find the value of x if the inverse of the matrix (((x+5 2)),((7 x)) ) is (((0 0)),((0 0)) )


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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

$t h e c a r a c t e r i s t i c p o l y n o m o f A i s$ $d e t (A - u I) = \| \begin{matrix} x + 5 - u 2 \\ 7 x - u \end{matrix} \|$ $= (x - u) (x - u + 5) - 14$ $= {(x - u)}^{2} + 5 x - 5 u - 14$ $= u^{2} - 2 x u + x^{2} + 5 x - 5 u - 14$ $= u^{2} - (2 x + 5) u + x^{2} + 5 x - 14 C a y l e y h a m i l t o n$ $t h e o r e m g i v e p_{c} (A) = 0 \Rightarrow$ $A^{2} - (2 x + 5) A + (x^{2} + 5 x - 14) I = 0 \Rightarrow$ $A^{2} - (2 x + 5) A = (x^{2} + 5 x - 14) I \Rightarrow$ $A (A - (2 x + 5) I) = (x^{2} + 5 x - 14) I \Rightarrow$ $A . \frac{A - (2 x + 5) I}{x^{2} + 5 x - 14} = I \Rightarrow A - (2 x + 5) I = 0 \Rightarrow$ $(\begin{matrix} x + 5 2 \\ 7 x \end{matrix}) - (\begin{matrix} 2 x + 5 0 \\ 0 2 x + 5 \end{matrix}) = 0 \Rightarrow$ $(\begin{matrix} - x 2 \\ 7 - x - 5 \end{matrix}) = 0 b u t t h a t i s i m p o s s i b l e t b e$ $v a l u e o f x d o n t e x i s t . . . .$
Commented by abdo mathsup 649 cc last updated on 19/May/18

$i f A^{- 1} = (\begin{matrix} 0 0 \\ 0 0 \end{matrix}) i s i n v e r s e o f A w e g e t$ $A . A^{- 1} = I \Rightarrow (\begin{matrix} 0 0 \\ 0 0 \end{matrix}) = (\begin{matrix} 1 0 \\ 0 1 \end{matrix})$ $a n d t h i s e q u a l i t y i s i m p o s s i b l e . . .$
	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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