If the system of equations 2 x + 3 y − z = 5 x + α y + 3 z = − 4 3 x − y + β z = 7 has infinitely many…

Given:

The system of equations,

$2x+3y-z=5$

$x+\alpha y+3z=-4$

$3x-y+\beta z=7$ have many solutions,

So, $△={△}_1={△}_2={△}_3$

Now, finding ${△}_2=\left|\begin{array}{ccc}2& 5& -1\\ 1& -4& 3\\ 3& 7& \beta \end{array}\right|=0$

$\Rightarrow 2\left(-4\beta -21\right)-5\left(\beta -9\right)-1\left(7+12\right)=0$

$\Rightarrow 13\beta =-16$

Now, finding ${△}_3=\left|\begin{array}{ccc}2& 3& 5\\ 1& \alpha & -4\\ 3& -1& 7\end{array}\right|=0$

$\Rightarrow 14\alpha -8-57-5-15\alpha =0$

$\Rightarrow \alpha =-70$

Hence, the value of $13\alpha \beta =70\times 16=1120$