Let the system of linear equations x + y + kz = 2 2x + 3y - z = 1...

Let the system of linear equations

$x + y + kz = 2$

$2x + 3y - z = 1$

$3x + 4y + 2z = k$

have infinitely many solutions. Then the system

\left(k + 1\right)x + \left(2k - 1\right)y = 7 $\left(2k + 1\right)x + \left|k + 5\right|y = 10$ has :

Options

Ainfinitely many solutions
Bunique solution satisfying $x - y = 1$
Cno solution
Dunique solution satisfying $x + y = 1$

Correct answer

D. unique solution satisfying $x + y = 1$

Step-by-step solution

Linear equations $x + y + kz = 2$ , $2x + 3y - z = 1 and 3x + 4y + 2z = k$ have infinitely many solutions.

Therefore,

\left| $\begin{cases}1 & 1 & k \\ 2 & 3 & - 1 \\ 3 & 4 & 2\end{cases}$ \right| = 0 $$ ⇒ 1\left(10\right) - 1\left(7\right) + k\left(- 1\right) = 0 $$ ⇒ 10 - 7 - k = 0 $$ ⇒ 3 - k = 0 $$ ⇒ k = 3 $For$ k = 3 $, the second system of equations is$ 4x + 5y = 7 \ldots \left(1\right) $$ 7x + 8y = 10 \ldots \left(2\right)

Clearly, they have a unique solution

Subtracting \left(1\right)\text{ $and }\left(2\right), we get$ 3x + 3y = 3 $$ ⇒ x + y = 1 $$

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