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Math 324, Test 2, November 10, 2004 Hints and Answers
Instructions. Do 10 of the following 11 questions. The last three questions are take-home questions on which you may use MathCAD.
- (a) Consider the matrix A =
. Find the eigenvalues of A. Find bases for
the eigenspaces corresponding to its eigenvalues. Is the matrix A diagonalizable? Explain.
(b) Same question with A =
Answer. (a) The eigenvalues are the numbers on the diagonal. Hence λ 1 = 3 and λ 2 = 2. A basis for Eλ 1 = {(1, 0 , 0), (0, 1 , −1)} and a basis for Eλ 2 = {(1, 1 , 0)}. Because dim(Eλ 1 ) + dim(Eλ 2 ) = 3, the matrix A is diagonalizable.
(b) The eigenvalue is λ = 3, a basis for its eigenspace is {(1, 0 , 0)}. The matrix A is not diagonalizable, because the dimensions of the eigenspaces add to 1 which is less than 3.
- Suppose A is a square matrix.
(a) Define what is meant by an eigenvalue with corresponding eigenvector for the matrix A.
(b) Explain or prove why the eigenvalues of A are found by solving the equation |A − λI| = 0.
(c) Suppose A is an invertible matrix. Is it possible for 0 to be an eigenvalue of A? Justify your answer.
Answer. (a) An eigenvalue is a scalar λ such that Av = λv for some nonzero vector v. The nonzero vector v is an eigenvector corresponding to λ.
(b) λ is an eigenvalue of A if and only if Av = λv for some v 6 = 0, if and only if (A − λI)v = 0 for some v 6 = 0, if and only if (A − λI) is not invertible if and only if |A − λI| = 0.
(c) No, if λ = 0, then |A − λI| = |A| 6 = 0 where we know |A| 6 = 0 because A is invertible. Therefore, λ = 0 is not a solution to |A − λI| = 0 if A is invertible.
- Suppose the determinant of M =
a b c d e f g h i
(^) is 5. Find the determinants of
(a) A =
a b c a b c 10 g 10 h 10 i
(^) (b) B =
d e f 10 a 10 b 10 c g + 10d h + 10e i + 10f
(c) C =
10 d 10 e 10 f 10 a 10 b 10 c 10 g 10 h 10 i
Answer. (a) |A| = 0 because it has tow identical rows.
(b) Notice that M → R 1 ↔ R 2 → 10 R 2 → B, therefore, |B| = (−1)(10)(5) = −50.
(c) Notice that M → R 1 ↔ R 2 → 10 R 1 → 10 R 2 → 10 R 3 → C. Therefore, |C| = (−1)(10^3 )(5) = −5000.
- Let A =
, find the inverse of A, and write A and A−^1 as products of
elementary matrices.
Answer. One sequence of elementary row operations that transforms A to I is: 2R 1 +R 2 → R 2 , R 1 +R 3 → R 3 , − 2 R 2 +R 3 → R 3 , 2R 3 +R 1 → R 1 , 5R 3 +R 2 → R 2 , −R 2 +R 1 → R 1 , −R 3 → R 3
thus A−^1 = E 7 E 6 E 5 E 4 E 3 E 2 E 1 =
(^) where
E 1 =
E 2 =
E 3 =
E 4 =
E 5 =
E 6 =
E 7 =
that is
A−^1 =
and A = E− 1 1 E 2 − 1 E 3 − 1 E 4 − 1 E− 5 1 E 6 − 1 E 7 − 1. where
E 1 − 1 =
E− 2 1 =
E− 3 1 =
E 4 − 1 =
E 5 − 1 =
E 6 − 1 =
E− 7 1 =
- Let A, B and C be n by n matrices where A = BCB−^1.
(a) Show that An^ = BCnB−^1.
(b) Suppose that B =
[
]
and C =
[
]
. Find a formula for An
[
]
Answer. (a) Intuitively,
An^ = (BCB−^1 )(BCB−^1 ) · · · (BCB−^1 ) = BC(B−^1 B)C(B−^1 B) · · · C(B−^1 B)CB−^1 = BCnB−^1
- Use Cramer’s rule to write formulas for the solutions x and y to the system { ax + by = e cx + dy = f
where ad − bc 6 = 0, and illustrate your formula with the system { 2 x − y = 3 5 x + 4 y = 4
(a) The formulas are
x =
e b f d
∣∣ a^ b c d
de − bf ad − bc
y =
a e c f
∣∣ a^ b c d
af − ce ad − bc
(b) Using the above formulas
x =
and y =
Part II. The following three questions may be completed with the help of MathCAD. They are due by 8am tomorrow morning. You may not ask anyone for assistance on these questions. You may use your notes, book or other resources.
- Consider the system of equations
x + 2 y + 2 z = 5 − 2 x − 5 y − 2 z = 8 2 x + 4 y + 7 z = 19
Solve this system using each of the following three methods:
(a) by row reducing the augemented matrix;
(b) using the inverse of the coefficient matrix;
(c) using Cramer’s rule.
- Find a formula for xn defined by the recursion relation xn = 5xn− 1 + 6xn− 2 where x 1 = 0 and x 2 = 1.
- (a) Verify that β = {(1, 1 , 1 , 1), (− 1 , 2 , 2 , 2), (− 2 , − 2 , 4 , 4), (− 3 , − 3 , − 3 , 6)} is a basis of IR^4.
(b) Given the ordered basis in (a), find x if its coordinate vector is [x]β =
(c) Given the ordered basis in (a), find [x]β when x = (− 6 , 3 , 15 , 24). Verify by hand that the coordinate vector you found is correct.
Answers to these questions are on a MathCAD file. The verification of 11(c) by hand is as follows.
4(1, 1 , 1 , 1) + 3(− 1 , 2 , 2 , 2) + 2(− 2 , − 2 , 4 , 4) + 1(− 3 , − 3 , − 3 , 6) = = (4 − 3 − 4 − 3 , 4 + 6 − 4 − 3 , 4 + 6 + 8 − 3 , 4 + 6 + 8 + 6) = (− 6 , 3 , 15 , 24)
as desired.
