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Solution
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Solution provided by AtoZmath.com
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Find the value of h,k for which the system of equations has a Unique solution calculator
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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
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Example1. 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 solutionSolution:Here x+3y-2z=-1 2x+5y+z=2 2x+6y-az=b =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)=(-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)=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)=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=4substitute 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=-2substitute 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=-2From (4), we get =>-b-2=0=>-b=2=>b=-2The system has unique solutions if D!=0, so a!=4The system has infinite solution if D=D_1=D_2=D_3=0, so a=4 and b=-2System 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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