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Math 2270-1 Final, May 4, 2009
Solutions
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
A =
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:
A−^1 =
Problem 4 (25 points). Consider the linear transformation from R^3 into R^3 given by the matrix
A =
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
A =
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.
