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College Algebra, 2013a

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604 Systems of Equations and Matrices<br />

⎡<br />

⎢<br />

⎣<br />

1 17<br />

1<br />

3<br />

0<br />

39<br />

− 2<br />

39<br />

− 2<br />

13<br />

0 1 0 − 10<br />

13<br />

− 8<br />

13<br />

0 0 1 − 2<br />

13<br />

1<br />

13<br />

15<br />

13<br />

3<br />

13<br />

⎤<br />

⎥<br />

⎦<br />

Replace R1 with<br />

−−−−−−−−−−→<br />

− 1 3 R2+R1<br />

⎡<br />

⎢<br />

⎣<br />

1 0 0<br />

9<br />

13<br />

0 1 0 − 10<br />

13<br />

− 8<br />

13<br />

0 0 1 − 2<br />

13<br />

2<br />

13<br />

− 7<br />

13<br />

15<br />

1<br />

13<br />

13<br />

3<br />

13<br />

⎤<br />

⎥<br />

⎦<br />

⎡<br />

We find A −1 ⎢<br />

= ⎣<br />

and<br />

A −1 A =<br />

9<br />

13<br />

− 10<br />

13<br />

− 8<br />

13<br />

− 2<br />

13<br />

⎡<br />

⎢<br />

⎣<br />

9<br />

13<br />

2<br />

13<br />

− 7<br />

13<br />

15<br />

1<br />

13<br />

− 10<br />

13<br />

− 8<br />

13<br />

− 2<br />

13<br />

13<br />

3<br />

13<br />

⎤<br />

2<br />

13<br />

− 7<br />

13<br />

15<br />

1<br />

13<br />

⎥<br />

⎦. To check our answer, we compute<br />

13<br />

3<br />

13<br />

⎤ ⎡<br />

⎤<br />

3 1 2<br />

⎥ ⎢<br />

⎦ ⎣ 0 −1 5<br />

2 1 4<br />

⎥<br />

⎦ =<br />

⎡ ⎤<br />

1 0 0<br />

⎢<br />

⎣ 0 1 0<br />

0 0 1<br />

⎥<br />

⎦ = I 3 ̌<br />

⎡<br />

3 1 2<br />

AA −1 ⎢<br />

= ⎣ 0 −1 5<br />

2 1 4<br />

⎤ ⎡<br />

⎥ ⎢<br />

⎦ ⎣<br />

9<br />

13<br />

− 10<br />

13<br />

− 8<br />

13<br />

− 2<br />

13<br />

2<br />

13<br />

− 7<br />

13<br />

15<br />

1<br />

13<br />

13<br />

3<br />

13<br />

⎤<br />

⎥<br />

⎦ =<br />

⎡ ⎤<br />

1 0 0<br />

⎢<br />

⎣ 0 1 0<br />

0 0 1<br />

⎥<br />

⎦ = I 3 ̌<br />

2. Each of the systems in this part has A as its coefficient matrix. The only difference between<br />

the systems is the constants which is the matrix B in the associated matrix equation AX = B.<br />

We solve each of them using the formula X = A −1 B.<br />

⎡<br />

(a) X = A −1 ⎢<br />

B = ⎣<br />

⎡<br />

(b) X = A −1 ⎢<br />

B = ⎣<br />

⎡<br />

(c) X = A −1 ⎢<br />

B = ⎣<br />

9<br />

13<br />

− 10<br />

13<br />

− 8<br />

13<br />

− 2<br />

13<br />

9<br />

13<br />

2<br />

13<br />

− 7<br />

13<br />

15<br />

1<br />

13<br />

− 10<br />

13<br />

− 8<br />

13<br />

− 2<br />

13<br />

9<br />

13<br />

13<br />

3<br />

13<br />

2<br />

13<br />

− 7<br />

13<br />

15<br />

1<br />

13<br />

− 10<br />

13<br />

− 8<br />

13<br />

− 2<br />

13<br />

13<br />

3<br />

13<br />

2<br />

13<br />

− 7<br />

13<br />

15<br />

1<br />

13<br />

13<br />

3<br />

13<br />

⎤ ⎡<br />

⎥ ⎢<br />

⎦ ⎣<br />

⎤ ⎡<br />

⎥ ⎢<br />

⎦ ⎣<br />

⎤ ⎡<br />

⎥ ⎢<br />

⎦ ⎣<br />

⎤ ⎡ ⎤<br />

26 −39<br />

⎥ ⎢<br />

39 ⎦ = ⎣ 91<br />

117 26<br />

⎤ ⎡ ⎤<br />

5<br />

4<br />

13<br />

⎥ ⎢ 19<br />

2 ⎦ = ⎣ 13<br />

9<br />

5<br />

13<br />

⎤ ⎡ ⎤<br />

9<br />

1<br />

13<br />

⎥ ⎢<br />

0 ⎦ = ⎣ − 10<br />

13<br />

0 − 2<br />

13<br />

⎥<br />

⎦. Our solution is (−39, 91, 26).<br />

⎥<br />

⎦. Weget ( 5<br />

13 , 19<br />

13 , 9<br />

13)<br />

.<br />

⎥<br />

⎦. We find ( 9<br />

13 , − 10<br />

13 , − 2 )<br />

13 .<br />

6<br />

In Example 8.4.1, we see that finding one inverse matrix can enable us to solve an entire family<br />

of systems of linear equations. There are many examples of where this comes in handy ‘in the<br />

wild’, and we chose our example for this section from the field of electronics. We also take this<br />

opportunity to introduce the student to how we can compute inverse matrices using the calculator.<br />

6 Note that the solution is the first column of the A −1 . The reader is encouraged to meditate on this ‘coincidence’.

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