Question

The number of values of \( \alpha \) for which the system of equations: \[ \begin{array}{l} x+y+z=\alpha \\ \alpha x+2 \alpha y+3 z=-1 \\ x+3 \alpha y+5 z=4 \end{array} \]

A. 0 

B. 1 

C. 2 

D. 3

Answer 1

Correct option is (B) 1

\(\begin{vmatrix}1&1&1\\\alpha&2\alpha&3\\1&3\alpha&5\end{vmatrix} = 0\)

⇒ \(1\begin{vmatrix}2\alpha&3\\3\alpha&5\end{vmatrix} - \begin{vmatrix}\alpha &3\\1&5\end{vmatrix} + \begin{vmatrix}\alpha &2\alpha\\1 &3\alpha\end{vmatrix} = 0\)

⇒ \((10\alpha - 9\alpha)- (5\alpha - 3) + 3\alpha^2 - 2\alpha = 0\)

⇒ \(3\alpha^2 - 6\alpha + 3 = 0\)

⇒ \(\alpha^2 - 2\alpha + 1 = 0\)

⇒ \((\alpha - 1)^2 = 0\)

⇒ \(\alpha = 1\)

If α = 1, then Adj A.B

\(=\begin{bmatrix}1&-2&1\\-2&4&-2\\1&-2&1\end{bmatrix} \begin{bmatrix}1\\-1\\4\end{bmatrix} = \begin{bmatrix}7\\-14\\7\end{bmatrix} \ne 0\)

∴ For α = 1, the system is in consistent.