Question

Consider the following system of linear equations
$\left[\begin{array}{ccc}2& 1& -4\\ 4& 3& -12\\ 1& 2& -8\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}a\\ 5\\ 7\end{array}\right]$
Number of values of $a$ for which system has infinitely many solutions.

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is A $1$

For the system $\left[\begin{array}{ccc}2& 1& -4\\ 4& 3& -12\\ 1& 2& -8\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}a\\ 5\\ 7\end{array}\right]$
has infinitely many solutions

$D=\left|\begin{array}{ccc}2& 1& -4\\ 4& 3& -12\\ 1& 2& -8\end{array}\right|=0$

$D_1=0$, for all values of $a$ as $C_2$ and $C_3$ are identical.

$D_2=0\Rightarrow \left|\begin{array}{ccc}2& a& -4\\ 4& 5& -12\\ 1& 7& -8\end{array}\right|=0$

Applying $R_2\to R_2-3R_1,R_2\to R_3-2R_1$
$\left|\begin{array}{ccc}2& a& -4\\ -2& 5-3a& 0\\ -3& 7-2a& 0\end{array}\right|=0\Rightarrow a=\frac{1}{5}$

Now, $D_3=0$
$\Rightarrow \left|\begin{array}{ccc}2& 1& a\\ 4& 3& 5\\ 1& 2& 7\end{array}\right|=0$

Applying $R_2\to R_2-2R_1,R_3\to R_3-0.5R_1$
$\Rightarrow \left|\begin{array}{ccc}2& 1& a\\ 0& 1& 5-2a\\ 0& 1.5& 7-0.5a\end{array}\right|=0\Rightarrow a=\frac{1}{5}$

Hence, there is only one value of $a$ for which the system has infinitetly many solution.