First slide

System of linear equations

Question

$The system of equations k x + y + z = 1, x + k y + z = k and x + y + z k = k^{2} has no solution if k is$

Moderate

A

$- 2$
B

$- 1$
C

0
D

1

Solution

$\begin{array}{l} The augmented matrix of the system of equations is [\begin{matrix} k & 1 & 1 & 1 \\ 1 & k & 1 & k \\ 1 & 1 & k & k^{2} \end{matrix}] \\ R_{2} \to k R_{2} - R_{1}, R_{3} \to k R_{3} - R_{1} \\ \approx [\begin{matrix} k & 1 & 1 & 1 \\ 0 & k^{2} - 1 & k - 1 & k^{2} - 1 \\ 1 & k - 1 & k^{2} - 1 & k^{3} - 1 \end{matrix}] \\ \approx {(k - 1)}^{2} [\begin{matrix} k & 1 & 1 & 1 \\ 0 & k + 1 & 1 & k + 1 \\ 0 & 1 & k + 1 & k^{2} + k + 1 \end{matrix}] \\ R_{3} \to (k + 1) R_{3} - R_{2} \\ \approx {(k - 1)}^{2} [\begin{matrix} k & 1 & 1 & 1 \\ 0 & k + 1 & 1 & k + 1 \\ 0 & 0 & k^{2} + 2 k & k (k^{2} + 2 k + 1) \end{matrix}] \\ \approx k {(k - 1)}^{2} [\begin{matrix} k & 1 & 1 & 1 \\ 0 & k + 1 & 1 & k + 1 \\ 0 & 0 & k + 2 & (k^{2} + 2 k + 1) \end{matrix}] \end{array}$
$T h i r d e q u a t i o n o f t h e s y s t e m o f e q u a t i o n s i s (k + 2) z = {(k + 1)}^{2} z = \frac{{(k + 1)}^{2}}{k + 2}$
$I f k + 2 = 0, s y s t e m h a s n o s o l u t i o n, i t m e a n s f o r k = - 2, s y s t e m h a s n o s o l u t i o n .$

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