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If the system of linear equations $2 x + y - z =7$ ; $x-3 y+2 z=1$ ; $x +4 y +\delta z = k$, where $\delta, k \in R$ has infinitely many solutions, then $\delta+ k$ is equal to
$-3$
$3$
$6$
$9$
Solution
$\quad\left|\begin{array}{ccc}2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta\end{array}\right|=0$
$\Rightarrow \delta=-3$
And $\left|\begin{array}{ccr}7 & 1 & -1 \\ 1 & -3 & 2 \\ K & 4 & -3\end{array}\right|=0 \Rightarrow K =6$
$\Rightarrow \delta+ K =3$
Alternate
$2 x + y – z =7$ $\dots(1)$
$x-3 y+2 z=1$ $\dots(2)$
$x +4 y +\delta z = k$ $\dots(3)$
Equation $(2) + (3)$
We get $2 x+y+(2+\delta) z=1+K$ $\dots(4)$
For infinitely solution
Form equation $(1)$ and $(4)$
$2+\delta=-1 \Rightarrow \delta=-3$
$1+ k =7 \Rightarrow k =6$
$\delta+ k =3$
Similar Questions
Let $\alpha, \beta$ and $\gamma$ be real numbers. consider the following system of linear equations
$x+2 y+z=7$
$x+\alpha z=11$
$2 x-3 y+\beta z=\gamma$
Match each entry in List – $I$ to the correct entries in List-$II$
| List – $I$ | List – $II$ |
| ($P$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has | ($1$) a unique solution |
| ($Q$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has | ($2$) no solution |
|
($R$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has |
($3$) infinitely many solutions |
| ($S$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has | ($4$) $x=11, y=-2$ and $z=0$ as a solution |
| ($5$) $x=-15, y=4$ and $z=0$ as a solution |
Then the system has
