Review exercise 1.19. Solve the following systems of equations using elimination and back substitution:
Answer: We start with the system
and subtract 1 times the first equation from the second and 3 times the first equation from the third to obtain the following system:
We can then subtract 2 times the second equation from the third to produce the following system:
We have reached a contradiction, so this system has no solution.
We next look at the system
We subtract 1 times the first equation from the second and 3 times the first equation from the third to obtain the following system:
From the second equation we have . Substituting into the third equation yields . Substituting and into the first equation yields . The solution is therefore
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.