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 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

(\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}

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