Question 2.11: Let A =[0 2 -1 1 ]. Do the same problems as in Example 1.
For \overrightarrow{x}=\left(x_{1},x_{2} \right) and \overrightarrow{b}=\left(b_{1},b_{2} \right) ,\overrightarrow{x}A=\overrightarrow{b} is equivalent to
\left \{ \begin{matrix} 0\cdot x_{1}-x_{2}=b_{1} \\ 2x_{1}+x_{2}=b_{2} \end{matrix} \right.
which can be easily solved as x_{1}=\frac{1}{2}\left(b_{1}+b_{2}\right) and x_{2}= -b_{1}.
The shortcoming of A, for the purpose of a general theory to be established later in this subsection, is that its leading diagonal entry is zero. To avoid this happening, we exchange the first row and the second row of A and the resulted matrix amounts to
B=\begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 2 \\ -1 & 1 \end{bmatrix}=E_{\left(1\right)\left(2\right) }A.
Then, perform column operations to
x_{2} x_{1}
\left [ \begin{matrix} E_{\left(1\right)\left(2\right) }A \\ \hdashline \overrightarrow{b} \end{matrix} \right ] =\left [ \begin{matrix} B \\ \hdashline \overrightarrow{b} \end{matrix} \right ]=\left [ \begin{matrix} -1 & 1 \\ 0 & 2 \\ \hdashline b_{1} & b_{2} \end{matrix} \right ]
\underset{\text{ }F_{-\left(1\right) } }{\longrightarrow} \left [ \begin{matrix} 1 & 1 \\ 0 & 2 \\ \hdashline -b_{1} & b_{2} \end{matrix} \right ]=\left [ \begin{matrix} B \\ \hdashline \overrightarrow{b} \end{matrix} \right ]F_{-\left(1\right) }
\underset{\text{ }F_{\left(2\right) -\left(1\right) } }{\longrightarrow} \left [ \begin{matrix} 1 & 0 \\ 0 & 2 \\ \hdashline -b_{1} & b_{1}+b_{2} \end{matrix} \right ]=\left [ \begin{matrix} B \\ \hdashline \overrightarrow{b} \end{matrix} \right ]F_{-\left(1\right) }F_{\left(2\right) -\left(1\right) }
\underset{\text{ }F_{\frac{1}{2} \left(2\right) } }{\longrightarrow} \left [ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \hdashline -b_{1} & \frac{b_{1}+b_{2}}{2} \end{matrix} \right ]=\left [ \begin{matrix} B \\ \hdashline \overrightarrow{b} \end{matrix} \right ]F_{-\left(1\right) }F_{\left(2\right) -\left(1\right) }F_{\frac{1}{2} \left(2\right) }.
Note that \overrightarrow{x}A=\left(\overrightarrow{x}E_{\left(1\right)\left(2\right) } \right)\left(E_{\left(1\right)\left(2\right) }A\right)= \left(\overrightarrow{x}E_{\left(1\right)\left(2\right) } \right)B=\left(x_{1 } x_{2} \right)B=\overrightarrow{b}
so that the first column corresponds to x_{2} while the second one to x_{1}.
Equivalently, we can perform row operations to
\left[\left(E_{\left(1\right)\left(2\right) }A\right)^{*} \mid \overrightarrow{b^{*} } \right]=\left[A^{*}F_{\left(1\right)\left(2\right) } \mid \overrightarrow{b^{*}} \right] =\left[B^{*}\mid \overrightarrow{b^{*}}\right]=\begin{array}{r c}\left [ \begin{matrix} -1 & 0 & \vdots b_{1} \\ \\ 1 & 2 & \vdots b_{2} \end{matrix} \right ] \begin{matrix} x_{2} \\ \\ x_{1} \end{matrix}\end{array}
\underset{\text{ }E_{-\left(1\right) } }{\longrightarrow}\left [ \begin{matrix} 1 & 0 & \vdots-b_{1} \\ 1 & 2 & \vdots b_{2}\end{matrix} \right ]\underset{\text{ }E_{\left(2\right)-\left(1\right) } }{\longrightarrow}\left [ \begin{matrix} 1 & 0 & \vdots -b_{1} \\ 0 & 2 & \vdots b_{1}+ b_{2}\end{matrix} \right ]
\underset{\text{ }E_{\frac{1}{2} \left(2\right) } }{\longrightarrow}\left [ \begin{matrix} 1 & 0 & \vdots-b_{1} \\ 0 & 1 & \vdots \frac{b_{1}+b_{2}}{2} \end{matrix} \right ].
In this case, the first row corresponds to x_{2} while the second one to x_{1}.
(a) The solution of \overrightarrow{x}A=\overrightarrow{b}
\left \{ \begin{matrix} x_{1}=\frac{b_{1}+b_{2}}{2} \\ \\ x_{2}=-b_{1} \end{matrix} \right. , or
\overrightarrow{x}=\left(x_{1},x_{2} \right)=\left(b_{1} b_{2}\right)\begin{bmatrix} \frac{1}{2} & -1 \\ \frac{1}{2} & 0 \end{bmatrix}= \overrightarrow{b} A^{-1}.
(b) The invertibility of A and its inverse A^{-1}
I_{2}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=BF_{-\left(1\right) } F_{\left(2\right) -\left(1\right) }F_{\frac{1}{2} \left(2\right) }
\Rightarrow B^{-1}= \left(E_{\left(1\right)\left(2\right) }A\right)^{-1} =A^{-1}E^{-1}_{\left(1\right)\left(2\right)} = F_{-\left(1\right) } F_{\left(2\right) -\left(1\right) }F_{\frac{1}{2} \left(2\right) }
\Rightarrow A^{-1}=F_{-\left(1\right) } F_{\left(2\right) -\left(1\right) }F_{\frac{1}{2} \left(2\right) } F_{\left(1\right) \left(2\right) }
=\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & -1 \\ \frac{1}{2} & 0 \end{bmatrix}.
(c) The elementary factorization of A
A=F^{-1}_{\left(1\right) \left(2\right)} F^{-1}_{\frac{1}{2} \left(2\right)}F^{-1}_{\left(2\right) -\left(1\right)}F^{-1}_{-\left(1\right)}
=F^{}_{\left(1\right) \left(2\right)}F^{}_{2\left(2\right)}F^{}_{\left(2\right)+ \left(1\right)}F^{}_{-\left(1\right)}
=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}
=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} (2.7.63)
\Rightarrow \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}A=\begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\0 & 1 \end{bmatrix}\begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}. (2.7.64)
(d) The determinants detA and det A^{-1}
detA = det\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\cdot det\begin{bmatrix} 1 & 0 \\0 & 2 \end{bmatrix}\cdot det\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\cdot det\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}
=\left(-1\right) \cdot 2 \cdot 1 \cdot \left(-1\right) =2 ;
detA^{-1}=det \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\cdot det\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\cdot det\begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix}\cdot det\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
=\left(-1\right) \cdot 1\cdot \left(\frac{1}{2} \right)\cdot \left(-1\right)=\frac{1}{2}.
See Fig. 2.62. Note that A does not have real eigenvalues.
