Question 2.10: Let A= [1 2 4 10] . (1) Solve the equation x^→A =  b^→ where...

Where written out, \overrightarrow{x}A=\overrightarrow{b} is equivalent to

\left(x_{1},x_{2} \right) \begin{bmatrix} 1 & 2 \\ 4 & 10 \end{bmatrix}=\left(b_{1},b_{2}\right)   or   \left \{ \begin{matrix} x_{1}+4x_{2}=b_{1} \\ 2x_{1}+10x_{2}=b_{2} \end{matrix} \right. .

A is called the coefficient matrix of the equations and the 3 × 2 matrix \left[\begin{matrix} A \\ \overrightarrow{b} \end{matrix} \right] _{3\times 2} its augmented matrix. Apply consecutive column operations to the augmented matrix as follows.

\left[\begin{matrix} A \\ \hdashline \overrightarrow{b} \end{matrix} \right] =\left [ \begin{matrix} 1 & 2 \\ 4 & 10 \\\hdashline b_{1} & b_{2} \end{matrix} \right ]\underset{\text{ }F_{\left(2\right)-2\left(1\right) } }{\longrightarrow} \left [ \begin{matrix} 1 & 0 \\ 4 & 2 \\\hdashline b_{1} & b_{2}-2b_{1} \end{matrix} \right ]=\left[\begin{matrix} A \\ \hdashline \overrightarrow{b} \end{matrix} \right]F_{\left(2\right)-2\left(1\right) }      \left( ^{*}1 \right)

\underset{\text{ }F_{\frac{1}{2} \left(2\right) } }{\longrightarrow} =\left [ \begin{matrix} 1 & 0 \\ 4 & 1 \\\hdashline b_{1} & \frac{b_{2}-2b_{1} }{2} \end{matrix} \right ]= \left[\begin{matrix} A \\ \hdashline \overrightarrow{b} \end{matrix} \right]F_{\left(2\right)-2\left(1\right) }F_{\frac{1}{2} \left(2\right) }      \left( ^{*}2 \right)

\underset{\text{ }F_{\left(1\right)-4\left(2\right) } }{\longrightarrow} \left [ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \hdashline  5b_{1}-2b_{2} & \frac{b_{2}-2b_{1}}{2} \end{matrix} \right ]

  = \left[\begin{matrix} A \\ \hdashline \overrightarrow{b} \end{matrix} \right] F_{\left(2\right)-2\left(1\right) }F_{\frac{1}{2} \left(2\right) } F_{\left(1\right)-4\left(2\right) }.

We can deduce some valuable information from the above process and the final result follows.

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

\left \{ \begin{matrix} x_{1}=5b_{1}-2b_{2} \\ x_{2}= -b_{1}+\frac{1}{2}b_{2} \end{matrix} \right.

is the solution of the equations. It is better to put in matrix form as

\overrightarrow{x} = \left(x_{1}   x_{2} \right) = \left(b_{1}   b_{2} \right) \begin{bmatrix} 5 & -1 \\ -2 & \frac{1}{2} \end{bmatrix}= \overrightarrow{b}A^{-1}.

In particular, we obtain the inverse A^{-1} of A.

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

I_{2}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=AF_{\left(2\right)-2\left(1\right) }F_{\frac{1}{2} \left(2\right) } F_{\left(1\right)-4\left(2\right) }

\Rightarrow A^{-1}= F_{\left(2\right)-2\left(1\right) }F_{\frac{1}{2} \left(2\right) } F_{\left(1\right)-4\left(2\right) }= \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix}\begin{bmatrix} 1 & 0 \\-4 & 1 \end{bmatrix}

=\frac{1}{2} \begin{bmatrix} 10 & -2 \\ -4 & 1 \end{bmatrix} =\begin{bmatrix} 5 & -1 \\ -2 & \frac{1}{2} \end{bmatrix}.

By the way,   A^{-1} can be written as a product of elementary matrices.

(c) A as a product of elementary matrices

A=F^{-1}_{\left(1\right)-4\left(2\right)}F^{-1}_{\frac{1}{2} \left(2\right)}F^{-1}_{\left(2\right)-2\left(1\right)} = F_{\left(1\right)+4\left(2\right)} F_{2\left(2\right)}F_{\left(2\right)+2\left(1\right)}

=\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}.    (2.7.61)

This factorization provides another way to investigate the geometric mapping properties of A than those presented in Sec. 2.7.4.

(d) The computations of detA and det   A^{-1}

  det A = det \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}\cdot det \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\cdot det \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} =1 \cdot 2 \cdot 1=2,

det A^{-1} = det \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}\cdot det \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{bmatrix}\cdot det \begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix} = 1 \cdot \frac{1}{2} \cdot 1= \frac{1}{2} .

Note When written in the transpose form A^{*}\overrightarrow{x^{*} }=\overrightarrow{b^{*} }, we can apply elementary row operations to the augmented matrix \left[ A^{*}\mid \overrightarrow{b^{*}} \right] _{2\times 3} and obtain exactly the same factorizations for A and A^{-1} and solve the equations. The only difference is that, when applying row operations to A^{*}, the corresponding elementary matrices should multiply A in the front stepwise, i.e.

E^{}_{\left(1\right)-4\left(2\right)}E^{}_{\frac{1}{2} \left(2\right)}E^{}_{\left(2\right)-2\left(1\right)}A^{*}= I_{2},

which is the same as A F_{\left(2\right)-2\left(1\right)} F_{\frac{1}{2} \left(2\right)}F_{\left(1\right)-4\left(2\right)}=I_{2} obtained when applying column operations.

Adopt the factorization in (2.7.61) and see Fig. 2.61 for geometric mapping properties step by step.

Note that another way to study the geometric mapping properties is to find the eigenvalues   \frac{11\pm \sqrt{113} }{2}   of A and model after Fig. 2.50 and Example 2.8 \left(\lambda _{1}\neq \lambda _{2} \right) in Sec. 2.7.2.

Note that (2.7.61) can also be written as

A=\begin{bmatrix} 1 & 0 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix} ,     (2.7.62)

which is a product of a lower triangular matrix before an upper triangular matrix. Readers are urged to explain how these factorizations affect the mapping properties in Fig. 2.61.

It is worthy to notice that the appearance of lower and upper triangular matrices factorization in (2.7.62) is not accidental and is not necessarily a consequence of the previous elementary matrices factorization in (2.7.61).

We get it in the algebraic operation process up to the steps (^{*}1 )  and (^{*}2) . Stop there and we deduce that

AF_{\left(2\right)-2\left(1\right) }= \begin{bmatrix} 1 & 0 \\ 4 & 2 \end{bmatrix}

\Rightarrow A=\begin{bmatrix} 1 & 0 \\ 4 & 2 \end{bmatrix}F^{-1}_{\left(2\right)-2\left(1\right)}= \begin{bmatrix} 1 & 0 \\ 4 & 2 \end{bmatrix}F^{}_{\left(2\right)+2\left(1\right)}=\begin{bmatrix} 1 & 0 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 1 & 2 \\0 & 1 \end{bmatrix},

and

AF_{\left(2\right)-2\left(1\right) }F_{\frac{1}{2}\left(2\right) } =\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}

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

Does factorization like (2.7.62) help in solving the equations \overrightarrow{x} A=\overrightarrow{b} ?

Yes, it does. Take the first factorization for example,

\overrightarrow{x}\begin{bmatrix} 1 & 2 \\ 4 & 10 \end{bmatrix}=\overrightarrow{b}

\Leftrightarrow \overrightarrow{x}\begin{bmatrix} 1 & 0 \\ 4 & 2 \end{bmatrix}=\overrightarrow{y} and \overrightarrow{y}\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}=\overrightarrow{b}  where \overrightarrow{y}=\left(y_{1},y_{2} \right).

Solve firstly

and secondly

\overrightarrow{x}\begin{bmatrix} 1 & 0 \\ 4 & 2 \end{bmatrix}=\left(b_{1},-2b_{1}+b_{2} \right)

\Rightarrow \overrightarrow{x}=\left(b_{1},-2b_{1}+b_{2} \right)\begin{bmatrix} 1 & 0 \\ -2 & \frac{1}{2} \end{bmatrix}=\left(5b_{1}-2b_{1},-b_{1}+\frac{1}{2}b_{2} \right).