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

Question

If the system of linear equations

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

has infinitely many solutions, then $12\alpha +13\beta $ is equal to

Accepted Answer

$\begin{gathered} D=|\begin{array}{ccc}1 \quad  -2 \quad  1 \ 2 \quad  \alpha  \quad  3 \ 3 \quad  -1 \quad  \beta \end{array}| \ =1(\alpha \beta +3)+2(2\beta -9)+1(-2-3\alpha ) \ =\alpha \beta +3+4\beta -18-2-3\alpha \end{gathered}$

For infinite solutions $D=0,{D}_{1}=0,{D}_{2}=0$ and

$\begin{gathered} {D}_{3}=0 \ D=0 \end{gathered}$

$\alpha \beta -3\alpha +4\beta =17$ ..... (1)

$\begin{gathered} {D}_{1}=|\begin{array}{ccc}-4 \quad  -2 \quad  1 \ 5 \quad  \alpha  \quad  3 \ 3 \quad  -1 \quad  \beta \end{array}|=0 \ {D}_{2}=|\begin{array}{ccc}1 \quad  -4 \quad  1 \ 2 \quad  5 \quad  3 \ 3 \quad  3 \quad  \beta \end{array}|=0 \ \Rightarrow 1(5\beta -9)+4(2\beta -9)+1(6-15)=0 \ 13\beta -9-36-9=0 \end{gathered}$

$13\beta =54,\beta =\frac{54}{13}$ put in (1)

$\frac{54}{13}\alpha -3\alpha +4(\frac{54}{13})=17$

$54\alpha -39\alpha +216=221$

$15\alpha =5\, \alpha =\frac{1}{3}$

Now, $12\alpha +13\beta =12\cdot \frac{1}{3}+13\cdot \frac{54}{13}$
 $=4+54=58$

Solutions · 1