06.09.2021 Views

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

492 ANSWERS TO SELECTED EXERCISES<br />

G.11 Exercises<br />

6.6<br />

1 The maximum is 7 and it occurs when x 1 =7,x 2 =<br />

0,x 3 =0,x 4 =3,x 5 =5,x 6 =0.<br />

2 Maximize and minimize the following if possible.<br />

All variables are nonnegative.<br />

(a) The minimum is −7 and it happens when x 1 =<br />

0,x 2 =7/2,x 3 =0.<br />

(b) The maximum is 7 and it occurs when x 1 =<br />

7,x 2 =0,x 3 =0.<br />

(c) The maximum is 14 and it happens when x 1 =<br />

7,x 2 = x 3 =0.<br />

(d) The minimum is 0 when x 1 = x 2 =0,x 3 =1.<br />

4 Find a solution to the following inequalities for x, y ≥<br />

0 if it is possible to do so. If it is not possible, prove<br />

it is not possible.<br />

(a) There is no solution to these inequalities with<br />

x 1 ,x 2 ≥ 0.<br />

(b) A solution is x 1 =8/5,x 2 = x 3 =0.<br />

(c) There will be no solution to these inequalities<br />

for which all the variables are nonnegative.<br />

(d) There is a solution when x 2 =2,x 3 =0,x 1 =<br />

0.<br />

(e) There is no solution to this system of inequalities<br />

because the minimum value of x 7 is not<br />

0.<br />

G.12 Exercises<br />

7.3<br />

1 Because the vectors which result are not parallel to<br />

the vector you begin with.<br />

3 λ → λ −1 and λ → λ m .<br />

5 Letx be the eigenvector. Then A m x = λ m x,A m x =<br />

Ax = λx and so<br />

λ m = λ<br />

Hence if λ ≠0, then<br />

and so |λ| =1.<br />

λ m−1 =1<br />

7<br />

9<br />

11<br />

⎛<br />

⎝<br />

⎧<br />

⎨<br />

−1 −1 7<br />

−1 0 4<br />

−1 −1 5<br />

⎫<br />

⎬<br />

⎞<br />

⎠, eigenvectors:<br />

⎫<br />

⎬<br />

⎧<br />

3 ⎨ 2<br />

1<br />

⎩ ⎭ ↔ 1, 1 ↔ 2. This is a defective matrix.<br />

⎩ ⎭<br />

1<br />

1<br />

⎛<br />

⎞<br />

−7 −12 30<br />

⎝ −3 −7 15 ⎠, eigenvectors:<br />

−3 −6 14<br />

⎧⎛<br />

⎨<br />

⎝ −2 ⎞ ⎛<br />

1 ⎠ , ⎝ 5 ⎞⎫<br />

⎧⎛<br />

⎬ ⎨<br />

0 ⎠<br />

⎩<br />

⎭ ↔−1, ⎝ 2 ⎞⎫<br />

⎬<br />

1 ⎠<br />

⎩ ⎭ ↔ 2<br />

0 1<br />

1<br />

This matrix is not defective because, even though<br />

λ = 1 is a repeated eigenvalue, it has a 2 dimensional<br />

eigenspace.<br />

⎛<br />

⎝<br />

3 −2 −1<br />

0 5 1<br />

0 2 4<br />

⎞<br />

⎠, eigenvectors:<br />

⎧⎛<br />

⎞ ⎛ ⎞⎫<br />

⎧⎛<br />

⎨ 1 0 ⎬ ⎨<br />

⎝ 0 ⎠ , ⎝ − 1 ⎠<br />

⎩<br />

2<br />

⎭ ↔ 3, ⎝<br />

⎩<br />

0 1<br />

This matrix is not defective.<br />

⎛<br />

13 ⎝ 5 2 −5 ⎞<br />

12 3 −10 ⎠, eigenvectors:<br />

12 4 −11<br />

⎧⎛<br />

⎨<br />

⎝ − ⎞ ⎛ ⎞⎫<br />

1 5<br />

3<br />

6<br />

⎬<br />

1 ⎠ , ⎝ 0 ⎠<br />

⎩<br />

⎭ ↔−1<br />

0 1<br />

15<br />

17<br />

−1<br />

1<br />

1<br />

⎞⎫<br />

⎬<br />

⎠<br />

⎭ ↔ 6<br />

This matrix is defective. In this case, there is only<br />

one eigenvalue, −1 of multiplicity 3 but the dimension<br />

of the eigenspace is only 2.<br />

⎛<br />

⎝<br />

1 26 −17<br />

4 −4 4<br />

−9 −18 9<br />

⎧⎛<br />

⎞⎫<br />

⎧⎛<br />

⎨ − 1 3<br />

⎬ ⎨<br />

⎝ 2 ⎠<br />

⎩<br />

3<br />

⎭ ↔ 0, ⎝<br />

⎩<br />

1<br />

⎧⎛<br />

⎨<br />

⎝ −1<br />

⎞⎫<br />

⎬<br />

0 ⎠<br />

⎩ ⎭ ↔ 18<br />

1<br />

⎞<br />

⎠, eigenvectors:<br />

−2<br />

1<br />

0<br />

⎞⎫<br />

⎬<br />

⎠<br />

⎭ ↔−12,<br />

⎛<br />

⎝ −2 1 2 ⎞<br />

⎧⎛<br />

⎨<br />

−11 −2 9 ⎠, eigenvectors: ⎝<br />

⎩<br />

−8 0 7<br />

This is defective.<br />

3<br />

4<br />

1<br />

4<br />

1<br />

⎞⎫<br />

⎬<br />

⎠<br />

⎭ ↔ 1

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!