4x5 Matrice Solution

[mammothrob]

soooo... I have these four equations and need to solve for four unknowns.

So I built a 4x5 matrix for it. Im useing (r1234) to denote the rows and bold font to seperate decemal numbers where they might look similar.

My ultimate goal here is to get this in reduced row ech. form



1 0 1 3 -1
2 1 0 6 3
0-1 2 0 -5
1-3 7 3 -16

-r1 + r3 = new r3

1 0 1 3 -1
2 1 0 6 3
0-3 6 0 -15
1-3 7 3 -16

r2 + r4 = new r4

1 0 1 3 -1
2 1 0 6 3
0-3 6 0 -15
2 0 2 6 -2

-(1/3)*r3

1 0 1 3-1
2 1 0 6 3
0 1-2 0-5
2 0 2 6-2

(1/2)*r4

1 0 1 3-1
2 1 0 6 3
0 1-2 0-5
1 0 1 3-1

moving r3 and r4 up

1 0 1 3-1
1 0 1 3-1
0 1-2 0-5
2 1 0 6 3


.5 * r4

1 0 1 3-1
1 0 1 3-1
0 1-2 0-5
1.5 0 31.5



this is where im starting to get lost. I have tried different routes to get
here but keep ending up with just an augmented matrix. My guess so far is that this may have infinite solutions but can get close to the reduced form that I want.

Rob

 

[mark07]

1 0 1 3 -1
2 1 0 6 3
0-1 2 0 -5
1-3 7 3 -16

-r1 + r3 = new r3

1 0 1 3 -1
2 1 0 6 3
0-3 6 0 -15
1-3 7 3 -16

Are you doing the row operations correctly? Adding (-1) times the 1st row to the 3rd row gives

1 0 1 3 -1
2 1 0 6 3
-1 -1 1 -3 -4
1 -3 7 3 -16

But the real question is, why would you start with that operation??

The goal is to reach to an Echelon form. Pivot on 1st row is the 1st entry. Try to make every other entry in that column 0 first. Then move to the 2nd row. Find pivot entry etc... I'll do a few so you see what I mean, but you can finish the rest.

1 0 1 3 -1
2 1 0 6 3
0 -1 2 0 -5
1 -3 7 3 -16

Add (-2) times 1st row to the 2nd (to make the 2 zero):

1 0 1 3 -1
0 1 -2 0 5
0 -1 2 0 -5
1 -3 7 3 -16

Now add (-1) times the 1st row to the last:

1 0 1 3 -1
0 1 -2 0 5
0 -1 2 0 -5
0 -3 6 0 -15

1st column done. Pivot on second row is 1 on the second entry. Use that to make all the entries below it zero.

Hope this helps.

 

[soroban]

Hello, mammothrob!

The matrix falls apart after a couple of steps . . .


. . . \(\displaystyle \begin{vmatrix}1 & 0 & 1 & 3 & | & -1 \\
2 & 1 & 0 & 6 & | & 3 \\ 0 & -1 & 2 & 0 & | & -5 \\ 1 & -3 & 7 & 3 & | & -16\end{vmatrix}\)


\(\displaystyle \begin{array}{cccccccc}. \\ \\ \\ R2-2R1 \\ \\ \\ \\. \\ \\ \\ \\ \\ R4-R1\end{array}\;
\begin{vmatrix}1 & 0 & 1 & 3 & | & -1 \\ 0 & 1 & -2 & 0 & | & 5 \\
0 & -1 & 2 & 0 & | & -5 \\ 0 & -3 & 6 & 0 & | & -15\end{vmatrix}\)


\(\displaystyle \begin{array}{cccccccccc}. \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ R3+R2 \\ R4+3R2\end{array}\;
\begin{vmatrix}1 & 0 & 1 & 3 & | & -1 \\ 0 & 1 & -2 & 0 & | & 5\\ 0 & 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & 0 & | & 0\end{vmatrix}\)

 

[mammothrob]

so I have my solution equations, which represent an infinite amount of solutions.

x + z + 3w = -1
y - 2z = 5

so to write my answer in a parametric form would this be correct?

Let z and y be free.

y=T
z=S

T-2S =5
x + S + 3w = -1

x + (T - 5)/2 + 3w = -1

Im kinda confused about how to do this with four vailables.

 

[mark07]

Looking at the 2nd equation, y AND z can not be free right? You can't use any y and any z for y-2z=5 to be true.

A systematic way of choosing the free variables is by looking at the reduced echelon form. First column corresponds to x, second to y, third to z, fourth to w. There are pivot entries on the first and second columns (see the 1's), so those are called pivot columns. Choose your free variables to be the ones that do NOT correspond to pivot columns. In this problem, z and w will be free.

After that I think you got the idea. Let z=s, w=t. Then you can write x and y in terms of s and t as well using the equations

x + z + 3w = -1
y - 2z = 5