Tutorial Worksheet. 1. Find all solutions to the linear system by following the given steps. x + 2y + 3z = 2 2x + 3y + z = 4.

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1 Mth 5 Tutoril Week 1 - Jnury 1 1 Nme Setion Tutoril Worksheet 1. Find ll solutions to the liner system by following the given steps x + y + z = x + y + z = 4. y + z = Step 1. Write down the rgumented mtrix (oeffiient nd onstnts) of the system Step. Reple the seond row of the mtrix with the seond row subtrted by times the first row. In our nottion, we would write R R R 1 to denote this opertion. This gives us the mtrix R R R 1 1 R R R Step. Keep performing the row opertions until you n solve the eqution. 1 1 R R 1 5 R R R This is in ehelon form, nd the orresponding liner system is lredy esy to solve, but we keep going on to row redued ehelon form, where it is even esier:

2 R 1 R 1 R R R 5R R 1 4 R R 1 R 1 + 7R This is in row redued ehelon form; the orresponding liner system is x =, y = 1, z = whih is the unique solution of the originl system.. As generliztion to the first question, n you find the solution to the system x + y + z = x + y + z = b y + z = for rbitrry rel numbers, b, nd? The sme row opertions s in 1) work: R R b R R R 1 b R 1 R 1 R R 1 4 R R R 5R R 1 R 1 + 7R R R R b b b b + b b ( b )/4 b (5 + b )/4 ( b )/4 ( + b 7)/4 (5 + b )/4 ( b )/4 b b + The unique solution is x = ( + b 7)/4, y = (5 + b )/4, z = ( b )/4.

3 . Whih mtries below re in ehelon form? Whih re in redued ehelon form? A = [ 1 5 B = [ C = [ D = [ Ehelon form :. Redued ehelon form :. It is esy to see tht ll re in ehelon form; leding entries of lower non-zero rows re further to the right, ll entries below eh leding entry re zero, nd ny zero rows our below non-zero rows. B nd D re not row-redued ehelon beuse there is non-zero entry bove the leding entry of their seond row. A nd C re row-redued ehelon sine they (re ehelon nd) hve ll leding entries 1 nd zeros everywhere bove nd below eh leding entry. Ehelon form : A, B, C, D. Redued ehelon form : A, C. 4. Whih olumns of the mtrix below re pivot nd whih re free? Remrk: A free olumn is olumn whih orresponds to free vrible. Row-redue R R R 1 7 R R R We puse the row redution here to mke to mke omment bout writing rowredutions omptly. In row redution, if you hve suessive steps involving dding multiples of fixed row to severl other rows, with experiene they n be displyed s single step to void unneessry repeted writing of rows. For exmple, the lst two steps whih involved dding multiples of R 1 to R nd R suessively n be represented in more ompt nottion R R R 1 R R R

4 This represents the effet of doing two row opertions one fter the other. Note the order in whih they re done does not mtter (doing R R R 1 then R R R 1 to the originl mtrix hs the sme effet s doing R R R 1 then R R R 1 ). On the other hnd, for other sequenes of row opertions, the order in whih they re done usully ffets the result, nd experiene shows tht trying to represent suh series of row opertions in single nottion often uses onfusion nd errors, nd so should be voided. In the reminder of the row-redution, we use the bove more ompt nottion R 1 R 1 R R R R R 1 R 1 + R The leding entries of the first, seond nd third rows re their first, seond nd fourth entries from the left respetively; these re the pivot entries of the originl mtrix. The pivot olumns re therefore the first, seond nd fourth olumns, while the third nd fifth olumns re non-pivot olumns. If this mtrix rose s the ugmented mtrix of liner system, the bound vribles (nd olumns) would be the first, seond nd fourth, while the only free vrible (or olumns) would be the third (sine the fifth nd lst olumn does not orrespond to vrible, but rther ontins the onstnt terms of the liner equtions). Note tht to find the pivot positions et bove, we put the mtrix in row-redued ehelon form. This is more work thn ws stritly neessry to find the pivot positions. It tully is enough to get it the mtrix to ehelon form nd tke the pivot positions s the positions of leding entries of the non-zero rows of the ehelon form mtrix (sine further row redution to row redued ehelon form does not hnge the positions of the leding entries).. 5. Find ll solutions to the liner system x y + z w = x 6y + z w = x 9y + 5z w = 4. Row-redue the ugmented mtrix of the liner system: R R R R R R R 1 R 1 R R R R 1 1 1

5 This is in row redued ehelon form. The orresponding liner system is x y w = z + w = = where the lst eqution = n obviously be omitted (note on the other hnd if we hd ended up with ny equtions = where is non-zero onstnt, then the originl system would be inonsistent i.e. hve no solutions). The pivot positions of the row redued ehelon mtrix re the first entry of the first row nd third entry of the seond row; the pivot olumns re the first nd third olumns, nd the orresponding vribles, x nd z, re bound. The other vribles, y nd w, re free. Rewrite the lst liner system (with = omitted) so the bound vribles re on the left nd free vribles re on the right: { x = + y + w z = w. The bove is one form of the solution; the free vribles y nd w n tke rbitrry vlues nd the bound vribles x nd z re then determined by the bove equtions. Letting the vlues of y nd w be s nd t respetively, nother form of the solution is x = + s + t y = s z = t w = t where s nd t re rbitrry slrs. Although we hdn t overed vetors t the time of the first tutoril, we hve by now nd note tht we n write the solution lso in the following prmetri vetor form x y z = w where s nd t re rbitrry slrs. + s 1 + t 1