Sample First Exam with Solution - Linear Algebra | MATH 3013, Exams of Linear Algebra

Material Type: Exam; Professor: Binegar; Class: LINEAR ALGEBRA; Subject: Mathematics ; University: Oklahoma State University - Stillwater; Term: Summer 2007;

Typology: Exams

Pre 2010

Uploaded on 03/11/2009

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Math 3013
SAMPLE FIRST EXAM
(with solutions)
1. (10 pts) Show that the points (1,0,1,1) , (1,1,0,1), and (1,2,3,1) form the vertices of a right triangle in
R4.
If A= (1,0,1,1) ,B= (1,1,0,1), and C= (1,2,3,1))are the vertices of a triangle, then the sides
of the triangle correspond to the vectors
AB =BA= (0,1,1,0)
AC =CA= (0,2,2,0)
BC =CB= (0,1,3,0)
Since
AB ·AC = (0)(0) + (1)(2) + (1)(2) + (0)(0) = 0 + 2 2 + 0 = 0
the first two sides must be perpendicular. Hence the points A,Band Cform the vertices of a
right triangle in R4
2. Consider the following linear system
x1x2+ 2x3= 1
2x1+x2+x3= 5
x1+x2+ 2x3=1
x2+x3= 1
(a) (5 pts) Write down the corresponding augmented matrix and reduce it to row-echelon form.
11 2
2 1 1
112
0 1 1
1
5
1
1
R2R22R1
R3R3+R1
11 2
0 3 3
0 0 4
0 1 1
1
3
0
1
R4R41
3R2
11 2
0 3 3
0 0 4
0 0 2
1
3
0
0
R4R41
2R2
11 2
0 3 3
0 0 4
0 0 0
1
3
0
0
(b) (5 pts) Reduce the augmented matrix further to reduced row-echelon form.
R21
3R2
R31
4R3
11 2
0 1 1
0 0 1
0 0 0
1
1
0
0
R2R2+R3
R1R1+R2
101
010
001
000
2
1
0
0
R1R1R3
100
010
001
000
2
1
0
0
(c) (5 pts) Write down the solution of the original linear system.
x1= 2 , x2= 1 , x3= 0
1
pf3
pf4

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Math 3013

SAMPLE FIRST EXAM

(with solutions)

  1. (10 pts) Show that the points (1,0,1,1) , (1,1,0,1), and (1,2,3,1) form the vertices of a right triangle in

R 4 .

  • If A = (1, 0 , 1 , 1) , B = (1, 1 , 0 , 1), and C = (1, 2 , 3 , 1))are the vertices of a triangle, then the sides

of the triangle correspond to the vectors

AB = B − A = (0, 1 , − 1 , 0)

AC = C − A = (0, 2 , 2 , 0)

BC = C − B = (0, 1 , 3 , 0)

Since

AB · AC = (0)(0) + (1)(2) + (−1)(2) + (0)(0) = 0 + 2 − 2 + 0 = 0

the first two sides must be perpendicular. Hence the points A, B and C form the vertices of a

right triangle in R 4 

  1. Consider the following linear system

x 1 − x 2 + 2x 3 = 1

2 x 1 + x 2 + x 3 = 5

−x 1 + x 2 + 2x 3 = − 1

x 2 + x 3 = 1

(a) (5 pts) Write down the corresponding augmented matrix and reduce it to row-echelon form.

R 2 → R 2 − 2 R 1

R 3 → R 3 + R 1

R 4 → R 4 −

R 2

R 4 → R 4 −

R 2

(b) (5 pts) Reduce the augmented matrix further to reduced row-echelon form.

R 2 →

1 3

R 2

R 3 →

1 4

R 3

R 2 → R 2 + R 3

R 1 → R 1 + R 2

R 1 → R 1 − R 3

(c) (5 pts) Write down the solution of the original linear system.

  • x 1 = 2 , x 2 = 1 , x 3 = 0 

1

  1. (15 pts) Compute the inverse of

A =

, inverse: and verify that you have the correct inverse by showing AA

− 1 = I.

⇒ A

− 1

 (^) and AA − 1 =

3in

  1. (10 pts) Use the fact that

A =

 ⇒ A−^1 =

to solve

x 1

x 2

x 3

  • If A is invertible then Ax = b implies x = A − 1 b. Taking b = (2, 1 , −1) we have

x =

x 1

x 2

x 3

 (^) = A−^1 b =

  1. (10 pts) Find a basis for the null space of

A =

  • Note that the matrix A is already in reduced row-echelon form and that columns 2 and 4 do not

contain pivots. The linear system corresponding to the matrix equation Ax = 0 is

x 1 + x 2 + 2x 4 = 0

x 3 + 3x 4 = 0

0 = 0

0 = 0

x 1 = −x 2 − 2 x 4

x 3 = − 3 x 4

x =

−x 2 − 2 x 4

x 2

− 3 x 4

x 4

= x 2

  • x 4

Therefore, {[− 1 , 1 , 0 , 0] , [− 2 , 0 , − 3 , 1]} will be a basis for the null space of A 

  1. (10 pts) Mark each of the following statements True or False. (Think carefully.)

F (a) If A and B are invertible n × n matrices, then AB = BA implies B = A

− 1 .

T (b) A square linear system Ax = b has a unique solution if and only if A is row-equivalent to the

identity matrix.

T (c) If A, B and C are invertible n × n matrices, then AC = BC implies A = B.

T (d) If a linear system Ax = b has infinitely many solutions, then it’s possible that another linear

Ax = b

′ (with b

′ 6 = b) will has no solution.

F (e) If a consistent linear system has more equations than unknowns, then there will be a unique

solution.