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A First Course in Linear Algebra, 2017a

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132 Determ<strong>in</strong>ants<br />

You can verify that AA −1 = I and hence our answer is correct.<br />

♠<br />

We will look at another example of how to use this formula to f<strong>in</strong>d A −1 .<br />

Example 3.42: F<strong>in</strong>d the Inverse From a Formula<br />

F<strong>in</strong>d the <strong>in</strong>verse of the matrix<br />

⎡<br />

A =<br />

⎢<br />

⎣<br />

− 1 6<br />

− 5 6<br />

1<br />

2<br />

0<br />

1<br />

2<br />

1<br />

3<br />

− 1 2<br />

2<br />

3<br />

− 1 2<br />

⎤<br />

⎥<br />

⎦<br />

us<strong>in</strong>g the formula given <strong>in</strong> Theorem 3.40.<br />

Solution. <strong>First</strong>weneedtof<strong>in</strong>ddet(A). This step is left as an exercise and you should verify that det(A)=<br />

1<br />

6 . The <strong>in</strong>verse is therefore equal to A −1 = 1<br />

(1/6) adj(A)=6adj(A)<br />

We cont<strong>in</strong>ue to calculate as follows. Here we show the 2×2 determ<strong>in</strong>ants needed to f<strong>in</strong>d the cofactors.<br />

⎡<br />

1<br />

3<br />

− 1 2<br />

− 1 6<br />

− 1 2<br />

− 1 1<br />

⎤T<br />

6 3<br />

2 3<br />

− 1 −<br />

2<br />

− 5 6<br />

− 1 2<br />

− 5 2<br />

6 3<br />

1<br />

A −1 0 1 1<br />

1<br />

= 6<br />

−<br />

2<br />

2 2<br />

2<br />

0<br />

2 3<br />

− 1 2<br />

− 5 6<br />

− 1 −<br />

2<br />

− 5 2<br />

6 3<br />

⎢<br />

1<br />

0<br />

1 1<br />

1<br />

2<br />

2 2<br />

2<br />

0<br />

⎥<br />

⎣<br />

1<br />

∣ 3<br />

− 1 −<br />

2 ∣ ∣ − 1 6<br />

− 1 2 ∣ ∣<br />

− 1 1<br />

⎦<br />

6 3 ∣<br />

Expand<strong>in</strong>g all the 2 × 2 determ<strong>in</strong>ants, this yields<br />

⎡<br />

A −1 = 6<br />

⎢<br />

⎣<br />

1<br />

6<br />

1<br />

3<br />

− 1 6<br />

1<br />

3<br />

1<br />

6<br />

1<br />

6<br />

− 1 3<br />

1<br />

6<br />

1<br />

6<br />

⎤<br />

⎥<br />

⎦<br />

T<br />

⎡<br />

= ⎣<br />

1 2 −1<br />

2 1 1<br />

1 −2 1<br />

⎤<br />

⎦<br />

Aga<strong>in</strong>, you can always check your work by multiply<strong>in</strong>g A −1 A and AA −1 and ensur<strong>in</strong>g these products<br />

equal I.<br />

⎡<br />

1 1<br />

⎤<br />

⎡<br />

⎤ 2<br />

0 2 ⎡ ⎤<br />

1 2 −1<br />

A −1 A = ⎣ 2 1 1 ⎦<br />

− 1 1<br />

6 3<br />

− 1 1 0 0<br />

2<br />

⎢<br />

⎥<br />

1 −2 1 ⎣<br />

⎦ = ⎣ 0 1 0 ⎦<br />

0 0 1<br />

− 5 6<br />

2<br />

3<br />

− 1 2

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