Question 2.13: Let A =[2 3 -4 -6 ].(1) Solve the equation Ax∗^→ = b∗^→ .(2)...

Write A\overrightarrow{x^{*} } =\overrightarrow{b^{*} } out as

\left \{ \begin{matrix} 2x_{1}+3 x_{2}=b_{1}  \\ -4x_{1}-6 x_{2}=b_{2} \end{matrix} \right..

The equations have solutions if and only if 2b_{1}+b_{2}=0 . In this case, the solutions are x_{1}=\frac{1}{2}\left(b_{1}  –  3x_{2}\right) with x_{2} arbitrary scalars.

Apply row operations to

\left[A\mid \overrightarrow{b^{*} } \right]_{}=\left [ \begin{matrix} 2& 3 & \vdots b_{1} \\-4 & -6 & \vdots b_{2} \end{matrix} \right ]

\underset{\text{ }E_{\frac{1}{2} \left(1\right) } }{\longrightarrow}\left [ \begin{matrix} 1 & \frac{3}{2} & \vdots \frac{b_{1}}{2} \\-4 & -6 & \vdots \ b_{2} \end{matrix} \right ]=E_{\frac{1}{2} \left(1\right) }\left[A\mid \overrightarrow{b^{*}}\right]

\underset{\text{ }E_{\left(2\right)+4\left(1\right) } }{\longrightarrow}\left [ \begin{matrix} 1 & \frac{3}{2} & \vdots               \frac{b_{1}}{2} \\0 & 0 & \vdots \ 2b_{1}+b_{2} \end{matrix} \right ]= E_{\left(2\right)+4\left(1\right) } E_{\frac{1}{2} \left(1\right) }\left[A\mid \overrightarrow{b^{*} } \right].

Since   r\left(A\right)=1  and elementary matrices preserve ranks, then

A\overrightarrow{x^{*} } =\overrightarrow{b^{*} } has a solution.

  ⇔r\left(A\right)= r\left(\left[A\mid \overrightarrow{b^{*} } \right]\right) ⇔ 2b_{1}+b_{2}=0 .

This constrained condition coincides with what we obtained via traditional method. On the other hand,

\begin{bmatrix} 1 & \frac{3}{2} \\ 0 & 0 \end{bmatrix} =E_{\left(2\right)+4\left(1\right) } E_{\frac{1}{2} \left(1\right) }A

\Rightarrow A=E^{-1}_{\frac{1}{2} \left(1\right)}E^{-1}_{\left(2\right)+4\left(1\right)}\begin{bmatrix} 1 & \frac{3}{2} \\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 2& 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix} \begin{bmatrix} 1 & \frac{3}{2} \\ 0 & 0 \end{bmatrix}.

See Fig. 2.64.