# solution sets in linear algebra

• Feb 12th 2011, 02:04 PM
situation
solution sets in linear algebra
hey ive been given a system of equations:

$w+x+y+z=5 ; w+x+y+2z=6 ; 2w+2x+2y+3z=11$

i put this into an augmented matrix and got w+x+y=4 so im guessing that this gives a solution set which is that plane..how do i write this mathematically and am i correct??

thank you
• Feb 12th 2011, 02:19 PM
dwsmith
Quote:

Originally Posted by situation
hey ive been given a system of equations:

$w+x+y+z=5 ; w+x+y+2z=6 ; 2w+2x+2y+3z=11$

i put this into an augmented matrix and got w+x+y=4 so im guessing that this gives a solution set which is that plane..how do i write this mathematically and am i correct??

thank you

You should have gotten:

$\displaystyle\begin{bmatrix}1&1&1&0&4\\0&0&0&1&1\\ 0&0&0&0&0\end{bmatrix}$

$w=4-x-y$

x and y are free

$z=1$

$\displaystyle\left\{\begin{bmatrix}4-x-y\\x\\y\\1\end{bmatrix}\right\}$
• Feb 12th 2011, 02:58 PM
topspin1617
Quote:

Originally Posted by situation
hey ive been given a system of equations:

$w+x+y+z=5 ; w+x+y+2z=6 ; 2w+2x+2y+3z=11$

i put this into an augmented matrix and got w+x+y=4 so im guessing that this gives a solution set which is that plane..how do i write this mathematically and am i correct??

thank you

In a nutshell, you are exactly right. From the first two equations alone, we can see $z=1$. Plugging this in to the equations, we just get the same equation repeated three times: $w+x+y=4$. One equation in three values means that two of the variables are free (can be anything), and the value of the third depends on the values of the first two. So if, for example, $x=s,y=t$, then the set of all solutions would look like vectors (or points) of the form

$\left(\begin{array}{c}{x\\y\\z\\w}\end{array}\righ t)=\left(\begin{array}{c}{4-s-t\\s\\t\\1}\end{array}\right)=\left(\begin{array}{ c}{4\\0\\0\\1}\end{array}\right)+s\left(\begin{arr ay}{c}{-1\\1\\0\\0}\end{array}\right)+t\left(\begin{array} {c}{-1\\0\\1\\0}\end{array}\right)$

(many linear algebra tests would write the solution in the latter form, separating pieces corresponding to the independent parameters. But you don't HAVE to; there is absolutely nothing wrong with writing it the other way).
• Feb 12th 2011, 03:12 PM
situation
Quote:

Originally Posted by topspin1617
In a nutshell, you are exactly right. From the first two equations alone, we can see $z=1$. Plugging this in to the equations, we just get the same equation repeated three times: $w+x+y=4$. One equation in three values means that two of the variables are free (can be anything), and the value of the third depends on the values of the first two. So if, for example, $x=s,y=t$, then the set of all solutions would look like vectors (or points) of the form

$\left(\begin{array}{c}{x\\y\\z\\w}\end{array}\righ t)=\left(\begin{array}{c}{4-s-t\\s\\t\\1}\end{array}\right)=\left(\begin{array}{ c}{4\\0\\0\\1}\end{array}\right)+s\left(\begin{arr ay}{c}{-1\\1\\0\\0}\end{array}\right)+t\left(\begin{array} {c}{-1\\0\\1\\0}\end{array}\right)$

(many linear algebra tests would write the solution in the latter form, separating pieces corresponding to the independent parameters. But you don't HAVE to; there is absolutely nothing wrong with writing it the other way).

thanks for that dude, the question in the text book is asking for the solution set in parametric form..also the next part asks me to solve the system with the same coefficient matrix but just augmented with the zero vector instead...the associated homogeneous system. not sure how i do that though :S

thanks
• Feb 12th 2011, 03:15 PM
dwsmith
Quote:

Originally Posted by situation
thanks for that dude, the question in the text book is asking for the solution set in parametric form..also the next part asks me to solve the system with the same coefficient matrix but just augmented with the zero vector instead...the associated homogeneous system. not sure how i do that though :S

thanks

Solve the matrix without the last column vector.
• Feb 12th 2011, 03:16 PM
situation
Quote:

Originally Posted by dwsmith
Solve the matrix without the last column vector.

ill give it a go now thanks.
• Feb 12th 2011, 03:52 PM
topspin1617
Quote:

Originally Posted by situation
thanks for that dude, the question in the text book is asking for the solution set in parametric form..also the next part asks me to solve the system with the same coefficient matrix but just augmented with the zero vector instead...the associated homogeneous system. not sure how i do that though :S

thanks

Yea.... basically means instead of =4, =5, =6, take the same equations with =0, =0 =0.