find-the-value-of-x-if-the-inverse-of-the-matrix-x-5-2-7-x-is-0-0-0-0-

Question Number 35426 by Rio Mike last updated on 18/May/18

f i n d t h e v a l u e o f x i f t h e i n v e r s e

o f t h e m a t r i x (\begin{matrix} x + 5 2 \\ 7 x \end{matrix}) i s

(\begin{matrix} 0 0 \\ 0 0 \end{matrix})

Commented by prof Abdo imad last updated on 19/May/18

t h e c a r a c t e r i s t i c p o l y n o m o f A i s

d e t (A - u I) = | \begin{matrix} x + 5 - u 2 \\ 7 x - u \end{matrix} |

= (x - u) (x - u + 5) - 14

= {(x - u)}^{2} + 5 x - 5 u - 14

= u^{2} - 2 x u + x^{2} + 5 x - 5 u - 14

= u^{2} - (2 x + 5) u + x^{2} + 5 x - 14 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 p_{c} (A) = 0 \Rightarrow

A^{2} - (2 x + 5) A + (x^{2} + 5 x - 14) I = 0 \Rightarrow

A^{2} - (2 x + 5) A = (x^{2} + 5 x - 14) I \Rightarrow

A (A - (2 x + 5) I) = (x^{2} + 5 x - 14) I \Rightarrow

A . \frac{A - (2 x + 5) I}{x^{2} + 5 x - 14} = I \Rightarrow A - (2 x + 5) I = 0 \Rightarrow

(\begin{matrix} x + 5 2 \\ 7 x \end{matrix}) - (\begin{matrix} 2 x + 5 0 \\ 0 2 x + 5 \end{matrix}) = 0 \Rightarrow

(\begin{matrix} - x 2 \\ 7 - x - 5 \end{matrix}) = 0 b u t t h a t i s i m p o s s i b l e t b e

v a l u e o f x d o n t e x i s t \dots .

Commented by abdo mathsup 649 cc last updated on 19/May/18

i f A^{- 1} = (\begin{matrix} 0 0 \\ 0 0 \end{matrix}) i s i n v e r s e o f A w e g e t

A . A^{- 1} = I \Rightarrow (\begin{matrix} 0 0 \\ 0 0 \end{matrix}) = (\begin{matrix} 1 0 \\ 0 1 \end{matrix})

a n d t h i s e q u a l i t y i s i m p o s s i b l e \dots

Answered by Rio Mike last updated on 19/May/18

i f t h e i n v e r s e o f a m a t r i x i s

(\begin{matrix} 0 0 \\ 0 0 \end{matrix}) t h e i t i s a s i n g u l a r m a t r i x

h e n c e

x (x + 5) - 2 (7) = 0

x^{2} + 5 x - 14 = 0

x^{2} - 2 x + 7 x - 14 = 0

x (x - 2) + 7 (x - 2)

(x - 2) (x + 7) = 0

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