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If the system of equations $x + y+ z = 5$ ; $x + 2y + 3z = 9$ ; $x + 3y + \alpha z = \beta $ has infinitely many solutions, then $\beta - \alpha $ equals
$21$
$8$
$18$
$5$
Solution
$x + y – z = 5$
$x + 2y + 3z = 9,$
$x + 3y + \alpha z = \beta $
$D = \left| {\begin{array}{*{20}{c}}
1&1&1\\
1&2&3\\
1&3&0
\end{array}} \right| = 0 \Rightarrow \left( {2\alpha – 9} \right) + \left( {3 – \alpha } \right) + \left( {3 – 2} \right) = 0 \Rightarrow \alpha = 5$
Now, ${D_3} = \left| {\begin{array}{*{20}{c}}
1&1&5\\
1&2&9\\
1&3&\beta
\end{array}} \right| = 0 \Rightarrow 2\beta – 27 + 9\beta – 5\left( {3 – 2} \right) = 0 \Rightarrow \beta = 13$
$ \Rightarrow $ at $\alpha = 5,b = 13$ above $3$ planes from common line
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
