Answer :

Solution :

$\left(\begin{array}{cc}\mathrm{x}& \mathrm{y}\end{array}\right)\left[\begin{array}{cc}2& 5-2\alpha \\ \alpha & 1\end{array}\right]=\left(\begin{array}{cc}0& 0\end{array}\right)$

for non-trivial infinite solution: |A| = 0

A = $\left[\begin{array}{cc}2& 5-2\alpha \\ \alpha & 1\end{array}\right]$ ⇒ |A| = 2 - α(5 - 2α) = 0

2α2 - 5α + 2 = 0

(α - 2) (2α - 1) = 0

α = 2, 1/2

For α = 2

A = $\left[\begin{array}{cc}2& 1\\ 2& 1\end{array}\right]$

$\left(\begin{array}{cc}\mathrm{x}& \mathrm{y}\end{array}\right)\left[\begin{array}{cc}2& 1\\ 2& 1\end{array}\right]=\left(\begin{array}{cc}0& 0\end{array}\right)$

2x + 1/2 y = 0

4x + y = 0

from both equations we get,

y = - 4x

Substitute options in equation.

We get,

option 1. x = 2, y = −2 and

option 2. x = −1, y = 4 satisfies equations.