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A First Course in Linear Algebra, 2017a

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96 Matrices<br />

Exercise 2.1.42 Let<br />

⎡<br />

A = ⎣<br />

1 2 3<br />

2 1 4<br />

4 5 10<br />

F<strong>in</strong>d A −1 if possible. If A −1 does not exist, expla<strong>in</strong> why.<br />

Exercise 2.1.43 Let<br />

A =<br />

⎡<br />

⎢<br />

⎣<br />

⎤<br />

⎦<br />

1 2 0 2<br />

1 1 2 0<br />

2 1 −3 2<br />

1 2 1 2<br />

F<strong>in</strong>d A −1 if possible. If A −1 does not exist, expla<strong>in</strong> why.<br />

Exercise 2.1.44 Us<strong>in</strong>g the <strong>in</strong>verse of the matrix, f<strong>in</strong>d the solution to the systems:<br />

⎤<br />

⎥<br />

⎦<br />

(a) [<br />

2 4<br />

1 1<br />

(b) [<br />

2 4<br />

1 1<br />

][<br />

x<br />

y<br />

][<br />

x<br />

y<br />

] [<br />

1<br />

=<br />

2<br />

] [<br />

2<br />

=<br />

0<br />

]<br />

]<br />

Now give the solution <strong>in</strong> terms of a and b to<br />

[<br />

2 4<br />

1 1<br />

][<br />

x<br />

y<br />

] [<br />

a<br />

=<br />

b<br />

]<br />

Exercise 2.1.45 Us<strong>in</strong>g the <strong>in</strong>verse of the matrix, f<strong>in</strong>d the solution to the systems:<br />

(a)<br />

(b)<br />

⎡<br />

⎣<br />

⎡<br />

⎣<br />

1 0 3<br />

2 3 4<br />

1 0 2<br />

1 0 3<br />

2 3 4<br />

1 0 2<br />

⎤⎡<br />

⎦⎣<br />

⎤⎡<br />

⎦⎣<br />

x<br />

y<br />

z<br />

x<br />

y<br />

z<br />

⎤<br />

⎡<br />

⎦ = ⎣<br />

⎤<br />

⎡<br />

⎦ = ⎣<br />

1<br />

0<br />

1<br />

3<br />

−1<br />

−2<br />

⎤<br />

⎦<br />

⎤<br />

⎦<br />

Now give the solution <strong>in</strong> terms of a,b, and c to the follow<strong>in</strong>g:<br />

⎡ ⎤⎡<br />

⎤ ⎡ ⎤<br />

1 0 3 x a<br />

⎣ 2 3 4 ⎦⎣<br />

y ⎦ = ⎣ b ⎦<br />

1 0 2 z c

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