For what real values of k, the system of equations
$x + 2 y + z = 1, x + 3 y + 4 z = k; x + 5 y + 10 z = k^{2}$ has solution? Find the solution in each case.

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Solution

$△ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & 1 1 & 3 & 4 1 & 5 & 10 \end{matrix} ∣ ∣ ∣ ∣ = 0$ after making two zeros
$∴$ Solution is not unique.
The system will have infinite solutions if $△_{1} = 0, △_{2} = 0$
$△_{3} = 0$
$△_{1} = 0 \Rightarrow k^{2} - 3 k + 2 = 0 ∴ k = 1, 2$
For these values $△_{2}$ and $△_{3}$ are also zero.
Now take $z = λ, k = 2$ then $x = - 1 + 5 λ, y = 1 - 3 λ$

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Similar questions

k

x + 2 y + z = 1; x + 3 y + 4 z = k; x + 5 y + 10 z = k^{2}

x - 2 y + 3 z = 1

- x + y - 2 z = k

x - 3 y + 4 z = 1

k \neq 3

∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & - 1 - 1 & - 2 & k 1 & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣ \neq 0

k \neq 3

x - 2 y + 3 z = - 1

- x + y - 2 z = k

x - 3 y + 4 z = 1

k \neq 3,

∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & - 1 - 1 & - 2 & k 1 & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣ \neq 0,

k \neq 3.

p^{2}

x + y - 2 z = 0, 2 x - 3 y + z = 0, x - 5 y + 4 z = k

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