what-procedure-will-you-use-to-find-the-inverse-of-A-2-1-9-1-5-1-3-0-3-

Question Number 85131 by Rio Michael last updated on 19/Mar/20

what procedure will you use to

find the inverse of

A = (\begin{matrix} 2 & 1 & 9 \\ 1 & 5 & 1 \\ 3 & 0 & 3 \end{matrix})

Commented by jagoll last updated on 19/Mar/20

(1) find ∣ A ∣

(2) find adj (A)

(3) if ∣ A ∣ \neq 0 then

A^{- 1} = \frac{1}{∣ A ∣} \times adj (A)

Commented by Rio Michael last updated on 19/Mar/20

thanks

Commented by mathmax by abdo last updated on 19/Mar/20

p (A) = d e t (A - x I) = | \begin{matrix} 2 - x 1 9 \\ 1 5 - x 1 \end{matrix} |

∣ 3 0 3 - x ∣

= (2 - x) | \begin{matrix} 5 - x 1 \\ 0 3 - x \end{matrix} | - | \begin{matrix} 1 9 \\ 0 3 - x \end{matrix} | + 3 | \begin{matrix} 1 9 \\ 5 - x 1 \end{matrix} |

= (2 - x) (5 - x) (3 - x) - (3 - x) + 3 (1 - 9 (5 - x))

= (10 - 7 x + x^{2}) (3 - x) - 3 + x + 3 (1 - 45 + 9 x)

= (x^{2} - 7 x + 10) (3 - x) + 28 x - 135

= 3 x^{2} - x^{3} - 21 x + 7 x^{2} + 30 - 10 x + 28 x - 135

= - x^{3} + 10 x^{2} - 31 x + 28 x - 105

= - x^{3} + 10 x^{2} - 3 x - 105

c a y l e y h a m i l t o n t h e o r e m g i v e A^{3} - 10 A^{2} + 3 A + 105 I = 0 \Rightarrow

A (A^{2} - 10 A + 3 I) = - 105 I \Rightarrow

A \times (- \frac{1}{105} (A^{2} - 10 A + 3 I)) = I \Rightarrow

A^{- 1} = - \frac{1}{105} (A^{2} - 10 A + 3 I)

Commented by mathmax by abdo last updated on 19/Mar/20

a l s o w e c a n u s e A^{- 1} = \frac{t (c o m A)}{d e t A} t h i s m e t h o d i s s i m p l e \dots