Linear Algebra and Its Applications, Review Exercise 2.22

Review exercise 2.22. a) Given

A = \begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \qquad b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}

what conditions must b satisfy in order for Ax = b to have a solution?

b) Find a basis for the nullspace of A.

c) Find the general solution for Ax = b for those cases when a solution exists.

d) Find a basis for the column space of A?

e) What is the rank of A^T?

Answer: a) Given the system

\begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}

from the second row we have

0 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 + 0 \cdot x_4 = b_2

This can be true only if b_2 = 0.

b) The system Ax = 0 corresponds to

\begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

Doing Gaussian elimination on A we subtract 2 times the first row from the third:

\begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 0&0&0&-5 \end{bmatrix}

Since there are pivots in columns 1 and 4 we have x_1 and x_4 as basic variables and x_2 and x_3 as basic variables. (Note that this would be true whether or not we do a row exchange to put the matrix in true echelon form.)

From the third equation we have x_4 = 0. Setting x_2 = 1 and x_3 = 0, from the first equation we have x_1 + 2x_2 + 3x_4 = x_1 + 2 + 0 = 0 or x_1 = -2. Setting x_2 = 0 and x_3 = 1, from the first equation we have x_1 + 2x_2 + 3x_4 = x_1 + 0 + 0 = 0 or x_1 = 0.

The following vectors are thus solutions to Ax = 0 and serve as basis vectors for the nullspace of A:

\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}

c) From (a) above, in order for Ax = b to have a solution we must have b_2 = 0. The system Ax = b then becomes

\begin{bmatrix} 1&2&0&3 \\ 0&0&0&0 \\ 2&4&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} b_1 \\ 0 \\ b_3 \end{bmatrix}

To find a particular solution we set the free variables x_2 = x_3 = 0. From the first equation we have x_1 + 2x_2 + 3x_4 = x_1 + 3x_4 = b_1 and from the third equation we have 2x_1 + 4x_2 + x_4 = 2x_1 + x_4 = b_3. Subtracting twice the first equation from the third equation we obtain -5x_4 = b_3-2b_1 or x_4 = \frac{1}{5}(2b_1-b_3).

Substituting the values of x_4 into the first equation we have

x_1 + 3x_4 = x_1 + \frac{3}{5}(2b_1-b_3)

= x_1 + \frac{6}{5}b_1 - \frac{3}{5}b_3 = b_1

or

x_1 = b_1 -\frac{6}{5}b_1 + \frac{3}{5}b_3

= -\frac{1}{5}b_1 + \frac{3}{5}b_3 = \frac{1}{5}(-b_1+3b_3).

The particular solution to Ax = b is therefore

\frac{1}{5} \begin{bmatrix} -b_1+3b_3 \\ 0 \\ 0 \\ 2b_1-b_3 \end{bmatrix}

Combining the particular solution with the homogeneous solution from (b) above gives the general solution

x = \frac{1}{5} \begin{bmatrix} -b_1+3b_3 \\ 0 \\ 0 \\ 2b_1-b_3 \end{bmatrix} + c_1 \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}

where c_1 and c_2 are arbitrary scalars.

d) From (b) above we have pivots in columns 1 and 4 of the echelon form of A. Columns 1 and 4 of A therefore serve as a basis for the column space of A:

\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} \qquad \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix}

e) The rank of A^T is the same as the rank of A, namely 2.

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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