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Solutions for Linear Algebra Problems - Final Examination | MATH 2270, Exams of Linear Algebra

Material Type: Exam; Class: Linear Algebra; Subject: Mathematics; University: University of Utah; Term: Spring 2009;

Typology: Exams

Pre 2010

Uploaded on 08/31/2009

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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+yz+t= 1
2x+y+2z+t=7
using Gauss elimination.
Solution: The augmented matrix of the system is:
1 1 1 1 3
1 1 1 1 1
2 1 2 1 7
.
By adding the first row to the second and subtracting its double from
the last we get:
1 1 1 1 3
0 2 0 2 2
01 0 11
.
By dividing the second row by 2 and adding it to the last we get
11113
01011
00002
.
Dividing the last row by 2 and adding it to the second and adding
its triple to the first we get:
11110
01010
00001
.
Subtracting the second row from the first we get the reduced row ech-
elon form:
10100
01010
00001
.
Hence, the system has no solutions.
1
pf3
pf4
pf5
pf8
pf9

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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.