Review exercise 2.22. a) Given

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

b) Find a basis for the nullspace of .

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

d) Find a basis for the column space of ?

e) What is the rank of ?

Answer: a) Given the system

from the second row we have

This can be true only if .

b) The system corresponds to

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

Since there are pivots in columns 1 and 4 we have and as basic variables and and 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 . Setting and , from the first equation we have or . Setting and , from the first equation we have or .

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

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

To find a particular solution we set the free variables . From the first equation we have and from the third equation we have . Subtracting twice the first equation from the third equation we obtain or .

Substituting the values of into the first equation we have

or

.

The particular solution to is therefore

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

where and are arbitrary scalars.

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

e) The rank of is the same as the rank of , 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.