Question

For $\alpha, \beta \in R$, suppose the system of linear equations $ x-y+z=5 $ $2 x+2 y+\alpha z=8 $ $3 x-y+4 z=\beta$ has infinitely many solutions Then $\alpha$ and $\beta$ are the roots of

• A. $x^2-18 x+56=0$
• B. $x^2+14 x+24=0$
• C. $x^2-10 x+16=0$
• D. $x^2+18 x+56=0$

Answer 1

The Correct Option is A

The system of equations is: \(\begin{pmatrix} 1 & -1 & 1 \\[0.3em] 2 & 2 & \alpha \\[0.3em] 3 & -1 & 4 \end{pmatrix} \begin{pmatrix} x  y  z \end{pmatrix} = \begin{pmatrix} 5  8  \beta \end{pmatrix}\)

For infinitely many solutions, the determinant must be zero: \(\text{det}\left( \begin{pmatrix} 1 & -1 & 1 \\[0.3em] 2 & 2 & \alpha  \\[0.3em]3 & -1 & 4 \end{pmatrix} \right) = 0\)

Expanding the determinant and solving, we get: 

\(8 + \alpha - 3 \alpha - 6 = 0 \Rightarrow \alpha = 4\)

 Now, substitute \(\alpha = 4\) into the system and solve for \(\beta\): \(8 + \alpha - 2 (4 + 1) + 3 (4 - 2) = 0 \Rightarrow \beta = 56\)

Thus, \(\alpha = 4\) and \(\beta = 56\).