Let , R be such that the system of linear equations array|c|c| x+2y+z & =5 2x+y+ z & =5 8x…

Question

Let $\alpha ,\beta \in R$ be such that the system of linear equations

$\begin{array}{|c|c|} x+2y+z & =5 \ 2x+y+\alpha z & =5 \ 8x+4y+\beta z & =18 \end{array}$

has no solution. Then $\frac{\beta }{\alpha }$ is equal to :

Accepted Answer

$\begin{gathered} x+2y+z=5 \ 2x+y+\alpha z=5 \ 8x+4y+\beta z=18 \end{gathered}$

For a system of three linear equations to have no solution, it is necessary that the coefficient matrix be singular, i.e. its determinant is zero: $\, \Delta=0$.

$\begin{gathered} \Delta=|\begin{array}{ccc}1 \quad  2 \quad  1 \ 2 \quad  1 \quad  \alpha \ 8 \quad  4 \quad  \beta \end{array}|=0 \ \Delta=1(\beta -4\alpha )-2(2\beta -8\alpha )+1(0)=0 \ =\beta -4\alpha -4\beta +16\alpha =0 \ \Rightarrow 4\alpha =\beta \end{gathered}$

Now substitute $\beta =4\alpha $ in the third equation. Then the left side of the third equation becomes $8x+4y+4\alpha z$, which is exactly $4$ times the left side of the second equation $2x+y+\alpha z$.

But the constants do not match: $4	imes 5=20$, whereas the third equation has $18$. So we get $4	imes (2x+y+\alpha z)=20$ but $8x+4y+4\alpha z=18$, which is a contradiction. Hence, the system is inconsistent and has no solution.

Therefore, $\frac{\beta }{\alpha }=\frac{4\alpha }{\alpha }=4$.

Solutions · 1