Eam 2 Solutions | Linear Algebra | MATH 4100, Exams of Linear Algebra

Material Type: Exam; Class: LINEAR ALGEBRA; Subject: Mathematics; University: Rensselaer Polytechnic Institute; Term: Fall 2007;

Typology: Exams

2011/2012

Uploaded on 02/17/2012

koofers-user-blk
koofers-user-blk 🇺🇸

5

(1)

10 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
MATH-4100 October 29, 2007 Exam #2 Solutions
Please answer all 6questions, showing your work in reasonable detail. Closed books; laptops
and calculators are not permitted. 1
2
3
4
5
6
1Suppose the n×nmatrix Anhas 5s along its main diagonal and 2s along the diagonal below
and in its (1, n) position:
A4=
5002
2500
0250
0025
Find by cofactors of row 1 or otherwise the determinant of A4and then detAnfor n > 4.
By cofactors 5 53223,similarly for n×n: 5 5n1+ (1)n122n1= 5n+ (1)n12n
1
pf3
pf4
pf5

Partial preview of the text

Download Eam 2 Solutions | Linear Algebra | MATH 4100 and more Exams Linear Algebra in PDF only on Docsity!

MATH-4100 October 29, 2007 Exam #2 Solutions Please answer all 6 questions, showing your work in reasonable detail. Closed books; laptops

and calculators are not permitted. 1

1 Suppose the n × n matrix An has 5s along its main diagonal and 2s along the diagonal below and in its (1, n) position:

A 4 =

   

   

Find by cofactors of row 1 or otherwise the determinant of A 4 and then det An for n > 4.

By cofactors 5 ∗ 53 − 2 ∗ 23 ,similarly for n × n : 5 ∗ 5 n−^1 + (−1)n−^1 2 ∗ 2 n−^1 = 5n^ + (−1)n−^1 2 n

2 If you know that det A = 7, what is det B?

A =

  

row 1 row 2 row 3

   ,^ B^ =

  

row 3 + row 2 + row 1 row 2 + row 1 row 1

  

det

  

row 1 row 2 row 3

   =^ −^ det

  

row 3 row 2 row 1

   =^ −^ det

  

row 3 row 2 + row 1 row 1

  

= − det

 

row 3 + row 2 + row 1 row 2 + row 1 row 1

 

Therefore det B = − det A = − 7

4 Find the matrix of projection onto the plane x + y − z = 0. Find the projection of b=[2, 1 , 2]T onto the plane. Hint: It is helpful to select a basis in the plane.

To select a basis solve [ 1 1 − 1

]

 

x y z

  =

 

 

one pivot; solutions

A =

  

  

AT^ A =

[ 2 − 1 − 1 2

] ,

( AT^ A

)− 1

[ 2 1 1 2

]

p =

 

 

[ 2 1 1 2

] [ 1 0 1 − 1 1 0

]

  

  

  

   =^ P

  

   =

  

  

An easier way to find p to project b onto the normal to the plane n = [1, 1 , −1]T^ first:

nnT^ b/nT^ n =

n

Then p = b − 13 n = [5/ 3 , 2 / 3 , 7 /3]T^.

If you start with an orthogonal basis, say

a = [1, − 1 , 0] , b = [1, 1 , 2]

then the matrix P of the projection on the plane is the sum of projections onto the basis vectors:

P = aaT^ aT^ a/(aT^ a) + bbT^ /bT^ b =

1 2

 

  +^1 6

 

  = 1 3

 

 

5 Given vectors

q 1 =

  

   , a 2 =

  

   , a 3 =

  

  

(a) Apply Gram-Schmidt to get orthonormal vectors q 1 , q 2 , and q 3

(b) After a little thinking find the projection of a 3 onto the orthogonal complement of q 1 and a 2.

Note that q 1 is already normalized; q 2 = a 2 /

2 , q˜ 3 = a 3 − (a 3 · q 1 ) q 1 − (a 3 · q 2 ) q 2 =

a 3 − ( 4 / 3 ) q 1 +

( 1 /

) q 2 =    

   

   

   

   

   

   

   

‖q˜ 3 ‖ = 19 (2 ∗ 7

2 4 + 14

18

(b) this projection is ˜q 3 itself.

(c) The matrix below has orthogonal columns

W =

   

   

Without much computation find W −^1.

W W T^ =

  

   ,

therefore

W −^1 =

  

  