S2009, Section 3.2: 10,14,18,24,26,28 3.2-10 ====== The 3 equations in 5 unknowns have infinitely many solutions. The general solution is obtained by applying the Last Frame Algorithm. Please recognize that you are already given the last frame, there are no toolkit steps of swap, combo, mult to perform! Expected solution steps: Last Frame Algorithm 0. Justify that this is indeed the last frame, by applying the Last Frame Test. 1. Identify the lead variables. 2. Identify the free variables. Assign invented symbols t1, t2, t3, ... 3. Back-substitute the free variables into the lead variable equations. 4. Report the solution as the list of variables, each followed by an equal sign. Each equal sign is followed by a constant or else an expression in the invented symbols. 3.2-14 ====== The 3 equations in 3 unknowns require a number of toolkit steps: combo, sqwap, mult. Write a frame sequence, one toolkit operation per frame, ending with the last frame, which has one free variable and two lead variables. Justify that this is indeed the last frame, that it passes the Last Frame Test. Then carry out the Last Frame Algorithm to display the solution. Last Frame Algorithm 1. Identify the lead variables. 2. Identify the free variables. Assign invented symbols t1, t2, t3, ... 3. Back-substitute the free variables into the lead variable equations. 4. Report the solution as the list of variables, in variable list order, each followed by an equal sign. Each equal sign is followed by a constant or else an expression in the invented symbols. 3.2-18 ====== There are 3 equations and 4 unknowns, therefore at most 3 lead variables and at least one free variable. This implies infinitely many solutions. Unlike 3.2-10, you must now display a frame sequence whose last frame is a reduced echelon system. Then proceed to apply the Last Frame Algorithm in order to write out the general solution, as in 3.2-10. 3.2-24 ====== Because the equations have symbol k appearing in the coefficients, there will in general be infinitely frame sequences, one for each value of k. The last frame of any such sequence can theoretically have a signal equation, or zero free variables, or at least one free variable, which classifies the problem as No Sol, Infinitely Many Solutions or a Unique Solution. The class the last frame falls into depends on the value of k, and not all classifications need be represented. Display as many complete frame sequences as necessary, each depending on a range of values of k, such that the last frame determines one of the three possibilities. Clearly state assumptions used for each frame sequence. See web references for similar problems, which give further hints on how to organize the work and draw conclusions. 3.2-26 ====== The advice for 3.2-24 applies, but this problem is considerably simpler, due to there being just one frame sequence reported, valid for all values of k. Conslusion: there are zero free variables and a unique solution. 3.2-28 ====== The basic advice given in 3.2-24 applies. There is just one frame sequence whose last frame has 2 lead variables and one free variable. However, it may have a signal equation if c-2a-3b is nonzero. The unique solution case can never happen.