Question

find the values of a for wich the linear system has

Answer 1

\[x+2y+3z=1\]
\[x+3y+(4a+3)z=5\]
\[2x+5y+3(a^2+6)z=a+2\]
a)has no solution
b)has unique solution
c)has infitely many solutions

Answer 2

I got as far as\[x+z(3-8a)=-7\]
\[y+4az=4\]
\[z(a^2-4a)=a-4\rightarrow z=\frac{ 1 }{ a }\]

Answer 3

so for no solution a=0
for infinite solutions a=4
for unique ???

Answer 4

@UnkleRhaukus

Answer 5

your problem as stated requires complex numbers. It makes more sense if the last row were
2 5 a^2+6 a+2

without the 3 in front of the (a^2+6)

Answer 6

I used complete elimination (Gaussian) to get to the simplified part

Answer 7

it dosent look complete

Answer 8

oh snap there is no 3 in front

Answer 9

as @phi said so is what can we do now

Answer 10

Yes, with no 3, this problem is doable

Answer 11

can we do it now

Answer 12

I HAVE TO GO NOW YOU CAN LEAVE ANY COMMENTS THANKS

Answer 13

[ 1 2 3 1]
[ 1 3 4*a + 3 5]
[ 2 5 a^2 + 6 a + 2]
sage: mm.echelon_form()
[ 1 0 0 (a - 4)*(8*a - 3)/(a^2 - 4*a) - 7 ]
[ 0 1 0 -4*(a - 4)*a/(a^2 - 4*a) + 4 ]
[ 0 0 1 (a - 4)/(a^2 - 4*a) ]

Answer 14

GREAT SO HOW DO WE RELATE TO THE THREE POSSIBILITIES

Answer 15

NO SOLUTION ,INFINITE ANDF UNIQUE

Answer 16

the last row becomes

0 0 (a^2 -4a) a-4

for an inf # of solutions you want this last row to be all zeros
for no solution you want the a^2-4a to be zero and a-4 non-zero (leads to a solution of z= #/0 which is undefined.)

all other a's will give particular solutions

Answer 17

THIS LOOKS GOOD YOU DID THE ECHELON VERY FAST THOUGH,MAKES SENSE THANKS

Answer 18

SORRY ABOUT THE 3

Answer 19

I would have been faster without the leading bogus 3...

Answer 20

THANKS ALOT TO UNCLE AND PHI