Question 1063521
.
Find the value of K for the following pair of linear equation have infinitely many solution :—
x+(k+1)y=5
(k+1)x+9y=8y-1
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<pre>
x      + (k+1)y = 5,         (1)
(k+1)x +     9y = 8y - 1.    (2)

is equivalent to 

x      + (k+1)y =  5,        (1')
(k+1)x +      y = -1.        (2')

The theory says: 

   - if the determinant of the matrix of the system is not zero, then the solution is unique.

   - if the determinant of the matrix of the system is zero,     then two options are possible: 

     1) there is NO solution,  OR  2) there are infinitely many solutions.

The determinant of the matrix is

     det {{{(matrix(2,2, 1, (k+1), (k+1), 1))}}} = {{{1*1 - (k+1)^2}}} = {{{1 - (k+1)^2}}}.

The condition det = 0 is this equation for "k"

{{{(k+1)^2}}} = 1,

which has these two solutions:  {{{k[1]}}} = 0,   {{{k[2]}}} = -2.


At k = 0 the system (1'), (2') is 

 x + y =  5,     
 x + y = -1

and has NO solutions. 


At k = -2 the system (1'), (2') is 

 x - y =  5,     
-x + y = -1

and has NO solutions, again. 
</pre>

<U>Answer</U>.  There is NO value of "k" such that the original system has infinitely many solutions.