The value of t for which the following system is consistent $x + y = 1, t x + y = t, (1 + t) x + 2 y = 3$ , is

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Solution

Given system of equation is consistent

$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & - 1 t & 1 & - t 1 + t & 2 & - 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow (t - 2) (1 - t) = 0$

$\Rightarrow t = 2$ or $t = 1$

for $t = 1$ system becomes

$x + y = 1, x + y = \frac{3}{2} ∴ t \neq 1$

So $t = 2$ is the only solution.

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