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College Algebra, 2013a

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8.4 Systems of Linear Equations: Matrix Inverses 603<br />

constants in B. This answers the question as to why we would bother doing row operations on<br />

a super-sized augmented matrix to find A −1 instead of an ordinary augmented matrix to solve a<br />

system; by finding A −1 we have done all of the row operations we ever need to do, once and for all,<br />

since we can quickly solve any equation AX = B using one multiplication, A −1 B.<br />

⎡<br />

Example 8.4.1. Let A = ⎣<br />

3 1 2<br />

0 −1 5<br />

2 1 4<br />

⎤<br />

⎦<br />

1. Use row operations to find A −1 . Check your answer by finding A −1 A and AA −1 .<br />

2. Use A −1 to solve the following systems of equations<br />

(a)<br />

⎧<br />

⎨<br />

⎩<br />

3x + y +2z = 26<br />

−y +5z = 39<br />

2x + y +4z = 117<br />

(b)<br />

⎧<br />

⎨<br />

⎩<br />

3x + y +2z = 4<br />

−y +5z = 2<br />

2x + y +4z = 5<br />

(c)<br />

⎧<br />

⎨<br />

⎩<br />

3x + y +2z = 1<br />

−y +5z = 0<br />

2x + y +4z = 0<br />

Solution.<br />

1. We begin with a super-sized augmented matrix and proceed with Gauss-Jordan elimination.<br />

⎡<br />

⎣<br />

⎡<br />

⎣<br />

⎡<br />

⎣<br />

⎡<br />

⎣<br />

⎡<br />

⎣<br />

⎡<br />

⎣<br />

3 1 2 1 0 0<br />

0 −1 5 0 1 0<br />

2 1 4 0 0 1<br />

1<br />

1<br />

3<br />

2<br />

3<br />

1<br />

3<br />

0 0<br />

0 −1 5 0 1 0<br />

2 1 4 0 0 1<br />

⎤<br />

⎦<br />

⎤<br />

⎦<br />

1 2 1<br />

1<br />

3 3 3<br />

0 0<br />

0 −1 5 0 1 0<br />

0<br />

1<br />

3<br />

8<br />

3<br />

− 2 3<br />

0 1<br />

Replace R1<br />

−−−−−−−→<br />

with 1 3 R1<br />

⎡<br />

Replace R3 with<br />

−−−−−−−−−−→<br />

−2R1+R3<br />

⎤<br />

⎦<br />

1 2 1<br />

1<br />

3 3 3<br />

0 0<br />

0 1 −5 0 −1 0<br />

0<br />

1<br />

3<br />

8<br />

3<br />

− 2 3<br />

0 1<br />

1 2 1<br />

1<br />

3 3 3<br />

0 0<br />

0 1 −5 0 −1 0<br />

0 0<br />

13<br />

3<br />

− 2 3<br />

1<br />

3<br />

1<br />

1 2 1<br />

1<br />

3 3 3<br />

0 0<br />

0 1 −5 0 −1 0<br />

0 0 1 − 2<br />

13<br />

1<br />

13<br />

3<br />

13<br />

Replace R2<br />

−−−−−−−→<br />

with (−1)R2<br />

⎤<br />

⎦<br />

⎤<br />

⎦<br />

⎤<br />

⎦<br />

2 1<br />

3 3<br />

0 0<br />

0 −1 5 0 1 0<br />

1<br />

1<br />

3<br />

⎤<br />

⎣<br />

⎦<br />

2 1 4 0 0 1<br />

⎡<br />

1 2 1<br />

1<br />

3 3 3<br />

0 0<br />

⎣ 0 −1 5 0 1 0<br />

1<br />

0<br />

3<br />

⎡<br />

⎣<br />

Replace R3 with<br />

−−−−−−−−−−→<br />

− 1 3 R2+R3<br />

Replace R3<br />

−−−−−−−→<br />

with 3 13 R3<br />

1<br />

1<br />

3<br />

8<br />

3<br />

− 2 3<br />

0 1<br />

⎤<br />

⎦<br />

2 1<br />

3 3<br />

0 0<br />

0 1 −5 0 −1 0<br />

1<br />

0<br />

3<br />

⎡<br />

⎣<br />

⎡<br />

⎣<br />

Replace R1 with<br />

− 2 3 R3+R1<br />

−−−−−−−−−−−−→<br />

Replace R2 with<br />

5R3+R2<br />

1<br />

1<br />

3<br />

8<br />

3<br />

− 2 3<br />

0 1<br />

⎤<br />

⎦<br />

2 1<br />

3 3<br />

0 0<br />

0 1 −5 0 −1 0<br />

0 0<br />

13<br />

3<br />

− 2 3<br />

1<br />

1<br />

3<br />

1<br />

3<br />

1<br />

2 1<br />

3 3<br />

0 0<br />

0 1 −5 0 −1 0<br />

0 0 1 − 2<br />

13<br />

⎡<br />

⎢<br />

⎣<br />

1<br />

13<br />

3<br />

13<br />

⎤<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 />

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