Consider the system of equations $x+y=2$, $ax+y=b$.Find the conditions on $a$ and $b$ under which the following holds

Question

Consider the system of equations $x+y=2$, $ax+y=b$.Find the conditions on $a$ and $b$ under which
(i) the system has exactly one solution;
(ii) the system has no solution;
(iii) the system has more than one solution.

I solved for $x$ and $y$ in terms of $a$ and $b$ and got some restricting conditions. But, it is difficult to solve that way. I was wondering if it could be solved like the determinants or matrices with Cramer's Rule or something similar because I read similar about similar conditions there. Please help!

Answer 1

By looking carefully at this system we can answer the questions by inspection , hardly doing any arithmetic at all.

$ \ \ x + y = 2 \\ ax + y = b $

The coefficients of y are equal therefore there are two cases.

Case 1

If a = 1 then the left hand sides of the equations are identical. If $b \ne 2 $ then the lines are parallel (not coinciding) , therefore no solution. If b = 2 then the lines (are parallel) coincide , therefore infinite number of solutions.

Case 2

If $ a \ne 1 $ then there is no way to make the left hand sides identical (parallel) because of the y coefficients therefore there will be a unique solution.

Summary

(i) Unique solution for $a \ne 1 $

(ii) No solution for $ a = 1 $ AND $ b \ne 2 $

(iii) Infinite number of solutions for $ a = 1 $ AND $ b = 2 $

Answer 2

for case 1 you have following matrix

 1, 1
 a, 1

unique solution would have if determinant of this matrix is not zero,in this case $1-a$ is not zero,or $a$ is not equal to $1$,

now

$x+y=2$

$a*x+y=b$

if rank of coeficient matrix is not equal to rank of augmented matrix rank,then it has not solution,

and finally if this rank is equal,but less then number ofvariables,then it has many solution,like determiant of

1   1
 a   1

is zero or $a=1$

http://en.wikipedia.org/wiki/System_of_linear_equations

Answer 3

to find inconsistency that means no solution of equations we must have rank of coefficient matrix A = rank of augmented matrix [A,C] now here augmented matrix

1  1  2      1    1      2
         =   
a  1  b      0  (a-1)  (2a-b)

now if a≠1 & 2a=b then rank(A)≠rank(C) system has no solution if a=1 & 2a=b then rank(A)=rank(C)=1 < no.of variables n i.e 2 system has infinitely many solution if a≠1 & 2a≠b then rank(A)=rank of(C)=2=n system has unique solution I hope u get the answer & process.GOOD QUESTION