If the system of linear equations
$7x + 11y + \alpha z = 13$
$5x + 4y + 7z = \beta$
$175x + 194y + 57z = 361$
has infinitely many solutions,then $\alpha + \beta + 2$ is equal to

  • A
    $4$
  • B
    $3$
  • C
    $5$
  • D
    $6$

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Similar Questions

For what value of $k$ does the following system of equations possess a non-trivial solution?
$x + ky + 3z = 0$
$3x + ky - 2z = 0$
$2x + 3y - 4z = 0$

Difficult
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If $A$ is a matrix such that $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] A \left[\begin{array}{ll} 1 & 1 \end{array}\right] = \left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$, then $A$ is equal to

If $x^a y^b=e^m, x^c y^d=e^n, \Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|, \Delta_2=\left|\begin{array}{ll}a & m \\ c & n\end{array}\right|, \Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$,then the values of $x$ and $y$ are respectively ($e$ is the base of natural logarithm).

The system of equations $\lambda x - y + (\cos\theta) z = 0$,$3x + y + 2z = 0$,and $(\cos\theta) x + y + 2z = 0$ for $0 < \theta < 2\pi$ has non-trivial solution$(s)$:

The set of real values of $\alpha$ for which the system of linear equations
$\begin{aligned}
& x+(\sin \alpha) y+(\cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& -x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}$
has a non-trivial solution is

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