Find value of α such that the following system has infinite many solutions x − 3z = −3 −2x − αy + z = 2 x + 2y + αz = 1


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 42375 by Joel578 last updated on 24/Aug/18

$Find value of α such that the following system$ $has infinite many solutions$ $x - 3 z = - 3$ $- 2 x - α y + z = 2$ $x + 2 y + α z = 1$
Commented by Joel578 last updated on 24/Aug/18

$What is the idea to solve this problem ?$
Commented by maxmathsup by imad last updated on 24/Aug/18
$first we must have det A =0 (if detA≠0 weget a unique solution) with A = (((1 0 −3)),((−2 −α 1)) ) (1 2 α) det A =−α^2 −2 +2(6) −3α =−α^2 −3α +10 det A =0 ⇒α^2 +3α −10 =0 Δ =9 +40 =49 ⇒α_1 = ((−3 +7)/2) =2 and α_2 = ((−3−7)/2) =−5 for α =2 we get the system { ((x−3z =−3)),((−2x−2y+z =2)) :} ⇒ {x+2y +2z =1 { ((x−3z =−3)),((−x+3z =3 and x+2y +2z =1 ⇒ { ((x−3z =−3)),((2y =1−x−2z)) :})) :} ⇒ { ((x =3z−3)),((y=(1/2){1−3z +3−2z}=(1/2){2−5z} ⇒)) :} (x,y,z) =(3z−3,1−(5/2)z ,z) infinite solution for α =−5 we follow the same method ....$
$f i r s t w e m u s t h a v e d e t A = 0 (i f d e t A \neq 0 w e g e t a u n i q u e s o l u t i o n)$ $w i t h A = (\begin{matrix} 1 0 - 3 \\ - 2 - α 1 \end{matrix})$ $(1 2 α)$ $d e t A = - α^{2} - 2 + 2 (6) - 3 α = - α^{2} - 3 α + 10$ $d e t A = 0 \Rightarrow α^{2} + 3 α - 10 = 0$ $Δ = 9 + 40 = 49 \Rightarrow α_{1} = \frac{- 3 + 7}{2} = 2 a n d α_{2} = \frac{- 3 - 7}{2} = - 5$ $f o r α = 2 w e g e t t h e s y s t e m {\begin{cases} x - 3 z = - 3 \\ - 2 x - 2 y + z = 2 \end{cases} \Rightarrow$ ${x + 2 y + 2 z = 1$ ${\begin{cases} x - 3 z = - 3 \\ - x + 3 z = 3 a n d x + 2 y + 2 z = 1 \Rightarrow {\begin{cases} x - 3 z = - 3 \\ 2 y = 1 - x - 2 z \end{cases} \end{cases}$ $\Rightarrow {\begin{cases} x = 3 z - 3 \\ y = \frac{1}{2} {1 - 3 z + 3 - 2 z} = \frac{1}{2} {2 - 5 z} \Rightarrow \end{cases}$ $(x, y, z) = (3 z - 3, 1 - \frac{5}{2} z, z) i n f i n i t e s o l u t i o n$ $f o r α = - 5 w e f o l l o w t h e s a m e m e t h o d . . . .$
Commented by Joel578 last updated on 25/Aug/18

$t h a n k y o u v e r y m u c h$
Commented by math khazana by abdo last updated on 26/Aug/18

$y o u a r e w e l c o m e .$
Commented by Joel578 last updated on 27/Aug/18

$Sir, if α = - 5, the system will have no solution$
	Answered by behi83417@gmail.com last updated on 24/Aug/18

	$1 x + 0 y - 3 z = - 3$ $- 2 x - α y + 1 z = 2$ $1 x + 2 y + α z = 1$ $▴ = 1 \times \| \begin{matrix} - α 1 \\ 2 α \end{matrix} \| - 0 \times \| \begin{matrix} - 2 1 \\ 1 α \end{matrix} \| - 3 \times \| \begin{matrix} - 2 - α \\ 1 2 \end{matrix} \| \neq 0$ $\Rightarrow - α^{2} - 2 - 3 (- 4 + α) \neq 0$ $\Rightarrow α^{2} + 3 α - 10 \neq 0$ $\Rightarrow α \neq \frac{- 3 \pm \sqrt{9 + 40}}{2} \neq \frac{- 3 \pm 7}{2} \neq 2, - 5 . ◼$
	Commented by maxmathsup by imad last updated on 24/Aug/18

	$s i r b e h i i f d e t M \neq 0 t h e s y s t e m h a v e o n e s o l u t i o n . . .$
	Commented by behi83417@gmail.com last updated on 24/Aug/18

	$y e s s i r .$ $w h e n ▴ = 0 \Rightarrow i . e : α = 2 o r - 5, t h e$ $s y s t e m h a v e m a n y s o l u t i o n s .$
	Commented by Joel578 last updated on 25/Aug/18

	$t h a n k y o u v e r y m u c h$
	Commented by Joel578 last updated on 27/Aug/18

	$If α = - 5, the system will have no solution$ $Pls check$
	Commented by Joel578 last updated on 27/Aug/18

	$(\begin{matrix} 1 & 0 & - 3 & - 3 \\ - 2 & 5 & 1 & 2 \\ 1 & 2 & - 5 & 1 \end{matrix})$ $R_{2} + 2 R_{1} \to R_{2} (\begin{matrix} 1 & 0 & - 3 & - 3 \\ 0 & 5 & - 5 & - 4 \\ 1 & 2 & - 5 & 1 \end{matrix})$ $R_{3} - R_{1} \to R_{3} (\begin{matrix} 1 & 0 & - 3 & - 3 \\ 0 & 5 & - 5 & - 4 \\ 0 & 2 & - 2 & 4 \end{matrix})$ $R_{2} \leftrightarrow R_{3} (\begin{matrix} 1 & 0 & - 3 & - 3 \\ 0 & 2 & - 2 & 4 \\ 0 & 5 & - 5 & - 4 \end{matrix})$ $\frac{1}{2} R_{2} \leftrightarrow R_{2} (\begin{matrix} 1 & 0 & - 3 & - 3 \\ 0 & 1 & - 1 & 2 \\ 0 & 5 & - 5 & - 4 \end{matrix})$ $R_{3} - 5 R_{2} \to R_{3} (\begin{matrix} 1 & 0 & - 3 & - 3 \\ 0 & 1 & - 1 & 2 \\ 0 & 0 & 0 & - 14 \end{matrix})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum