91影视

For a linear system $\mathit{A}\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{=}\mathit{b}$, a vector${\stackrel{\mathbf{鈫�}}{\mathbf{x}}}^{\mathbf{鈭�}}$ in ${\mathit{鈩�}}^{\mathbf{n}}$is considered as least-squares solution of the given linear system if $\mathbf{鈥�}\stackrel{\mathbf{鈫�}}{\mathbf{b}}\mathbf{鈭�}\mathbf{A}{\stackrel{\mathbf{鈫�}}{\mathbf{x}}}^{\mathbf{鈭�}}\mathbf{鈥�}\mathbf{鈮�}\mathbf{鈥�}\stackrel{\mathbf{鈫�}}{\mathbf{b}}\mathbf{鈭�}\mathbf{A}\stackrel{\mathbf{鈫�}}{\mathbf{x}}\mathbf{鈥�}$for every$\stackrel{\mathbf{鈫�}}{\mathbf{x}}$ in ${\mathit{鈩�}}^{\mathbf{m}}$.

(a)

Here, we have a liner system$A\stackrel{鈫�}{x}=\stackrel{鈫�}{b}$where$A=\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]$and $\stackrel{鈫�}{b}=\left[\begin{array}{c}10\\ 20\end{array}\right]$.It is known that $\ker \left(A\right)=A\stackrel{鈫�}{x}$which implies . $\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=0$So, the system of equations become$x_1+3x_2=0$
and $2x_1+6x_2=0$.

The$\ker \left(\stackrel{鈫�}{A}\right)$is a straight line whose equation is $x_2=鈭�\frac{1}{3}{x}_1$, this passes through origin having slope$鈭�\frac{1}{3}$.This implies $\ker \left(\stackrel{鈫�}{A}\right)=span\left(\left[\begin{array}{c}鈭�3\\ 1\end{array}\right]\right)$.

Now, the transpose of $A$is$A^T=\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]$. This implies:

$\begin{array}{c}\text{im}\left(A^T\right)=span(\left[\begin{array}{c}1\\ 3\end{array}\right],\left[\begin{array}{c}2\\ 6\end{array}\right])\\ =span\left(\left[\begin{array}{c}1\\ 3\end{array}\right]\right)\end{array}$

The sketch of the equations is shown below:

It is known that the image of$A^T$is the perpendicular line to $\ker \left(A\right)$that is$3x_1鈭�x_2=0$. So, the solution set is given by:

$\begin{array}{c}{x}_1=\frac{1}{3}{x}_2\\ \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\frac{1}{3}{x}_2\left[\begin{array}{c}1\\ 3\end{array}\right]\end{array}$

This implies ${\left(\ker A\right)}^{鈯�}=span\left(\left[\begin{array}{c}1\\ 3\end{array}\right]\right)$. So, sketch of the image of$A^T$is shown below:

Now, find the solution set of the system as follows:

$\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}10\\ 20\end{array}\right]$

The above equation gives the below system of equations:

$\begin{array}{c}{x}_1+3x_2=10\\ 2x_1+6x_2=20\end{array}$

The equivalent equation the above system is $x_1=10鈭�3x_2$. So, the solution is:

$\begin{array}{c}\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}10鈭�3x_2\\ x_2\end{array}\right]\\ =\left[\begin{array}{c}10\\ 0\end{array}\right]+x_2\left[\begin{array}{c}鈭�3\\ 1\end{array}\right]\end{array}$

So, the solution set$S$is $s鈭�\{\left[\begin{array}{c}10\\ 0\end{array}\right]+a\left[\begin{array}{c}鈭�3\\ 1\end{array}\right]\}$where$a鈭�\mathrm{鈩�}$
.

The solution set of $A\stackrel{鈫�}{x}=0$is parallel to the system$A\stackrel{鈫�}{x}=\stackrel{鈫�}{b}$.Therefore, the sketch of the solution set$S$is shown below:

(b)

From part (a), we have$im\left(A^T\right)={\left(\ker A\right)}^{鈯�}$ . So, we can say that$\text{im}\left(A^T\right)$ is always perpendicular to$\left(\ker A\right)$ .

Thus, the $\left(\ker A\right)$is always perpendicular to$\text{im}\left(A^T\right)$ .

(c)

The solution set of $A\stackrel{鈫�}{x}=0$is parallel to$S$ . This implies$S$ is parallel to$\left(\ker A\right)$ .

Thus, the$S$ is parallel to$\left(\ker A\right)$ .

(d)

The system $A\stackrel{鈫�}{x}=\stackrel{鈫�}{b}$equivalent to $\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}10\\ 20\end{array}\right]$gives $x_1+3x_2=10$.

Now,$S$and ${\left(\ker A\right)}^{鈯�}$are perpendicular. So, the corresponding equation in ${\left(\ker A\right)}^{鈯�}$is $3x_1鈭�x_2=0$.

Now, the intersection point is the solution of the equations $x_1+3x_2=10$and$3x_1鈭�x_2=0$. The solution is $x_1=1$and$x_2=3$.

Thus, the unique vector$x_2=3$is the intersection of $S$and${\left(\ker A\right)}^{鈯�}$.

The sketch of the unique vector is shown below:

(e)

Here, the vector${\stackrel{鈫�}{x}}_0$ is perpendicular to both$\left(\ker A\right)$ and $S$. This statement implies that the length of ${\stackrel{鈫�}{x}}_0$is the smallest of all the vectors.

Thus, the length of vector${\stackrel{鈫�}{x}}_0$is smallest of all vectors in $S$.