Determine the value of k such that the following system has (i) a Unique solution (ii) No solution (iii) More than one solution (a) Kx + y + z = 1 x +Ky + z = 1 x + y + Kz = 1


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 180682 by Mastermind last updated on 15/Nov/22

$Determine the value of k such that the$ $following system has (i) a Unique$ $solution (ii) No solution (iii) More than$ $one solution$ $(a) Kx + y + z = 1$ $x + Ky + z = 1$ $x + y + Kz = 1$
	Answered by manolex last updated on 16/Nov/22
	$((k,1,1,1),(1,k,1,1),(1,1,k,1) ) ((1,1,k,1),(1,k,1,1),(k,1,1,1) ) −f_1 +f_2 −kf_1 +f_3 ((1,1,k,1),(0,(k−1),(1−k),0),(0,(−k+1),(−k^2 +1),(−k+1)) ) f_2 +f_1 ((1,1,k,1),(0,(k−1),(1−k),0),(0,0,(2−k−k^2 ),(−k+1)) ) k≠−2;1 z=(((−k+1))/((k+2)(−k+1)))=(1/(k+2)) (k−1)y+(1−k)[(1/(k+2))]=0 y−[(1/(k+2))]=0⇒y=(1/(k+2)) x+[(1/(k+2))]+k[(1/(k+2))]=1 x=1−((k+1)/(k+2)) x=(1/(k+2)) C.S={(1/(k+2));(1/(k+2));(1/(k+2))} if k=1 ((1,1,1,1),(0,0,0,(0])),(0,0,0,0) ) 1eq,3 incog; 0+0+0=0 compatible indeterminado z=t y=s x=1−t−s C.S={1−t−s;s;t} if k=−2 ((1,1,(−2),1),(0,(−3),3,0),(0,0,0,3) ) 0+0+0=3⇒incompatible .no solution resumen determinant ((k,x,y,z,(comentario)),((≠−2;1),( (1/(k+2))),(1/(k+2)),(1/(k+2)),(all equal)),(( 1),(1−t−s),( s),( t),(infinitas solution)),((−2),,,,(incompatible solution)))$
	$(\begin{matrix} k & 1 & 1 & 1 \\ 1 & k & 1 & 1 \\ 1 & 1 & k & 1 \end{matrix})$ $(\begin{matrix} 1 & 1 & k & 1 \\ 1 & k & 1 & 1 \\ k & 1 & 1 & 1 \end{matrix})$ $- f_{1} + f_{2}$ $- {k f}_{1} + f_{3}$ $(\begin{matrix} 1 & 1 & k & 1 \\ 0 & k - 1 & 1 - k & 0 \\ 0 & - k + 1 & - k^{2} + 1 & - k + 1 \end{matrix})$ $f_{2} + f_{1}$ $(\begin{matrix} 1 & 1 & k & 1 \\ 0 & k - 1 & 1 - k & 0 \\ 0 & 0 & 2 - k - k^{2} & - k + 1 \end{matrix})$ $k \neq - 2; 1$ $z = \frac{(- k + 1)}{(k + 2) (- k + 1)} = \frac{1}{k + 2}$ $(k - 1) y + (1 - k) [\frac{1}{k + 2}] = 0$ $y - [\frac{1}{k + 2}] = 0 \Rightarrow y = \frac{1}{k + 2}$ $x + [\frac{1}{k + 2}] + k [\frac{1}{k + 2}] = 1$ $x = 1 - \frac{k + 1}{k + 2}$ $x = \frac{1}{k + 2}$ $C . S = {\frac{1}{k + 2}; \frac{1}{k + 2}; \frac{1}{k + 2}}$ $i f k = 1$ $(\begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0] \\ 0 & 0 & 0 & 0 \end{matrix})$ $1 e q, 3 i n c o g;$ $0 + 0 + 0 = 0 c o m p a t i b l e i n d e t e r m i n a d o$ $z = t$ $y = s$ $x = 1 - t - s$ $C . S = {1 - t - s; s; t}$ $i f k = - 2$ $(\begin{matrix} 1 & 1 & - 2 & 1 \\ 0 & - 3 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{matrix})$ $0 + 0 + 0 = 3 \Rightarrow i n c o m p a t i b l e . n o s o l u t i o n$ $r e s u m e n$ $\begin{array}{cccc} k & x & y & z & c o m e n t a r i o \\ \neq - 2; 1 & \frac{1}{k + 2} & \frac{1}{k + 2} & \frac{1}{k + 2} & a l l e q u a l \\ 1 & 1 - t - s & s & t & i n f i n i t a s s o l u t i o n \\ - 2 & i n c o m p a t i b l e s o l u t i o n \end{array}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum