Previously we saw how to get echelon form, pivot/free column vectors of a matrix say
\(A\)
also saw how to solve \(A\vec{x}=0\)
.Now let's push it little further and see the reduce row echelon form of our matrix
\(A\)
.Reduced Row Echelon Form
In reduced row echelon form we use elimination and make all elements\(0\)
above and below pivot elements.We saw the echelon form of
\(A=\begin{bmatrix} 1 & 2 & 2 & 2\\ 2 & 4 & 6 & 8\\ 3 & 6 & 8 & 10\\ \end{bmatrix}\)
is say \(U\)
\(U=\begin{bmatrix} \fbox{1} & 2 & 2 & 2\\ 0 & 0 & \fbox{2} & 4\\ 0 & 0 & 0 & 0\\ \end{bmatrix}\)
Now we make element above pivot
\(=0\)
So let's use elimination to do it.
\(U=\begin{bmatrix} \fbox{1} & 2 & 2 & 2\\ 0 & 0 & \fbox{2} & 4\\ 0 & 0 & 0 & 0\\ \end{bmatrix}\)
\(r_1\leftarrow r_1-r_2\)
\(U'=\begin{bmatrix} \fbox{1} & 2 & 0 & -2\\ 0 & 0 & \fbox{2} & 4\\ 0 & 0 & 0 & 0\\ \end{bmatrix}\)
Now make pivot elements
\(=1\)
\(r_2\leftarrow r_2/2\)
\(U'=\begin{bmatrix} \fbox{1} & 2 & 0 & -2\\ 0 & 0 & \fbox{1} & 2\\ 0 & 0 & 0 & 0\\ \end{bmatrix}\)
Say this matrix is
\(R\)
.\(R=\begin{bmatrix} \color{red}{\fbox{1}} & \color{blue}{2} & \color{red}{0} & \color{blue}{-2} \\ \color{red}{0} & \color{blue}{0} & \color{red}{\fbox{1}} & \color{blue}{2} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}\)
Here you can see that there is a
\(2\times 2\)
Identity matrix shown in \(\color{red}{\text{red}}\)
and one more
\(2\times 2\)
matrix shown in \(\color{blue}{\text{blue}}\)
So
\(R\)
is something like:\(R'=\begin{bmatrix} \color{red}{\text{I}} & \color{blue}{\text{F}} \\ 0 & 0 \\ \end{bmatrix}\)
and here
\(\color{red}{ I=\begin{bmatrix} 1&0\\0&1\\ \end{bmatrix} }\)
and \(\color{blue}{ F=\begin{bmatrix} 2&-2\\0&2\\ \end{bmatrix} }\)
Here you can see that there is some reordering of columns but it's fine.Remember that number of special solution is
Because we are striving for\(\vec{x}\)that solves\(R\vec{x}=0\)\(R\vec{x}=\begin{bmatrix} \color{red}{\fbox{1}} & \color{blue}{2} & \color{red}{0} & \color{blue}{-2} \\ \color{red}{0} & \color{blue}{0} & \color{red}{\fbox{1}} & \color{blue}{2} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} \color{red}{x_1}\\\color{blue}{x_2}\\\color{red}{x_3}\\\color{blue}{x_4}\\ \end{bmatrix} =0\)
We can also write it as:\(R'\vec{x}'=\begin{bmatrix} \color{red}{\fbox{1}} & \color{red}{0} & \color{blue}{2} & \color{blue}{-2} \\ \color{red}{0} & \color{red}{\fbox{1}} & \color{blue}{0} & \color{blue}{2} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} \color{red}{x_1}\\\color{red}{x_3}\\\color{blue}{x_2}\\\color{blue}{x_4} \\ \end{bmatrix} =0\)
Now you can see\(R\)as:\(R'=\begin{bmatrix} \color{red}{\text{I}} & \color{blue}{\text{F}} \\ 0 & 0 \\ \end{bmatrix}\)
\(\color{blue}{2}\)
because we have 2 free column vectors.So we know that there are
\(\color{blue}{2}\)
special solution possible.Say that special solutions are:
\(\vec{x}^{(1)}=\begin{bmatrix} \color{red}{x^{(1)}_1}\\\color{blue}{x^{(1)}_2}\\\color{red}{x^{(1)}_3}\\\color{blue}{x^{(1)}_4}\\ \end{bmatrix}\)
and \(\vec{x}^{(2)}=\begin{bmatrix} \color{red}{x^{(2)}_1}\\\color{blue}{x^{(2)}_2}\\\color{red}{x^{(2)}_3}\\\color{blue}{x^{(2)}_4}\\ \end{bmatrix}\)
Now rearrange then
\(\vec{x}'^{(1)}=\begin{bmatrix} \color{red}{x^{(1)}_1}\\\color{red}{x^{(1)}_3}\\\color{blue}{x^{(1)}_2}\\\color{blue}{x^{(1)}_4}\\ \end{bmatrix}\)
and \(\vec{x}'^{(2)}=\begin{bmatrix} \color{red}{x^{(2)}_1}\\\color{red}{x^{(2)}_3}\\\color{blue}{x^{(2)}_2}\\\color{blue}{x^{(2)}_4}\\ \end{bmatrix}\)
Now stack then in a matrix say (
\(N\)
)\(N=\begin{bmatrix} \color{red}{x^{(1)}_1} & \color{red}{x^{(2)}_1} \\ \color{red}{x^{(1)}_3} & \color{red}{x^{(2)}_3} \\ \color{blue}{x^{(1)}_2} & \color{blue}{x^{(2)}_2} \\ \color{blue}{x^{(1)}_4} & \color{blue}{x^{(2)}_4} \\ \end{bmatrix};\quad\)
and say \(x_{\text{pivot}}=\begin{bmatrix} \color{red}{x^{(1)}_1} & \color{red}{x^{(2)}_1} \\ \color{red}{x^{(1)}_3} & \color{red}{x^{(2)}_3} \\ \end{bmatrix}\)
\(x_{\text{free}} =\begin{bmatrix} \color{blue}{x^{(1)}_2} & \color{blue}{x^{(2)}_2} \\ \color{blue}{x^{(1)}_4} & \color{blue}{x^{(2)}_4} \\ \end{bmatrix}\)
And we are free to choose free variables so let
\(x_{\text{free}} =\begin{bmatrix} \color{blue}{1} & \color{blue}{0} \\ \color{blue}{0} & \color{blue}{1} \\ \end{bmatrix}\)
Now
\(N=\begin{bmatrix} \color{red}{x_\text{pivot}}\\\color{blue}{I_{2\times2}} \end{bmatrix}\)
We want
\(R\vec{x}^{(1)}=0\)
and \(R\vec{x}^{(2)}=0\)
\(\Rightarrow R'\vec{x}'^{(1)}=0\)
and \(R'\vec{x}'^{(2)}=0\)
\(\Rightarrow R'N=0\)
So
\(R'N=\begin{bmatrix} \color{red}{I_{2\times2}} & \color{blue}{\text{F}} \\ 0 & 0 \\ \end{bmatrix} \begin{bmatrix} \color{red}{x_\text{pivot}}\\\color{blue}{I_{2\times2}} \end{bmatrix} = 0 \)
\(\Rightarrow \color{red}{x_\text{pivot}} \color{red}{I_{2\times2}} + \color{blue}{\text{F}} \color{blue}{I_{2\times2}} =0 \)
\(\Rightarrow \color{red}{x_\text{pivot}} = - \color{blue}{\text{F}}\)
So
\(N=\begin{bmatrix} - \color{blue}{\text{F}} \\ \color{blue}{I_{2\times2}}\\ \end{bmatrix} \)
And we know that
\(\color{blue}{ F=\begin{bmatrix} 2&-2\\0&2\\ \end{bmatrix} }\)
So
\(N=\begin{bmatrix} \color{red}{-2} & \color{red}{2} \\ \color{red}{0} & \color{red}{-2} \\ \color{blue}{1} & \color{blue}{0} \\ \color{blue}{0} & \color{blue}{1} \\ \end{bmatrix};\quad\)
\(\Rightarrow\)
\(\vec{x}'^{(1)}=\begin{bmatrix} \color{red}{x^{(1)}_1}\\\color{red}{x^{(1)}_3}\\\color{blue}{x^{(1)}_2}\\\color{blue}{x^{(1)}_4}\\ \end{bmatrix} = \begin{bmatrix} \color{red}{-2}\\\color{red}{0}\\\color{blue}{1}\\\color{blue}{0}\\ \end{bmatrix}\)
and \(\vec{x}'^{(2)}=\begin{bmatrix} \color{red}{x^{(2)}_1}\\\color{red}{x^{(2)}_3}\\\color{blue}{x^{(2)}_2}\\\color{blue}{x^{(2)}_4}\\ \end{bmatrix} = \begin{bmatrix} \color{red}{2}\\\color{red}{-2}\\\color{blue}{0}\\\color{blue}{1}\\ \end{bmatrix}\)
\(\Rightarrow\)
\(\vec{x}^{(1)}=\begin{bmatrix} \color{red}{x^{(1)}_1}\\\color{blue}{x^{(1)}_2}\\\color{red}{x^{(1)}_3}\\\color{blue}{x^{(1)}_4}\\ \end{bmatrix} = \begin{bmatrix} \color{red}{-2}\\\color{red}{1}\\\color{blue}{0}\\\color{blue}{0}\\ \end{bmatrix}\)
and \(\vec{x}^{(2)}=\begin{bmatrix} \color{red}{x^{(2)}_1}\\\color{blue}{x^{(2)}_2}\\\color{red}{x^{(2)}_3}\\\color{blue}{x^{(2)}_4}\\ \end{bmatrix} = \begin{bmatrix} \color{red}{2}\\\color{red}{0}\\\color{blue}{-2}\\\color{blue}{1}\\ \end{bmatrix}\)
So our Null space is space of all
\(\vec{x}\)
that are linear combinations of \(\begin{bmatrix} -2\\1\\0\\0\\ \end{bmatrix} \)
and \(\begin{bmatrix} 2\\0\\-2\\1\\ \end{bmatrix}\)
\(x=\alpha \begin{bmatrix} -2\\1\\0\\0\\ \end{bmatrix} + \beta \begin{bmatrix} 2\\0\\-2\\1\\ \end{bmatrix};\quad\)
\(\alpha,\beta\in\mathbb{R}\)