for-what-value-of-k-the-system-of-equation-has-no-solution-x-2y-3z-1-2x-ky-5z-1-3x-4y-7z-1-

Question Number 5963 by Ashis last updated on 07/Jun/16

for what value of k the system of equation has no solution

x + 2 y + 3 z = 1

2 x + ky + 5 z = 1

3 x + 4 y + 7 z = 1

Commented by Yozzii last updated on 07/Jun/16

k = 3 ?

Commented by prakash jain last updated on 07/Jun/16

D = | \begin{matrix} 1 & 2 & 3 \\ 2 & k & 5 \\ 3 & 4 & 7 \end{matrix} | = 0

D = | \begin{matrix} 1 & 2 & 3 \\ 1 & k - 2 & 2 \\ 1 & 4 - k & 2 \end{matrix} | = 0 (R 2 \to R 2 - R 1, R 3 \to R 3 - R 2)

D = | \begin{matrix} 1 & 2 & 3 \\ 0 & k - 4 & - 1 \\ 0 & 2 - k & - 1 \end{matrix} | = 0 (R 2 \to R 2 - R 1, R 3 \to R 3 - R 1)

= - (k - 4) + (2 - k) = 0

= - k + 4 + 2 + - k = 0 \Rightarrow k = 3

This itself does not guarantee that equation

has no solution we also need to check if

a) at least one of D_{1}, D_{2} or D_{3} is non zero

b) if D_{1}, D_{2}, D_{3} are all zero then verify that no

solution exist by solving the equation .

D_{1} = | \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 1 & 4 & 7 \end{matrix} |

D_{1} = | \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \end{matrix} | = 0

D_{2} = | \begin{matrix} 1 & 1 & 3 \\ 2 & 1 & 5 \\ 3 & 1 & 7 \end{matrix} | = | \begin{matrix} 1 & 1 & 3 \\ 1 & 0 & 2 \\ 1 & 0 & 2 \end{matrix} | = 0

D_{3} = | \begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 4 & 1 \end{matrix} | = | \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 0 \end{matrix} | = 0

x + 2 y + 3 z = 1 \Rightarrow 2 x + 4 y + 6 z = 2

2 x + 3 y + 5 z = 1

f r o m t h e a b o v e t w o y + z = 1 \Rightarrow y = 1 - z

x = 1 - 2 y - 3 z \Rightarrow x = 1 - 2 (1 - z) - 3 z = - 1 - 5 z

3 x + 4 y + 7 z = 1

3 (- 1 - 5 z) + 4 (1 - z) + 7 z = 1

- 3 - 15 z + 4 - 4 z + 7 z = 1 \Rightarrow z = 0

y = 1, x = - 1

c h e c k s o l u t i o n (- 1, 1, 0) f o r k = 3

- 1 + 2 = 1 (eqn 1)

- 2 + 3 = 1 (eqn 2)

- 3 + 4 = 1 (eqn 3)

k = 3 gives unique solution .

Answered by Ashis last updated on 07/Jun/16

y e s 3 b u t h o w ?

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