Method and examples
Method
Find the value of h,k for which the system of equations has a Unique solution
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Find the value of h,k for which the system of equations has a Unique solution calculator

1. 7x-7y+6z=-4,-8x+7y+3z=6,-37x+35y+hz=k
2. x+y+z=6,x+2y+3z=10,x+2y+hz=k
3. 2x+y+z=5,3x+5y+2z=15,Ax+y+4z=8
4. x+2y=3,5x+ky=-7
5. 2x+3y=5,4x+ky=10
6. 3x+5y=0,kx+10y=0


Example
1. Find the value of h,k for which the system of equations x+3y-2z=-1,2x+5y+z=2,2x+6y-az=b has No solution

Solution:
Here x+3y-2z=-1
2x+5y+z=2
2x+6y-az=b


|D| = 
 1  3  -2 
 2  5  1 
 2  6  -a 


 =
 1 × 
 5  1 
 6  -a 
 -3 × 
 2  1 
 2  -a 
 -2 × 
 2  5 
 2  6 


=1 xx (5 × (-a) - 1 × 6) -3 xx (2 × (-a) - 1 × 2) -2 xx (2 × 6 - 5 × 2)

=1 xx (-5a -6) -3 xx (-2a -2) -2 xx (12 -10)

=1 xx (-5a-6) -3 xx (-2a-2) -2 xx (2)

= -5a-6 +6a+6 -4

=a-4 ->(1)


|D_1| = 
 -1  3  -2 
 2  5  1 
 b  6  -a 


 =
 -1 × 
 5  1 
 6  -a 
 -3 × 
 2  1 
 b  -a 
 -2 × 
 2  5 
 b  6 


=(-1) xx (5 × (-a) - 1 × 6) -3 xx (2 × (-a) - 1 × b) -2 xx (2 × 6 - 5 × b)

=(-1) xx (-5a -6) -3 xx (-2a -b) -2 xx (12 -5b)

=(-1) xx (-5a-6) -3 xx (-2a-b) -2 xx (-5b+12)

= 5a+6 +6a+3b +10b-24

=11a+13b-18 ->(2)


|D_2| = 
 1  -1  -2 
 2  2  1 
 2  b  -a 


 =
 1 × 
 2  1 
 b  -a 
 +1 × 
 2  1 
 2  -a 
 -2 × 
 2  2 
 2  b 


=1 xx (2 × (-a) - 1 × b) +1 xx (2 × (-a) - 1 × 2) -2 xx (2 × b - 2 × 2)

=1 xx (-2a -b) +1 xx (-2a -2) -2 xx (2b -4)

=1 xx (-2a-b) +1 xx (-2a-2) -2 xx (2b-4)

= -2a-b -2a-2 -4b+8

=-4a-5b+6 ->(3)


|D_3| = 
 1  3  -1 
 2  5  2 
 2  6  b 


 =
 1 × 
 5  2 
 6  b 
 -3 × 
 2  2 
 2  b 
 -1 × 
 2  5 
 2  6 


=1 xx (5 × b - 2 × 6) -3 xx (2 × b - 2 × 2) -1 xx (2 × 6 - 5 × 2)

=1 xx (5b -12) -3 xx (2b -4) -1 xx (12 -10)

=1 xx (5b-12) -3 xx (2b-4) -1 xx (2)

= 5b-12 -6b+12 -2

=-b-2 ->(4)


From (1), we get

=>a-4=0

=>a=4


substitute a=4 in equation (2), we get

=>13b+11*4-18=0

=>13b=-11*4+18

=>13b=-44+18

=>13b=-26

=>b=(-26)/13

=>b=-2


substitute a=4 in equation (3), we get

=>-5b-4*4+6=0

=>-5b=4*4-6

=>-5b=16-6

=>-5b=10

=>b=10/(-5)

=>b=-2


From (4), we get

=>-b-2=0

=>-b=2

=>b=-2


The system has unique solutions if D!=0, so a!=4

The system has infinite solution if D=D_1=D_2=D_3=0, so a=4 and b=-2

System has no solution if D=0 and at least one of D_1,D_2,D_3 is nonzero, so a=4 and b!=-2




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