Determine the values of b so that the system of linear equations { ((x+2y+z=1)),((2x+by+2z=2)),((4x+8y+b^2 z=2b)) :} has (a) no solution (b) a unique solution (c) infinitely many solutions


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 182900 by greougoury555 last updated on 16/Dec/22
$Determine the values of b so that the system of linear equations { ((x+2y+z=1)),((2x+by+2z=2)),((4x+8y+b^2 z=2b)) :} has (a) no solution (b) a unique solution (c) infinitely many solutions$
$D e t e r m i n e t h e v a l u e s o f b s o t h a t$ $t h e s y s t e m o f l i n e a r e q u a t i o n s$ ${\begin{cases} x + 2 y + z = 1 \\ 2 x + b y + 2 z = 2 \\ 4 x + 8 y + b^{2} z = 2 b \end{cases}$ $h a s (a) n o s o l u t i o n$ $(b) a u n i q u e s o l u t i o n$ $(c) i n f i n i t e l y m a n y s o l u t i o n s$
	Answered by cortano1 last updated on 16/Dec/22
	$(a) (((1 2 1 1)),((0 b−4 0 0)),((0 0 b^2 −4 2b−4)) ) { ((x=1−2y−z)),(((b−4)y = 0)),(((b^2 −4)z = 2b−4)) :} has no solution if { ((b^2 −4 =0)),((2b−4 ≠ 0)) :} i.e b =− 2$
	$(a) (\begin{matrix} 1 2 1 1 \\ 0 b - 4 0 0 \\ 0 0 b^{2} - 4 2 b - 4 \end{matrix})$ ${\begin{cases} x = 1 - 2 y - z \\ (b - 4) y = 0 \\ (b^{2} - 4) z = 2 b - 4 \end{cases}$ $h a s n o s o l u t i o n i f {\begin{cases} b^{2} - 4 = 0 \\ 2 b - 4 \neq 0 \end{cases}$ $i . e b = - 2$
	Commented by manolex last updated on 16/Dec/22

	$a n a l i z a n d o [r e s u l t a d o s} d e s i r C o r t a n o$ $b = 4 (\begin{matrix} 1 & 2 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 12 & 4 \end{matrix})$ $12 z = 4 \Rightarrow z = 3$ $x + 2 y + 3 = 4$ $y = t$ $x = 1 - 2 t$ $s o l u t i o n b = 4 \Rightarrow (1 - 2 t; t; 3) i n f i n i t e s o l u t i o n$ $b = 2$ $(\begin{matrix} 1 & 2 & 1 & 1 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})$ $- 2 y = 0 \Rightarrow y = 0$ $x + z = 1$ $z = t$ $x = 1 - t$ $s o l u t i o n b = 2 \Rightarrow (1 - t; 0; t) i n f i n i t e s o l u t i o n$ $b = - 2$ $(\begin{matrix} 1 & 2 & 1 & 1 \\ 0 & - 6 & 0 & 0 \\ 0 & 0 & 0 & 8 \end{matrix})$ $0 = 8 \Rightarrow n o s o l u t i o n$ $r e s u l t b = - 2 n o s o l u t i o n$ $b \neq 2, - 2, 4$ $(\begin{matrix} 1 & 2 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & b + 2 & 2 \end{matrix})$ $(b + 2) z = 2 \Rightarrow z = \frac{2}{b + 2}$ $y = 0$ $x + 2 y + z = 1$ $x = 1 - \frac{2}{b + 2}$ $r e s u l t (1 - \frac{2}{b + 2}; 0; \frac{2}{b + 2}) u n i q u e s o l u t i o n$ $c o n c l u s i o n$ $\begin{array}{ccccc} b & x & y & z & c o n c l u s i o n \\ 4 & 1 - 2 t & t & 3 & i n f i n i t e s o l u t i o n \\ 2 & 1 - t & 0 & t & i n f i n i t e s o l u t i o n \\ - 2 & 0 = 8 & n o s o l u t i o n \\ \neq 2, - 2, 4 & 1 - \frac{2}{b + 2} & 0 & \frac{2}{b + 2} & u n i q u e s o l u t i o n \end{array}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum