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A First Course in Linear Algebra, 2017a

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3.2. Applications of the Determ<strong>in</strong>ant 131<br />

Example 3.41: F<strong>in</strong>d Inverse Us<strong>in</strong>g the Determ<strong>in</strong>ant<br />

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

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

⎡<br />

A = ⎣<br />

1 2 3<br />

3 0 1<br />

1 2 1<br />

⎤<br />

⎦<br />

Solution. Accord<strong>in</strong>g to Theorem 3.40,<br />

A −1 = 1<br />

det(A) adj(A)<br />

<strong>First</strong> we will f<strong>in</strong>d the determ<strong>in</strong>ant of this matrix. Us<strong>in</strong>g Theorems 3.16, 3.18, and3.21, we can first<br />

simplify the matrix through row operations. <strong>First</strong>, add −3 times the first row to the second row. Then add<br />

−1 times the first row to the third row to obta<strong>in</strong><br />

⎡<br />

B = ⎣<br />

1 2 3<br />

0 −6 −8<br />

0 0 −2<br />

By Theorem 3.21,det(A)=det(B). By Theorem 3.13,det(B)=1 ×−6 ×−2 = 12. Hence, det(A)=12.<br />

Now, we need to f<strong>in</strong>d adj(A). To do so, first we will f<strong>in</strong>d the cofactor matrix of A.Thisisgivenby<br />

⎡<br />

−2 −2<br />

⎤<br />

6<br />

cof(A)= ⎣ 4 −2 0 ⎦<br />

2 8 −6<br />

Here, the ij th entry is the ij th cofactor of the orig<strong>in</strong>al matrix A which you can verify. Therefore, from<br />

Theorem 3.40, the<strong>in</strong>verseofA is given by<br />

⎡<br />

⎤<br />

⎡<br />

A −1 = 1 ⎣<br />

12<br />

−2 −2 6<br />

4 −2 0<br />

2 8 −6<br />

⎤<br />

⎦<br />

T<br />

⎤<br />

⎦<br />

− 1 6<br />

=<br />

− 1<br />

⎢ 6<br />

− 1 6<br />

⎣<br />

1<br />

3<br />

1<br />

6<br />

2<br />

3<br />

1<br />

2<br />

0 − 1 2<br />

⎥<br />

⎦<br />

Remember that we can always verify our answer for A −1 . Compute the product AA −1 and A −1 A and<br />

make sure each product is equal to I.<br />

Compute A −1 A as follows<br />

⎡<br />

− 1 6<br />

1<br />

3<br />

A −1 A =<br />

− 1<br />

⎢ 6<br />

− 1 6<br />

⎣<br />

1<br />

6<br />

2<br />

3<br />

1<br />

2<br />

0 − 1 2<br />

⎤<br />

⎡<br />

⎣<br />

⎥<br />

⎦<br />

1 2 3<br />

3 0 1<br />

1 2 1<br />

⎤<br />

⎡<br />

⎦ = ⎣<br />

1 0 0<br />

0 1 0<br />

0 0 1<br />

⎤<br />

⎦ = I

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