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MATH 323.502
Exam 1 Solutions
March 3, 2015
1.
Consider the system of equations
4x + 8y = 4
3x + 6y − z = 0
3x + 6y − 2z = −3
(a) Write down the augmented matrix for this system of equations with the variables
ordered x, y, z.
(b) Carry out the row reduction of the augmented matrix in (a) and transform it into
reduced row echelon form. Show all of the steps you are taking.
(c) Determine the complete set of solutions of this system of equations.
(d) Is the coefficient matrix of this system invertible? Why or why not?
Solution: (a) Reading off the coefficients of the variables, the augmented matrix is


4 8 0
4
 3 6 −1 0  .
3 6 −2 −3
(b) We perform the following row operations: (there are several ways to do do this
correctly)




1 2 0
4
1
4 8 0
(i)
 3 6 −1 0 
 3 6 −1 0  −→
3 6 −2 −3
3 6 −2 −3




1 2 0
1
1 2 0 1
(ii)
(iii)
−→  0 0 −1 −3  −→  0 0 1 3  .
0 0 −2 −6
0 0 0 0
Here in (i) we divided the top row by 4; in (ii) we subtracted 3 times the first row
from the second row and from the third row; in (iii) we multiplied the second row by
−1 and added twice it to the third.
(c) In reduced row echelon form our system is equivalent to x + 2y = 1 and z = 3.
Thus the full set of solutions is



 1 − 2y 
 y y∈R .


3
1
(d) No the coefficient matrix is not invertible . There are a couple legitimate ways
to show this. One way is the check that the determinant is non-zero:


4 8 0
det 3 6 −1 = −48 − 24 + 24 + 48 = 0.
3 6 −2
Another way to see this is that if the matrix were invertible, then the system of
equations would have a unique solution. Since there are infinitely many solutions,
the matrix must not be invertible.
2.
For each statement below, write down whether it is true or false.
(a) A system of 6 linear equations in 3 variables must be inconsistent.
(b) A homogeneous system of 3 linear equations in 6 variables must have more than
one solution.
(c) If F1 and F2 are elementary matrices of type II, then F1 F2 = F2 F1 .
(d) For n × n matrices A and B, (AB)2 = A2 B 2 .
(e) For n × n matrices A and B, if AB is singular, then at least one of A and B must
be singular.
3
−1
0
2 (f) The set W = a
+b
+
∈ R a, b ∈ R is a subspace of R2 .
2
1
1
Solution: Credit was given for the correct answers. The explanations were not necessary, but the ones below may answer some of your questions about these problems.
(a) False : They could be consistent if, for example, all six equations are the same
equation.
(b) True : Such a linear system is always consistent (since 0 is a solution). In row
echelon form, the matrix will have at most 3 leading variables, so at least 3 free
variables. Having free variables gives infinitely many solutions.
(c) True : The easiest way to see this is recall that type II matrices correspond to the
row operation of multiplying a row by a nonzero constant. Since multiplying rows by
constants can be done in either order, the matrices must commute.
(d) False : Generally speaking, (AB)2 = ABAB, which may not be the same as
A2 B 2 .
(e) True : If AB is singular, then det(AB) = 0. But det(AB) = det(A) det(B), so it
must be that det(A) = 0 or det(B) = 0. Thus one of them has to be singular.
(f) False : W fails to be closed under addition and scalar multiplication. Also, it
does not contain the zero vector.
2
3.
Consider the matrix


1 0 −3
A = 0 1 0  .
0 2 1
Find two elementary matrices E1 and E2 of type III so that E2 E1 A = I.
Solution: We want to analyze which row operations of type III applied to A will
yield I. We see that if we replace Row 3 with its sum with −2 times Row 2, and
then replace Row 1 with its sum with 3 times Row 3, we obtain the identity. Thus,
we take




1 0 0
1 0 3
E1 = 1 0 0 , E2 = 0 1 0 .
0 −2 1
0 0 1
4.
Determine if the following sets of vectors are or are not vector spaces. In each case,
if the set is not a vector space, explain why not.
(a) V = the solution set of the equations x − 2y + 3z − 4w = 0 and x + 3z = 0 in R4 .
(b) W = the closed unit disk = {[ xy ] ∈ R2 | x2 + y 2 ≤ 1}.
Solution: (a) V is a vector space , since solutions of homogeneous systems of equations are always vector spaces.
(b) W is not a vector space . There are various ways to check this, but the most
straightforward ones are to show that it is not closed under addition or scalar multiplication. For example, [ 10 ] ∈ W , but
1
3
3·
=
∈
/ W.
0
0
Thus W is not closed under scalar multiplication.
5.
Let A and B be square matrices. We say that A is similar to B if there exists an
invertible matrix P so that A = P BP −1 .
(a) Prove that if A is similar to B, then B is similar to A.
(b) Prove that if A is similar to B, then det(A) = det(B).
Solution: (a) Suppose A is similar to B. Then we can pick a matrix P so that
A = P BP −1 . If we multiply on the left by P −1 and on the right by P , we find
P −1 AP = P −1 P BP −1 P = B.
Therefore,
B = P −1 A(P −1 )−1 ,
(since P = (P −1 )−1 ),
so taking P to be P −1 in the definition of similarity, we see that B is similar to A.
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(b) Suppose A is similar to B. Then as above we can pick P so that A = P BP −1 .
Therefore,
det(A) = det(P BP −1 )
= det(P ) det(B) det(P −1 )
= det(P ) det(B) det(P )−1 (since det(P −1 ) = 1/ det(P ))
= det(B).
4