Linear Algebra Answers

Prove that the dot product between two vectors is commutative and not associative

Let ~u =< 3, −1, −2 >, ~v =< −1, 0, 2 > and w~ =< −6, 1, 4 >.

Compute the expressions below.

(2.1) ~v + 3w~

(2.2) 3~u − 2~v

(2.3) −(~v + 3~u).

Find a nonzero vector ~u with initial point P(−6, −7, 0) such that

(1.1) ~u has the same direction as ~v =< −1, 2, 4 >.

(1.2) ~u is oppositely directed to ~v =< −1, 2, 4 >

(9.1)Determine for which value (s) of k will the matrix below be non-singular.

2-k -3

A= 2 k+1

(9.2)Determine for which value (s) of k will the matrix below be non-singular.

2 2 1

A = 3 1 3

1 3 k

Assume that A is a 3 by 3 matrix such that det(A) = −10. Let B be a matrix obtained from A using the following elementary row operations:

R3 + 2R1 → R3,

5R1 → R1,

−2R2 → R2

R2 ↔ R3.

Find the determinant of B obtained from the resulting operations, i.e., det(B)

(7.1)Compute the product AB for

0 4 0 1 0 3

A=2 3 1 , B= 1 1 5

3 0 1 2 3 -1

(7.2) Use your answer in (7.1) to evaluate det(AB) and compare it to det(A) det(B).

(7.3)Determine whether or not if det(A + B) is related to det(A) + det(B).

Use Cramer’s rule to solve for x, y and z

2x + y − 3z = 0

4x + 5y + z = 4

x + y − 4z = −1 

(5.1)Find det(−2A) and compare it to det(A) for

A= 1 1

3 -1

(5.2)Find det(−2A) and compare it to det(A) for

-2 1 3

A = 1 4 5

2 3 1

(4.1)Use the reduced row echelon form to determine

2 4 6

0 0 2

2 -1 5

(4.2)Use the reduced row echelon form to determine

1 2 -4

2 -3 1

0 0 2

(3.1)Without calculating the determinant, inspect the following:

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 -2

(3.2)Without calculating the determinant, inspect the following:

1 0 0 0

0 1 0 0

0 0 0 1

0 0 1/4 0