# F10HW_6 - M341. F 2010. Homework 6. Section 3.1. I. 5.b)...

This preview shows pages 1–3. Sign up to view the full content.

M341. F 2010. Homework 6. Section 3.1. I. 5.b) The answer for a) is at the end of the book. b) 0 × ( 1) 4+1 A 4 , 1 (it’s useless to compute or even to explicitly write A 4 , 1 since it is multiplied by 0) +2 0 4 0 4 2 7 1 3 0 (the sign is ( 1) 6 = 1, but we may use the determinant of signs or just remember that the signs are alternatively + and -, the previous one was -, this one will be +) 0 5 0 4 1 7 1 0 0 + 5 0 5 4 4 1 2 1 0 3 . Anyway, the determinant we want to compute is equal to: +2 0 4 0 4 2 7 1 3 0 0 5 0 4 1 7 1 0 0 + 5 0 5 4 4 1 2 1 0 3 The ﬁrst of these three 3 × 3 determinants is easier to compute by using theorem 3.3 (3): if we interchange row 1 and row 3, we just change the sign of the determinant. (You might also use theorem 3.10 and make an expansion along the ﬁrst row) +2 0 4 0 4 2 7 1 3 0 = 2 1 3 0 4 2 7 0 4 0 = 8 ( 1 0 4 7 ) = 56 . 0 5 0 4 1 7 1 0 0 = ( 1)( 1) ( 5 7 1 0 ) = 35 5 0 5 4 4 1 2 1 0 3 = 5[ ( 5 4 1 2 ) + 3 ( 0 5 4 1 ) ] = 230 . The determinant is 251. You could check with a graphic calculator or a software. 9. I just give the results. About the method, see theorem 3.1 (1). a) 7, b) 18 (we take the absolute value) c) 12, d) 0 (this is because the two vectors are co-linear: the parallelogram is ﬂat. 10. There is a small inconvenience, not diﬃculty but an inconvenience. We would like the determinant, not its absolute value, to be the area of the parallelogram. This is possible if we give an orientation to the parallelogram. If we consider x as the ﬁrst vector and y as the second vector, the couple ( x , y ) and the parallelogram is positive if we go from 1

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

View Full Document
x to y , turning counter clockwise (as we do when running from i to j in physics or in high school). ( x , y ) and the parallelogram are negative if we go from x to y , turning clockwise. with this deﬁnition, the oriented area of the parallelogram is equal to the determinant. We did in class problem 15 from 3.2. If we take this deﬁnition for the determinant, to prove part 1 of theorem 3.1, it is suﬃcient to observe that if we multiply the length of either x or y by c , the area is multiplied by | c | , for the (not- oriented area)it does not matter which vector is ﬁrst, which vector is second, and ﬁnally it is easy to see that if we replace one of the vectors, say y by y + k x , the area of the parallelogram is unchanged (neither the basis nor the height is changed). Conclusion: the area of the parallelogram on
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/05/2011 for the course M 341 taught by Professor Hietmann during the Spring '08 term at University of Texas.

### Page1 / 6

F10HW_6 - M341. F 2010. Homework 6. Section 3.1. I. 5.b)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online