Arman G. answered • 11/10/19
Aerospace Engineer
Form the augmented matrix [1 3 7 12; 1 4 10 15; 2 3 5 X] and keep row reducing. Eventually you should get the bottom row is [ 0 0 0 X+c] where c is some constant. This could only be true if x = c, whatever that constant is that you get by row reduction
Hint: you should get b3 is between 14 and 17.
Extra extra hint, you're welcome :) : form the augmented matrix, then follow these row reduction steps:
- replace (row 2) with (row 2 - row 1)
- row 3 with (row 3 - 2*row 1)
- row 3 ---> row3 +3*row2
Notice then that the bottom row is [0 0 0 X + c] so 0*x + 0*y + 0*z = X+c if and only if X = c.
AKA the system is only consistent for that value
Michael A.
I am looking for what b3 is. But in this scenario I would also like to understand what exactly xy and z are because these values must be the same for each row. Basically the question is if I bought 1 water , 3 peices of bread , and 7 tissues, then the values of xyz times those amounts of A should equal 12 dollars11/11/19
Arman G.
when you finish row reducing, z is a free variable, so x and y are written in terms of z. you get x = 2z+3, y = -3z+3, z = 3. so you have to be given z, if z is 7 then x is 17, y is -18. This is the only solution that will satisfy x + 3y + 7z = 12, if z has to equal 7. This, therefore does not represent a physical system like you've described because you cant have negative 18 pieces of bread.11/11/19
Arman G.
I guess maybe what ur asking for since I told u x and y in terms of z multiply those like x plus 3y +z =12 and you’ll get just an equation of z = 12 (because u plug in x =3z+3, y =-3z+3) and u can just solve for z then11/11/19
Michael A.
I’ll keep trying to get what I need, the first part that you said makes sense to solve for b3 which is what I was trying to do. But like I was saying the question is about different amounts of the same item but the dollar smount should stay consistent with each row , 1 bottle of water is a dollar then 2 bottles 2. But let me give it another try to solve for b311/11/19
Arman G.
A is a 3x3 matrix, but the augmented matrix is a 3x4... the only way this matrix is consistent is if the last column is not a pivot column which could only happen if b3 = 0... also z would then be a free variable, but that doesn't have to do with this problem, you're only looking for b311/10/19