Review exercise 2.7. Find the most general solution to the following system of linear equations:
Answer: This system corresponds to the system where
The general solution to this system is a combination of a particular solution to plus the homogeneous solution to . We first solve the homogeneous system.
We start by multiplying the first row of by 1 and subtracting it from the second row:
The resulting matrix is in echelon form with pivots in columns 1 and 2, and as basic variables, and as a free variable. Setting to 1 and solving the homogeneous system, from the second row we have or . From the first row we then have or . So is a solution for the homogeneous system, as is any multiple of that vector.
To find a particular solution we set and go back to the original system. From the second equation we have or . From the first equation we have or . So is a particular solution.
The general solution to the original system is thus all vectors of the form
where can take on any value.
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.