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Material Type: Assignment; Class: Linear Algebra; Subject: Mathematics; University: University of Michigan - Ann Arbor; Term: Winter 2009;
Typology: Assignments
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Math 217 Winter 2009 - Group Homework 1 Solutions
a 11 x 1 + a 12 x 2 + · · · + a 1 nxn = b 1 a 21 x 1 + a 22 x 2 + · · · + a 2 nxn = b 2 .. . am 1 x 1 + am 2 x 2 + · · · + amnxn = bm
This system has m equations in n variables. Prove that any of the elementary row operations applied to this system results in an equivalent system. Solution: We must show that, for each of the three basic row opera- tions, when the operation is applied to the above system, the resulting system has the same solution set as the above system. First, for the operation of interchanging two rows, the resulting system of equations literally consists of the same set of equations, just in a different order. It follows immediately that they have the same solution sets. Let c be a nonzero real number, let k be an integer with 1 ≤ k ≤ m, and consider the operation Rk −→ c · Rk of multiplying row k by c. If
~x =
x 1 .. . xn
is a solution of the original system, then multiplying the kth^ equation by c preserves the equality (since the same thing has been done to both sides of that equation). It follows that ~x is a solution to the transformed system. Conversely, if ~x is a solution to the transformed system, then multiplying the kth^ equation by c−^1 (this exists since c 6 = 0) again preserves the equality and results in the kth^ equation in the original system, so ~x is a solution to the original system. Thus we have shown that ~x is a solution to the original system if and only if ~x is a solution to the transformed system, which means that these solutions have the same solution set, and are therefore equivalent by definition. Lastly, let k, l be distinct integers with 1 ≤ l, k ≤ m, let c ∈ R be any real number, and consider the operation Rk −→ Rk + c · Rl of
adding c times row l to row k. If ~x is a solution to the original system of equations. Multiplying equation l be c preserves the equality (but in this case might result in a pretty stupid equation if c = 0). Also, one may add equation k to the equation that results from multiplying equation l by c and one gets another valid equation (here we’re using the fact that one can adding equations also preserves equality). But this results in the kth^ equation in the transformed system, so ~x is a solution to the transformed system. Conversely, suppose that we start with a solution ~x to the transformed system of equations. Here’s the key: if we multiply the lth^ equation in this system (which is the same as the lth equation in the original system) by c (as usual, this preserves equality) and subtract this resulting equation from equation k (again preserving equality) you end up with equation k from the original system. In effect, the operation Rk −→ Rk − c · Rl undoes the row operation Rk −→ Rk + c · Rl. At any rate, the result is that ~x is also a solution to the original system of equations. Thus ~x is a solution to the original system if and only if it’s a solution to the transformed system, so they have the same solution sets and are equivalent.
A(~x 1 + c(~x 1 − ~x 2 )) = A~x 1 + cA(~x 1 − ~x 2 ) = ~b + c · ~0 = ~b
so each of these infinitely many vectors is a solution to the equation A~x = ~b, so the equation has infinitely many solutions.
lies in span{~v 1 , ~v 2 } if an only if
a 11 a 12 c 1 a 21 a 22 c 1 a 31 a 32 c 3
is the augmented matrix of a consistent system of linear equations. A row echelon form of the matrix (not necessarily the reduced one) must have two zeros to begin the last row since the leading entries of the rows must move to the right as you go down the rows. Thus a row echelon form must be in the form
∗ ∗ d 1 ∗ ∗ d 1 0 0 d 3
Of course, one can say more about the form, but this is all we need. Now the system is consistent if and only if d 3 = 0. Now here’s the key observation: the sequence of row operations used to arrive at this form can be run in reverse. We can undo each row operation by another row operation and proceed backwards successively undoing each row operation. It follows that we can start with any column of d’s we wish, run the sequence in reverse to arrive at a matrix of the form (2) that has the property that it yields (3) when you do the sequence of row operations in the original order. In particular, using d 1 = d 2 = 0, d 3 = 1 we get values of c 1 , c 2 , and c 3 that form an inconsistent system, so the vector (^)
c 1 c 2 c 3
is not in span{~v 1 , ~v 2 }.
Of course, Theorem 4 from the book can be used to give a simpler proof, but I wanted to give a self-contained argument.