Question

Let $S$ be the set of all real values of $a$ for which the following system of linear equations
$ax+2y+5z=1$
$2x+y+3z=1$
$3y+7z=1$
is consistent. Then the set $S$ is:

• A. equal to $R$
• B. equal to $R-\left[1\right]$
• C. equal to $\left[1\right]$
• D. an empty set

Answer 1

Answer: (A)

equal to $R$

Given system of linear equations are

$ax+2y+5z=1$

$2x+y+3z=1$

$3y+7z=1$

Their coefficient determinant is given by

$\Delta =\left|\begin{array}{ccc}a& 2& 5\\ 2& 1& 3\\ 0& 3& 7\end{array}\right|=a(-2)-2\left(14\right)+5\left(6\right)=-2a+2=-2(a-1)$

${\Delta }_x=\left|\begin{array}{ccc}1& 2& 5\\ 1& 1& 3\\ 1& 3& 7\end{array}\right|=1(-2)-2\left(4\right)+5\left(2\right)=0$

${\Delta }_y=\left|\begin{array}{ccc}a& 1& 5\\ 2& 1& 3\\ 0& 1& 7\end{array}\right|=a\left(2\right)-1\left(14\right)+5\left(2\right)=4a-4$

${\Delta }_z=\left|\begin{array}{ccc}a& 2& 1\\ 2& 1& 1\\ 0& 3& 1\end{array}\right|=a(-2)-2\left(2\right)+1\left(6\right)=-2(a-1)$

$\because$ system is consistent

$\therefore$ either $a\ne 1\Rightarrow$ unique solution

Now, if $a=1$, system of equation becomes

$x+2y+5z=1$

$2x+y+3z=1$

$3y+7z=1$

i.e there are only two equations

$x+2y+5z=1$

$3y+7z=1$

Which are not parallel

$\therefore$ system of equation is consistent.

So, the system of equations are consistent for all real values.