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If the system of linear equations
$7 x+11 y+\alpha z=13$
$5 x+4 y+7 z=\beta$
$175 x+194 y+57 z=361$
has infinitely many solutions, then $\alpha+\beta+2$ is equal to
$4$
$3$
$5$
$6$
Solution
Sol. $7 x+11 y+\alpha z=13$
$5 x+4 y+7 z=\beta$
$175 x+194 y+57 z=361$
$\text { (i) } \times 10+(\text { ii }) \times 21-(\text { iii) }$
$z (10 \alpha+147-57)=130+21 \beta-361$
$\therefore 10 \alpha+90=0$
$\alpha=-9$
$130-361+21 \beta=0$
$\beta=11$
$\alpha+\beta+2=4$
Similar Questions
Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2+x-1=0$. Consider the set $T=\{1, \alpha, \beta\}$. For a $3 \times 3$ matrix $M=\left(a_{\ell}\right) 3 \times 3_3$, define $R_l=a_{l 1}+a_{l 2}+a_\beta$ and $C_j=a_{1 j}+a_{2 l}+a_{3 j}$ for $i=1,2,3$ and $j=1,2,3$
Match each entry in $List-I$ to the correct entry in $List-II$.
| $List-I$ | $List-II$ |
| ($P$) The number of matrices $M=\left(a_{i j}\right)_3 \times 3$ with all entries in $T$ such that $R_i=C_j=0$ for all $i, j$ is | ($1$) ($1$) |
| ($Q$) The number of symmetric matrices $M=\left(a_{i j}\right) 3 \times 3$ with all entries in $T$ such that $C_j=0$ for all $j$ is | ($2$) ($2$) |
| ($R$) Let $M=\left(a_{i j}\right) 3 \times 3$ be a skew symmetric matrix such that $a_{i j} \in T$ for $i>j$. Then the number of elements in the set $\left\{\left(\begin{array}{l}x \\ y \\ z\end{array}\right): x, y \cdot z \in R, M\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}a_{12} \\ 0 \\ -a_{23}\end{array}\right)\right\}$ is is | ($3$) Infinite |
| ($S$) Let $M=\left(a_{i j}\right)_3 \times 3$ be a matrix with all entries in $T$ such that $R_i=0$ for all $i$. Then the absolute value of the determinant of $M$ is | ($4$) ($6$) |
| ($5$) ($0$) |
The correct option is
