Question 2.12: Let A = [1 2 2 -7]. (1) Solve the equation Ax ∗^→ = b ∗^→ wh...

As against \overrightarrow{x}A=\overrightarrow{b} in Examples 2.10 and 2.11, here we use column vector \overrightarrow{x^{*} }=\left[\begin{matrix} x_{1} \\ x_{2} \end{matrix} \right] as unknown vector and \overrightarrow{b^{*} }=\left[\begin{matrix} b_{1} \\ b_{2} \end{matrix} \right] as constant vector in A\overrightarrow{x^{*} }= \overrightarrow{b^{*} }  with A as coefficient matrix and \left[A\mid \overrightarrow{b^{*} } \right]_{2\times 3} as augmented matrix.

Now, apply consecutive row operations to

  \left[A\mid \overrightarrow{b^{*} } \right]_{}=\left [ \begin{matrix} 1 & 2 & \vdots b_{1} \\2 & -7& \vdots b_{2} \end{matrix} \right ]

\underset{\text{ }E_{\left(2\right)-2\left(1\right) } }{\longrightarrow}\left [ \begin{matrix} 1 & 2 & \vdots                 b_{1} \\0 & -11& \vdots \ b_{2}-2b_{1} \end{matrix} \right ]=E_{\left(2\right)-2\left(1\right) }\left[A\mid \overrightarrow{b^{*} } \right]_{}

\underset{\text{ }E_{-\frac{1}{11}\left(2\right) } }{\longrightarrow}\left [ \begin{matrix} 1 & 2 & \vdots             b_{1} \\0 & 1& \vdots \ \frac{2b_{1}-b_{2}}{11} \end{matrix} \right ]=E_{-\frac{1}{11}\left(2\right) }E_{\left(2\right)-2\left(1\right) }\left[A\mid \overrightarrow{b^{*} } \right]_{}

\underset{\text{ }E_{\left(1\right)-2\left(2\right) } }{\longrightarrow}\left [ \begin{matrix} 1 & 0 & \vdots \frac{7b_{1}+2b_{2}}{11}\\0 & 1& \vdots \ \frac{2b_{1}-b_{2}}{11} \end{matrix} \right ]=E_{\left(1\right)-2\left(2\right) }E_{-\frac{1}{11}\left(2\right) }E_{\left(2\right)-2\left(1\right) }\left[A\mid \overrightarrow{b^{*} } \right]_{}.

(a) The solution of A\overrightarrow{x^{*} }= \overrightarrow{b^{*} }

\left \{ \begin{matrix} x_{1}=\frac{1}{11}\left(7b_{1}+2b_{2}\right)  \\ \\ x_{2}=\frac{1}{11}\left(2b_{1}-b_{2}\right)\end{matrix} \right.  , or

\overrightarrow{x^{*} }=\left[\begin{matrix} x_{1} \\ x_{2} \end{matrix} \right] =\frac{1}{11} \begin{bmatrix} 7 & 2 \\ 2 & -1 \end{bmatrix} \left[\begin{matrix} b_{1} \\ b_{2} \end{matrix} \right] =A^{-1}\overrightarrow{b^{*} }.

(b) The invertibility of A and its inverse A^{-1}

I_{2}= E_{\left(1\right)-2\left(2\right)}E_{-\frac{1}{11}\left(2\right) }E_{\left(2\right)-2\left(1\right) }A

\Rightarrow A^{-1}=E_{\left(1\right)-2\left(2\right)}E_{-\frac{1}{11}\left(2\right) }E_{\left(2\right)-2\left(1\right) }=\begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -\frac{1}{11} \end{bmatrix}\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}

=\frac{1}{11}\begin{bmatrix} 7 & 2 \\ 2 & -1 \end{bmatrix}.

(c) The elementary factorization of A

A= E^{-1}_{\left(2\right)-2\left(1\right) } E^{-1}_{-\frac{1}{11}\left(2\right) } E^{-1}_{\left(1\right)-2\left(2\right) } =E_{\left(2\right)+2\left(1\right) }E_{-11\left(2\right) }E_{\left(1\right)+2\left(2\right) }

=\begin{bmatrix} 1 & 0\\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & -11 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}.    (2.7.65)

(d) The determinants det A and det A^{-1}

det A = 1\cdot (−11) \cdot 1 = −11,

det A^{-1}= 1 \cdot \left(-\frac{1}{11} \right)\cdot 1 =-\frac{1}{11}.

See Fig. 2.63.

Note that A has two distinct real eigenvalues -3\pm \sqrt{20}.