





Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Material Type: Exam; Class: Linear Algebra; Subject: Mathematics; University: University of Utah; Term: Spring 2009;
Typology: Exams
1 / 9
This page cannot be seen from the preview
Don't miss anything!
Problem 1 (25 points). Find all solutions of the linear system
x +y +z +t = − 3 −x +y −z +t = 1 2 x +y +2z +t = − 7
using Gauss elimination.
Solution: The augmented matrix of the system is:
By adding the first row to the second and subtracting its double from the last we get: (^)
By dividing the second row by 2 and adding it to the last we get
Dividing the last row by −2 and adding it to the second and adding its triple to the first we get:
Subtracting the second row from the first we get the reduced row ech- elon form: (^)
Hence, the system has no solutions.
1
Problem 2 (25 points). Calculate the inverse A−^1 of the matrix
Solution: Consider the matrix
By subtracting the first row from the third and its double from the second we get (^)
By multiplying the third row by −1 and switching it with the second we get: (^)
Multiplying the second row by 3 and adding it to the last we get:
Subtracting the last row from the second and its double from the first we get: (^)
Subtracting the double of the second row from the first we finally get:
Hence the inverse matrix is:
Problem 4 (25 points). Consider the linear transformation from R^3 into R^3 given by the matrix
Find:
a) A basis of the image of this linear transformation. b) The rank of this linear transformation. Solution: We have to find the reduced row echelon form of A. If we switch the first two rows we get:
If we subtract the first row from the second and its double from the third, we get: (^)
Hence, but subtracting the second row from the third, we get:
By dividing the second row by 2, we get the reduced row echelon form:
Hence, the image is spanned by the vectors
(^) and
and the rank of A is equal to 2.
Problem 5 (25 points). Find the matrix A of the orthogonal projection onto the plane in R^3 spanned by the vectors
u =
(^) and v =
Find the kernel of A.
Solution: The subspace spanned by u and v is equal to the subspace spanned by the standard basis vectors:
(^) and
Therefore, the projection is given by the matrix
Its kernel is the subspace spanned by
Problem 6 (25 points). Find the determinant ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 1 − 1 1 1 1 1 − 1 − 1 1 1 1 1 − 1 1 1
Solution: By Gauss elimination, we have ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 1 − 1 1 1 1 1 − 1 − 1 1 1 1 1 − 1 1 1
by subtracting the first row from the second and the fourth and adding it to the third. By taking the common factor 2 from the last three rows
Hence, y = 0 and x and z are arbitrary. It follows that the eigenspace for the eigenvalue 1 is spanned by
(^) and
Hence, the algebraic multiplicity of 1 is 3, and the geometric multiplic- ity is 2. It follows that the eigenbasis of A doesn’t exist and A is not similar to a diagonal matrix.
Problem 8 (25 points). Consider the linear transformation from R^3 into R^3 given by the symmetric matrix
Find
a) the eigenvalues of A; b) an orthonormal basis consisting of eigenvectors of A; c) the “change of basis” matrix S from the standard basis to the basis above; d) the diagonal matrix D = S−^1 AS.
Solution: The characteristic polynomial is: ∣ ∣ ∣ ∣ ∣ ∣ 4 − λ − 2 0 − 2 3 − λ − 2 0 − 2 2 − λ
= (4 − λ)(3 − λ)(2 − λ) − 4(4 − λ) − 4(2 − λ)
= (4 − λ)(3 − λ)(2 − λ) − 4(6 − 2 λ) = (4 − λ)(3 − λ)(2 − λ) − 8(3 − λ) = (3 − λ)
(4 − λ)(2 − λ) − 8
= (3 − λ)(λ^2 − 6 λ) = −λ(λ − 3)(λ − 6).
Hence, the eigenvalues are 0, 3 and 6. The eigenvectors for 0 satisfy
x y z
Hence, 2x = y and y = z, i.e., the solutions are proportional to
The corresponding normalized vector is
1 3
The eigenvectors for 3 satisfy
x y z
Hence, x = 2y and x = −z, i.e., the solutions are proportional to
The corresponding normalized vector is
1 3
The eigenvectors for 6 satisfy
x y z
Hence, x = −y and 2z = −y, i.e., the solutions are proportional to
The corresponding normalized vector is
1 3
Hence, the “change of basis” matrix S is
1 3
Since it is orthogonal and symmetric, S−^1 = S.