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MATH 3260 Exam 3 (Version 2) Solutions July 7, 2008 S. F. Ellermeyer Name
Instructions. Remember to include all important details of your work. You will not get full credit (or perhaps even any partial credit) if I see gaps in your reasoning. Also, use correct notation and write in complete sentences where appropriate. You may use a calculator on this exam but you may not use any books or notes.
- (DeÖnitions) Use complete sentences to write the following deÖnitions.
(a) Let
A =
a b c d
be a 2 2 matrix. What is meant by the determinant of A? (b) What is the n n identity matrix? (c) What does it mean for an n n matrix, A, to be invertible? (d) What is meant by the inverse of an invertible n n matrix, A? (e) What is an elementary matrix?
(a) Suppose that A is a 3 4 matrix whose columns span V 3 and suppose that C is a 3 3 matrix. Explain how to construct a 4 3 matrix, B, such that AB = C. (This is an essay question. Write carefully and use complete sentences.) (b) For the matrices
A =
(^5) and C =
use the procedure that you described in part a to Önd a 4 3 matrix, B, such that AB = C.
Solution: For the unknown matrix,
B = [b 1 b 2 b 3 ] ,
it must be true that
AB = [Ab 1 Ab 2 Ab 3 ] = [c 1 c 2 c 3 ] = C.
Thus, to Önd B, we must solve all of the matrix equations Ab 1 = c 1 , Ab 2 = c 2 , and Ab 3 = c 3. (Solutions to all of these equations exist because the columns of A span V 3 .) Since all of these matrix equations have the same coe¢ cient matrix
A = [a 1 a 2 a 3 a 4 ] ,
we can Önd B in ìone fell swoopîby rowñreducing the ìtripleñaugmentedîmatrix
[a 1 a 2 a 3 a 4 c 1 c 2 c 3 ].
Example: For the matrices, A and C, given in part b above we have
[A C] =
and we see from this that one possible B such that AB = C is given by
B =
This B was obtained by assigning x 4 = 0 in each equation Abi = ci. There are inÖnitely many other B such that AB = C. For example, note that
B =
also works.
- Use the algorithm [A I]
I A ^1
to Önd the inverse of the matrix
A =
if it exists. If A ^1 does not exist, then explain how the algorithm tells you this. You may not use determinants or your calculator in doing this problem. You must use the algorithm! Solution:
[A I] =
Thus
A ^1 =
7 2
3 2
1 2
- Compute the determinant 5 1 4 1 1 2 4 0 2 by performing a cofactor expansion along the Örst row. Then compute the determi- nant by performing a cofactor expansion along the Örst column. Show all of your computations. Solution: Expanding along the Örst row we obtain
5 1 4 1 1 2 4 0 2
Expanding along the Örst column we obtain
5 1 4 1 1 2 4 0 2
- Let S be the parallelogram determined by the vectors
b 1 =
and b 2 =
and let A =
Compute the area of the image of S under the linear transformation x 7! Ax. Solution: Since det (A) = 5, this linear transformation multiplies all areas by 5. The area of the parallelogram determined by b 1 and b 2 is
jdet ([b 1 b 2 ])j = 4
so the area of the image of S under the linear transformation x 7! Ax is 5 (4) = 20.
- Decide whether each of the following statements is true or false.
(a) If A, B, and C are all square matrices of the same size and AB = AC, then it must be true that B = C. (True, False ) (b) If A and B are square matrices of the same size, then it must be true that
(A + B) (A B) = A^2 B^2.
(True, False ) (c) The matrix (^2)
4
is an elementary matrix. ( True , False) (d) If A and B are square matrices of the same size and A is invertible and AB = BA, then A ^1 B = BA ^1. ( True , False) (e) If A is an invertible matrix, then det (A ^1 ) = det (A). (True, False ) (f) If A is an n n matrix and det (A) = 0, then the columns of A are linearly independent. (True, False ) (g) If A is a 4 4 matrix, then det (5A) = 5 det (A). (True, False ) (h) If A is a square matrix and det (A) 6 = 0, then the matrix equation Ax = b can be solved by using Cramerís Rule. ( True , False) (i) If A is a 2 2 matrix and det (A) = 5 , then the area of the parallelogram determined by the columns of A is 5. ( True , False) (j) Every invertible matrix can be written as a product of elementary matrices. ( True , False)
