Exercise 2.2.5. Consider the system of linear equations represented by the following matrix:
(This is the transpose of the matrix from exercise 2.2.4.) Find the echelon matrix , a set of basic variables, a set of free variables, and the general solution to . Then use elimination to find when the system has a solution, and express that solution as the sum of a particular solution and the general solution to . Finally, find the rank of .
Answer: We must do a row exchange to exchange the first and second rows. This is equivalent to multiplying by a permutation matrix as follows:
We then perform elimination on by subtracting 4 times the first row from the third row (i.e., using the multiplier ):
This completes elimination, and leaves us with the echelon matrix
and the factorization :
We now solve for
which is equivalent to
The pivot in is in column 1, so the (only) basic variable is and the free variable is .
From the first row of we have and thus . We then have
In considering the inhomogeneous system or
the elimination sequence above (including the initial row exchange) would produce the system or
From the second and fourth equations we must have . From the third equation we must have or . Thus for to have a solution the vector must have the form .
Taking produces the system
which after row exchange and elimination becomes
The first equation gives or . Setting the free variable to zero produces the particular solution to the system
We can combine the particular solution to this system with the solution to to produce the general solution for the system
We can check this solution as follows:
The rank of is 1, the number of basic variables (or pivots).
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.