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Unformatted text preview: the advantage of always producing what we need,
but it can also be ineﬃcient at times. For example, when solving 2 above, it is clear after we
eliminated the x’s in the second step to get the system 1 (E 1) x + 3 y − 1 z =
2 (E 2)
−15y + 4z = −3 (E 3)
−15y + 4z =
that equations E 2 and E 3 when taken together form a contradiction since we have identical left hand
sides and diﬀerent right hand sides. The algorithm takes two more steps to reach this contradiction.
We also note that substitution in Gaussian Elimination is delayed until all the elimination is done,
thus it gets called back-substitution. This may also be ineﬃcient in many cases. Rest assured,
the technique of substitution as you may have learned it in Intermediate Algebra will once again
take center stage in Section 8.7. Lastly, we note that the system in 3 above is underdetermined,
and as it is consistent, we have free variables in our answer. We close this section with a standard
‘mixture’ type application of systems of linear equations.
Example 8.1.3. Lucas needs to create a 500 milliliters (mL) of a 40% acid solution. He has...
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