{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# sol6 - SOLUTIONS TO HOMEWORK 6 Section 3.2 4(a We already...

This preview shows pages 1–4. Sign up to view the full content.

SOLUTIONS TO HOMEWORK 6 Section 3.2 4. (a) We already have 1 in the top left corner. First turn the rest of the ±rst row into zeroes, by the elementary column operations C 2 m→ C 2 C 1 , C 3 m→ C 3 C 1 , and C 4 m→ C 4 C 1 , we arrive at the matrix 1 0 0 0 2 2 3 2 1 0 0 0 Next, use elementary row operations R 2 m→ R 2 2 R 1 and R 3 m→ R 3 R 1 to obtain 1 0 0 0 0 2 3 2 0 0 0 0 To obtain 1 in the (2 , 2) position, use C 2 m→ ( 1 / 2) C 2 . This yields 1 0 0 0 0 1 3 2 0 0 0 0 Now, “clear away” the rest of the non-zero entries via C 3 m→ C 3 +3 C 2 and C 4 m→ C 4 C 2 . As a result, obtain 1 0 0 0 0 1 0 0 0 0 0 0 Thus, the rank of our matrix equals 2 . 5. (d) A is invertible (rank A = 3). Compute A 1 using elementary row operations on the augmented matrix ( A | I 3 ): 0 2 4 1 0 0 1 1 1 0 1 0 2 4 5 0 0 1 −→ R 1 R 2 1 1 1 0 1 0 0 2 4 1 0 0 2 4 5 0 0 1 R 3 R 3 2 R 1 1 1 1 0 1 0 0 2 4 1 0 0 0 2 3 0 2 1 R 2 →− R 2 / 2 1 1 1 0 1 0 0 1 2 1 / 2 0 0 0 2 3 0 2 1 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 SOLUTIONS TO HOMEWORK 6 −→ R 1 R 1 R 2 ,R 3 R 3 2 R 2 1 0 1 1 / 2 1 0 0 1 2 1 / 2 0 0 0 0 1 1 2 1 R 1 R 1 R 3 ,R 2 R 2 +2 R 3 1 0 0 1 / 2 3 1 0 1 0 3 / 2 4 2 0 0 1 1 2 1 Thus, A 1 = 1 / 2 3 1 3 / 2 4 2 1 2 1 (f) rank A = 2, hence A is not invertible. Once again, we apply elementary row operations to ( A | I 3 ). 1 2 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 R 2 R 2 R 1 ,R 3 R 3 R 1 1 2 1 1 0 0 0 2 0 1 1 0 0 1 0 1 0 1 R 2 R 3 1 2 1 1 0 0 0 1 0 1 0 1 0 2 0 1 1 0 R 2 →− R 2 / 2 1 2 1 1 0 0 0 1 0 1 0 1 0 2 0 1 1 0 R 3 R 3 2 R 2 1 2 1 1 0 0 0 1 0 1 0 1 0 0 0 3 1 2 We have arrived at an augmented matrix whose left side has a zero row. Thus, A cannot be invertible: its rank (the number of linearly independent rows) is less than 3 (recall that elementary transformations are rank-preserving). On the other hand, the two non- zero rows of the left side of the augmented matrix are linearly independent (they are not multiples of each other), hence rank A = 2. 6. (a) Let β = { p 0 ,p 1 2 } be the standard basis in P 2 ( R ) (here, p k ( x ) = x k ). Then T ( p 0 ) = p 0 , T ( p 1 ) = 2 p 0 p 1 , and T ( p 2 ) = 4 p 0 + 2 p 1 p 2 . Let A = [ T ] β = 1 2 4 0 1 2 0 0 1 To compute A 1 , we apply row transformations to the augmented matrix: 1 2 4 1 0 0 0 1 2 0 1 0 0 0 1 0 0 1 R 1 →− R 1 ,R 2 →− R 2 ,R 3 →− R 3 1 2 4 1 0 0 0 1 2 0 1 0 0 0 1 0 0 1 R 1 R 1 +2 R 2 1 0 8 1 2 0 0 1 2 0 1 0 0 0 1 0 0 1
SOLUTIONS TO HOMEWORK 6 3 −→ R 1 R 1 +8 R 3 ,R 2 R 2 +2 R 3 1 0 0 1 2 8 0 1 0 0 1 2 0 0 1 0 0 1 Therefore, [ T 1 ] β = A 1 = 1 2 8 0 1 2 0 0

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

sol6 - SOLUTIONS TO HOMEWORK 6 Section 3.2 4(a We already...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online