If S is the set of distinct values of 'b' for which the following system of linear equations

$ x + y + z = 1$

$x + ay + z = 1$

$ax + by + z = 0$

has no solution, then S is:

a)an empty set.

b)an infinite set.

c)a finite set containing two or more elements.

d)a singleton.

This question was asked in 2017 JEE Main.

This question can be solved by Cramer's rule.

For this set of equation to have no solution,

$$ \begin{vmatrix} 1&1&1\\ 1&a&1\\ a&b&1\\ \end{vmatrix}$$

This determinant needs to be $0$. Which is only true for $a = 1$.

But for$ a=1 $, All of these three determinants are also $0$, regardless of the value of $b$.

$$ \begin{vmatrix} 1&1&1\\ 1&a&1\\ 0&b&1\\ \end{vmatrix}$$

$$ \begin{vmatrix} 1&1&1\\ 1&1&1\\ a&0&1\\ \end{vmatrix}$$

$$ \begin{vmatrix} 1&1&1\\ 1&a&1\\ a&b&1\\ \end{vmatrix}$$ For $ a = 1 $, the equations have infinitely many solution, independent of the value of $ b $. And for $ a \neq 1 $, equations have a unique solution.

Thus my answer is a.

However, at b = 1 and a = 1, equations become this :

$ x + y + z = 1$

$x + y + z = 1$

$x + y + z = 0$

They have no solutions.

Thus my question is, what is wrong in my understanding of Cramer's rule?


2 Answers 2


From near the bottom of the linked Wikipedia article: “For $3\times3$ or higher systems, the only thing one can say when the coefficient determinant equals zero is that if any of the numerator determinants are nonzero, then the system must be incompatible. However, having all determinants zero does not imply that the system is indeterminate.”

This is exactly the situation you find yourself in. All of the numerator determinants are zero, but, unlike the $2\times2$ case, you can’t decide whether there are zero or an infinite number of solutions from that fact.


Since $\Delta=0$ gives $a=1$, we obtain the following system: $$x+y+z=1,$$ $$x+y+z=1$$ and $$x+by+z=0,$$ which gives $$(b-1)y=-1,$$ and we got that for $$(a,b)=(1,1)$$ only our system has no solutions.


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