If the system of equations gathered 2x+3y-z=5 x+ y+3z=-4 3x-y+ z=7 gathered has infinitely…

Question

If the system of equations

$\begin{gathered} 2x+3y-z=5 \ x+\alpha y+3z=-4 \ 3x-y+\beta z=7 \end{gathered}$

has infinitely many solutions, then $13\alpha \beta $ is equal to :

Accepted Answer

$\begin{gathered} 2x+3y-z=5 \ x+\alpha y+3z=-4 \ 3x-y+\beta z=7 \ {\Delta}_{2}=|\begin{array}{ccc}2 \quad  -1 \quad  5 \ 1 \quad  3 \quad  -4 \ 3 \quad  \beta  \quad  7\end{array}| \ {\Delta}_{2}=2(21+4\beta )+1(7+12)+5(\beta -9) \ {\Delta}_{2}=42+8\beta +19+5\beta -45 \ {\Delta}_{2}=13\beta +16 \ {\Delta}_{2}=0 \end{gathered}$

$\begin{gathered} 	herefore \beta =-\frac{16}{13} \ {\Delta}_{3}=|\begin{array}{ccc}2 \quad  3 \quad  5 \ 1 \quad  \alpha  \quad  -4 \ 3 \quad  -1 \quad  7\end{array}| \ {\Delta}_{3}=2(7\alpha -4)-3(7+12)+5(-1-3\alpha ) \ {\Delta}_{3}=14\alpha -8-57-5-15\alpha \ {\Delta}_{3}=-\alpha -70 \end{gathered}$

$\begin{gathered} {\Delta}_{3}=0 \ \alpha =-70 \ 13\alpha \beta =(13)(-70)(-\frac{16}{13}) \ =+1120 \end{gathered}$

Solutions · 1