Calculating Linear Combinations of Vectors: Step-by-Step Guide

[Mark53]

Homework Statement

Consider the four vectors (1, 1, 1), (2, −1, 3), (1, 7, −1) and (1, 4, 0). Calculate how many ways you can write (1, 1, 1) as a linear combination of the other three, explaining your reasoning.

The Attempt at a Solution

wouldn't any of these combinations give the correct answer if you multiply the vectors by the correct constant

 

[Delta2]

You got to solve a system of three linear equations with three unknowns (the three constants by which we multiply each vector in the linear combination of them). If the determinant of this system of equations is not zero then the system has unique solution, that is there are only three constants, so one possible way.

 

[nomadreid]

No. In fact, there is only one solution. Thankfully, the question does not actually ask you to calculate the answers. All you have to do is to either show how you that the system of linear equations
2x-y+3z =1 & x+7-z=1&x+4y=1
has exactly one solution. (If you've had matrices, that is a good way to go. For starters, you can check that the determinant of the corresponding matrix is non-zero.) You needn't actually calculate x,y,z, because I think the solutions are pretty big, like
353291421251867.000
-88322855312966.500
-264968565938900.000

 

[Ray Vickson]

No. In fact, there is only one solution. Thankfully, the question does not actually ask you to calculate the answers. All you have to do is to either show how you that the system of linear equations
2x-y+3z =1 & x+7-z=1&x+4y=1
has exactly one solution. (If you've had matrices, that is a good way to go. For starters, you can check that the determinant of the corresponding matrix is non-zero.) You needn't actually calculate x,y,z, because I think the solutions are pretty big, like
353291421251867.000
-88322855312966.500
-264968565938900.000

Sorry, but there is more than one solution. Check the value of the determinant of the system of equations.

 

[nomadreid]

Oops. I miscalculated the determinant. I stand corrected. Thanks for the correction, Ray Vickson.

 

[Mark53]

Oops. I miscalculated the determinant. I stand corrected. Thanks for the correction, Ray Vickson.

So would I have to calculate the determinant for each possible combination?

 

[nomadreid]

The idea is that x,y,z represent the factors that will allow x(2,-1,3)+y(1,3,-1)+z(1,4,0)=(1,1,1)
Giving you the three equations
2x+y+z=1
-x+3y+4z=1
3x-y=1
(Ah, sorry for my earlier incorrect equations. I should not answer these questions just before bedtime.)
In matrix form, this is
2 .. 1.. 1 ... .x...1
-1 .. 3 ..4 ... y... = 1
3 .. -1...0... z.... 1
The determinant of the left-hand matrix is 0. This means that there is not a unique solution. Therefore either there are no solutions or there are infinitely many solutions. Your next step is to determine which case it is: infinite or none. There are various ways to do this, but the easiest is just to try to solve it directionly without matrices. Label the first equation A, the second B, the third C, and you can use 4A-B to get an equation in only x and y, now combine that with C to find x and y, and with that try to find z in both A and B.

 

[Mark53]

The idea is that x,y,z represent the factors that will allow x(2,-1,3)+y(1,3,-1)+z(1,4,0)=(1,1,1)
Giving you the three equations
2x+y+z=1
-x+3y+4z=1
3x-y=1
(Ah, sorry for my earlier incorrect equations. I should not answer these questions just before bedtime.)
In matrix form, this is
2 .. 1.. 1 ... .x...1
-1 .. 3 ..4 ... y... = 1
3 .. -1...0... z.... 1
The determinant of the left-hand matrix is 0. This means that there is not a unique solution. Therefore either there are no solutions or there are infinitely many solutions. Your next step is to determine which case it is: infinite or none. There are various ways to do this, but the easiest is just to try to solve it directionly without matrices. Label the first equation A, the second B, the third C, and you can use 4A-B to get an equation in only x and y, now combine that with C to find x and y, and with that try to find z in both A and B.

when I use row operations on the matrix i get a unique solution

 

[Ray Vickson]

when I use row operations on the matrix i get a unique solution

OK, but that is not the correct system of equations. You should be looking at
$$\begin{array}{rl}
2x + 1y + 1z& = 1\\
-1x +7y +4z &= 1 \\
3x -1y+ 0 z& = 1
\end{array}
$$
The three vectors under consideration are (2, −1, 3), (1, 7, −1) and (1, 4, 0), not (2, -1, 3), (1, 3, -1) and (1, 4, 0).

 

[nomadreid]

Oops, typos can do horrible things. Until someone legislates that 3=7...

 

[Mark53]

Oops, typos can do horrible things. Until someone legislates that 3=7...

From my calculations I get infinitely many solutions

so would i just write my answer as

x=2/5-(1/5)t
y=1/5-(3/5)t
z=t

 

[nomadreid]

Your instructor certainly wants more than that: you show him/her how you got to those three equations, with a remark that these solutions are valid for all t (include domain).