 can then be expressed in terms of the free variables v and w as follows:

x_{homogeneous} = v \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}

By setting v = 1 and w = 0 we obtain one vector (-2, 1, 0, 0) in the nullspace of A and by setting v = 0 and w = 1 we obtain another vector (0, 0, 1, 0) also in the nullspace.

We can obtain a particular solution to Ax = b by going back to the system Ux = c derived above:

\begin{bmatrix} 1&2&0&3 \\ 0&0&0&1 \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 have y = b_2 - 2b_1. Substituting for y into the first equation we have u + 2v + 3(b_2 - 2b_1) = u + 2v +3b_2 - 6b_1 = b_1 or u = -2v + 7b_1 - 3b_2. Setting the free variables v and w to zero gives the particular solution

x_{particular} = \begin{bmatrix} 7b_1 - 3b_2 \\ 0 \\ 0 \\ -2b_1 + b_2 \end{bmatrix}

The general solution to Ax = b is then

x = x_{particular} + x_{homogeneous} = \begin{bmatrix} 7b_1 - 3b_2 \\ 0 \\ 0 \\ -2b_1 + b_2 \end{bmatrix} + v \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}

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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