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A First Course in Linear Algebra, 2017a

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539<br />

2.1.7 0A =(0 + 0)A = 0A + 0A.Nowadd−(0A) to both sides. Then 0 = 0A.<br />

2.1.8 A +(−1)A =(1 +(−1))A = 0A = 0. Therefore, from the uniqueness of the additive <strong>in</strong>verse proved<br />

<strong>in</strong> the above Problem 2.1.5, it follows that −A =(−1)A.<br />

[ ]<br />

−3 −6 −9<br />

2.1.9 (a)<br />

−6 −3 −21<br />

[<br />

]<br />

8 −5 3<br />

(b)<br />

−11 5 −4<br />

(c) Not possible<br />

[ ]<br />

−3 3 4<br />

(d)<br />

6 −1 7<br />

(e) Not possible<br />

(f) Not possible<br />

⎡<br />

−3 −6<br />

2.1.10 (a) ⎣ −9 −6<br />

−3 3<br />

⎤<br />

⎦<br />

(b) Not possible.<br />

⎡<br />

11<br />

⎤<br />

2<br />

(c) ⎣ 13 6 ⎦<br />

−4 2<br />

(d) Not possible.<br />

⎡ ⎤<br />

7<br />

(e) ⎣ 9 ⎦<br />

−2<br />

(f) Not possible.<br />

(g) Not possible.<br />

[ ]<br />

2<br />

(h)<br />

−5<br />

2.1.11 (a)<br />

(b)<br />

[<br />

⎡<br />

⎣<br />

1 −2<br />

−2 −3<br />

3 0 −4<br />

−4 1 6<br />

5 1 −6<br />

]<br />

(c) Not possible<br />

⎤<br />

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