Download 54 Quiz 3 Solutions GSI: Morgan Weiler Problem 0 (1 pt/ea). (a

54 Quiz 3 Solutions GSI: Morgan Weiler
Problem 0 (1 pt/ea).
(a). True or false: if A and B are invertible, then AB is invertible.
Solution: This problem is going to be taken off the quiz, because I did not specify the dimensions of
A and B.
(b). True or false: if A and B are n × n matrices and AB is invertible, A is invertible.
Solution: True – this was a homework problem.
(c). True or false: if A is invertible, then A is a product of elementary matrices.
Solution: True, because A can be row reduced to the identity, which is the same thing as being a
product of elementary matrices (take the product of the inverses of the elementary matrices used to
reduce A to the identity).
(d). If det A = 3 then what is det A−1 ?
Solution: Since (det A)(det A−1 ) = det(I) = 1, det A−1 =
1
1
detA
= 13 .
Problem 1 (2 pts/ea).
(a). Give two statements equivalent to the following statement: “An n × n matrix A is invertible.”
Solution: Any two statements from the Invertible Matrix Theorem, or one statement from the
Invertible Matrix Theorem and “detA =
6 0.”
(b). This question asks for TWO elementary matrices. Give a 3 × 3 elementary matrix E1 which, for
any 3 × 3 matrix A, has the property that E1 A is A with the first and third rows switched. Also give a
3 × 3 elementary matrix E2 which has the property that E2 A is A with two times the second row
added to the first row.
Solution:




0 0 1
1 2 0
E1 = 0 1 0 and E2 = 0 1 0
1 0 0
0 0 1
2


1 0 2 −1
3 0 0 5 

Problem 2 (6 pts). Let A = 
2 1 4 −3. Calculate the determinant and inverse of A. If you
1 0 5 0
need more space, feel free to use the top of the back page. Use any method you want.
Solution: For the determinant, it’s easiest to expand along the second column, remembering the signs
(checkerboard, starting with a + in the upper left). After that, no expansions are obviously better, so
here is one method:
1 2 −1
3 5 1 −1
detA = −1 3 0 5 = (−1) −2 − 5 3 5 = 2(−5) + 5(5 + 3) = −10 + 40 = 30
1
0
1 5 0 For the inverse, I chose to reduce [A I] to [I A−1 ]. I think it might be more illuminating for you to
check your own row reduction rather than to give mine, because it will be completely different, so I’ll
just give the inverse:


25
5 0 −10

1 
−75 3 30 6 
A−1 =
8 
30  −5 −1 0
−15 3 0
6
−1
Instead of trying to row reduce to [I A ] it might be easier to reduce to [30I 30A−1 ], just to get
integers instead of fractions.
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Problem 3 (6 pts). Suppose A is an n × n matrix with the property that the equation Ax = 0 has
only the trivial solution. WITHOUT using the Invertible Matrix Theorem, explain directly why the
equation Ax = b must have a solution for each b in Rn .
Solution: One way to see this is that since Ax = 0 has only the trivial solution, A is row equivalent
to the identity matrix. It is always possible to do the same row operations to the augmented matrix
for Ax = b, which will give a matrix with a pivot in every column except the final column, providing a
solution to Ax = b.
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