Linear Algebra: Solving for a vector x in Ax=b

[kankerfist]

I have to find a vector in terms of k1,k2,k3 from the following system:

\(\displaystyle \left[ {\begin{array}
1 & 2 & 1 \\
6 & { - 1} & 0 \\
{ - 1} & { - 2} & { - 1} \\
\end{array}} \right]\left[ {\begin{array}
{k1} \\
{k1} \\
{k2} \\
{k3} \\
\end{array}} \right] = \left[ {\begin{array}
0 \\
0 \\
0 \\
\end{array}} \right]\)

So I began by row reducing this matrix:

\(\displaystyle \left[ {\begin{array}
1 & 2 & 1 & 0 \\
6 & { - 1} & 0 & 0 \\
{ - 1} & { - 2} & { - 1} & 0 \\
\end{array}} \right]\)

Which results in the following reduced matrix:

\(\displaystyle \left[ {\begin{array}
1 & 0 & {1/13} & 0 \\
0 & 1 & {6/13} & 0 \\
0 & 0 & 0 & 0 \\
\end{array}} \right]\)

Now I am in diff EQ and I forget how to use linear algebra to use the above matrix to find the vector:

\(\displaystyle \left[ {\begin{array}
{k1} \\
{k1} \\
{k2} \\
{k3} \\
\end{array}} \right]\)

My study guide says the vector [k1;k2;k3] is:

\(\displaystyle \left[ {\begin{array}
1 \\
6 \\
-13 \\
\end{array}} \right]\)

Could someone point out how the [1;6;-13] can be gotten from my row reduced matrix? Thanks a lot!

 

[stapel]

Your solution is of the form [k₁, k₂, k₃] = [-(1/13)k₃, -(6/13)k₃, k₃], with k₃ arbitrary.

Pick k₃ = -13.

Eliz.

 

[kankerfist]

Ok now I remember! I forgot about pivots and the arbitrary constant. Thanks

 

[galactus]

See the bottom row of 0's in your ref. That means there are infinite solutions.

Except, I get a little different ref than you

1 -1/6 0
0 1 6/13
0 0 0


x-(y/6)=0
y-(6z/13)

Letting x=1 and solving the system results in your solutions of 1,6,-13.


If you let x=2 and solve, resulting in y=12 and z=-26, you also arrive at

[0;0;0]

\(\displaystyle \left[{\begin{array}{cc}1&2&1\\6&-1&0\\-1&-2&-1\end{array}}\right]\left[{\begin{array}{cc}2\\12\\26\end{array}}\right]=\left[{\begin{array}{cc}0\\0\\0\end{array}}\right]\)



Try with other numbers. Let x=3, y=18, z=-39

You get 0.