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A First Course in Linear Algebra, 2017a

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540 Selected Exercise Answers<br />

(d)<br />

(e)<br />

⎡<br />

⎣<br />

−4<br />

−5<br />

−1<br />

⎤<br />

−6<br />

−3 ⎦<br />

−2<br />

]<br />

[ 8 1 −3<br />

7 6 −6<br />

2.1.12<br />

[ −1 −1<br />

3 3<br />

][ x y<br />

]<br />

z w<br />

[<br />

Solution is: w = −y,x = −z so the matrices are of the form<br />

⎡<br />

2.1.13 X T Y = ⎣<br />

0 −1 −2<br />

0 −1 −2<br />

0 1 2<br />

⎤<br />

⎦,XY T = 1<br />

=<br />

=<br />

[ −x − z −w − y<br />

3x + 3z 3w + 3y<br />

[ ] 0 0<br />

0 0<br />

x<br />

−x<br />

y<br />

−y<br />

]<br />

.<br />

]<br />

2.1.14<br />

[ ][ ]<br />

1 2 1 2<br />

3 4 3 k<br />

[ ][ ]<br />

1 2 1 2<br />

3 k 3 4<br />

=<br />

=<br />

[ ]<br />

7 2k + 2<br />

15 4k + 6<br />

[<br />

]<br />

7 10<br />

3k + 3 4k + 6<br />

Thus you must have<br />

3k + 3 = 15<br />

2k + 2 = 10<br />

, Solution is: [k = 4]<br />

2.1.15<br />

[ ][ ]<br />

1 2 1 2<br />

3 4 1 k<br />

[ ][ ]<br />

1 2 1 2<br />

1 k 3 4<br />

=<br />

=<br />

[ ] 3 2k + 2<br />

7 4k + 6<br />

[<br />

]<br />

7 10<br />

3k + 1 4k + 2<br />

However, 7 ≠ 3 and so there is no possible choice of k which will make these matrices commute.<br />

[<br />

2.1.16 Let A =<br />

1 −1<br />

−1 1<br />

] [ 1 1<br />

,B =<br />

1 1<br />

] [ 2 2<br />

,C =<br />

2 2<br />

]<br />

.<br />

[<br />

[<br />

1 −1<br />

−1 1<br />

1 −1<br />

−1 1<br />

][ ] 1 1<br />

1 1<br />

][ ] 2 2<br />

2 2<br />

=<br />

=<br />

[ ] 0 0<br />

0 0<br />

[ ] 0 0<br />

0 0

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