Question

If the system of linear equations x + 2y + 3z = 6, x + 3y + 5z = 9 and 2x + 5y + az = b are consistent and have infinite number of solutions, then

• A. a = 8, b can be any real number
• B. b = 15, a can be any real number
• C. $a\in R-8$ and $b\in R-15$
• D. a = 8, b = 15
  

Answer 1

The correct option is D a = 8, b = 15


Given equations are consistent and have infinitely many solutions
$\therefore \left|\begin{array}{ccc}1& 2& 3\\ 1& 3& 5\\ 2& 5& a\end{array}\right|=0$ and $\left|\begin{array}{ccc}1& 2& 6\\ 1& 3& 9\\ 2& 5& b\end{array}\right|=0$
$\Rightarrow$ a =8 and b =15