91影视

Given a matrix

$\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 5\\ 1& 3& 7\end{array}\right]$

The column vectors of the matrix A are:

${\stackrel{鈫�}{v}}_1=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$, ${\stackrel{鈫�}{v}}_2=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]$and${\stackrel{鈫�}{v}}_3=\left[\begin{array}{l}1\\ 5\\ 7\end{array}\right]$

Here, ${\stackrel{鈫�}{v}}_1$and${\stackrel{鈫�}{v}}_2$ both are non-zero vectors.

Also,${\stackrel{鈫�}{v}}_2$ is not a scaler multiple of ${\stackrel{鈫�}{v}}_1$.

So, the vectors ${\stackrel{鈫�}{v}}_1$and${\stackrel{鈫�}{v}}_2$ are not redundant vectors.

Assume that${\stackrel{鈫�}{v}}_3$ is redundant.

So, it can be written as a linear combination of ${\stackrel{鈫�}{v}}_1$and${\stackrel{鈫�}{v}}_2$.

Let${\stackrel{鈫�}{v}}_3=c_1{\stackrel{鈫�}{v}}_1+c_2{\stackrel{鈫�}{v}}_2$

Consider the augmented matrix

$\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}|\begin{array}{l}1\\ 5\\ 7\end{array}\right]$

Apply Row operations as follows:

$\begin{array}{l}\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}|\begin{array}{l}1\\ 5\\ 7\end{array}\right]\underset{鈫�}{R_2:R_2鈭�R_1}\left[\begin{array}{cc}1& 1\\ 0& 1\\ 1& 3\end{array}|\begin{array}{l}1\\ 4\\ 7\end{array}\right]\\ \left[\begin{array}{cc}1& 1\\ 0& 1\\ 1& 3\end{array}|\begin{array}{l}1\\ 4\\ 7\end{array}\right]\underset{鈫�}{R_3:R_3鈭�R_1}\left[\begin{array}{cc}1& 1\\ 0& 1\\ 0& 2\end{array}|\begin{array}{l}1\\ 4\\ 6\end{array}\right]\\ \left[\begin{array}{cc}1& 1\\ 0& 1\\ 0& 2\end{array}|\begin{array}{l}1\\ 4\\ 6\end{array}\right]\underset{鈫�}{R_3:R_3鈭�2R_2}\left[\begin{array}{cc}1& 1\\ 0& 1\\ 0& 0\end{array}|\begin{array}{l}1\\ 4\\ 鈭�2\end{array}\right]\end{array}$

Thus the matrix implies:

$\begin{array}{c}{c}_1+c_2=1\\ 0鈰�c_1+c_2=4\\ 0鈰�c_1+0鈰�c_2=鈭�2\end{array}$

The last equation implies$0=鈭�2$ which is clearly false.

Hence, our assumption was wrong.

Thus, ${\stackrel{鈫�}{v}}_3$is not redundant.

Since, the three vectors are linearly independent, therefore the basis of the image of matrix A is

$\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right],\text{鈥�}\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right],\text{鈥�}\left[\begin{array}{l}1\\ 5\\ 7\end{array}\right]$