91影视

Consider the matrix A and let B= rref(A).

a) The objective is to verify whether ker(A)=ker(B) or not.

Here, matrix B is obtained by reducing matrix A to row reduced echelon form. Note that Gauss-Jordan elimination does not change the solution of a system. Therefore, the systems$A\stackrel{鈫�}{x}=\stackrel{鈫�}{0}\mathrm{and}B\stackrel{鈫�}{x}=\stackrel{鈫�}{0}$ have the same solutions. Also, note that the kernel of a linear system is the solution space of the system $A\stackrel{鈫�}{x}=\stackrel{鈫�}{0}$. As the solution spaces of the system$A\stackrel{鈫�}{x}=\stackrel{鈫�}{0}$ and$B\stackrel{鈫�}{x}=\stackrel{鈫�}{0}$ are equal, therefore, ker(A) = ker(B).

b) the objective is to verify whether im(A)=im(B) or not. The image of a matrix A is defined as,$im\left(A\right)=\left\{\stackrel{鈫�}{b}:A\stackrel{鈫�}{x}=\stackrel{鈫�}{b}\right\}$

Here, matrix B is obtained by reducing matrix A to row reduced echelon form. Note that im(A)鈮爄m(B).

Because let$A=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$ Then the row reduced echelon form of A is$\mathrm{B}=\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$

From the matrix A note that the image of A is,

$$\begin{gathered} im\left(A\right)=span\left\{\left[\begin{array}{c}1\\ 1\end{array}\right]\right\} \\ \mathrm{From}\mathrm{the}\mathrm{matrix}\mathrm{B}\mathrm{note}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{of}\mathrm{B}\mathrm{is}, \\ \mathrm{im}\left(\mathrm{B}\right)=\mathrm{span}\left\{\left[\begin{array}{c}1\\ 0\end{array}\right]\right\} \\ \mathrm{Thus},\mathrm{im}\left(\mathrm{A}\right)鈮�\mathrm{im}\left(\mathrm{B}\right)\mathrm{even}\mathrm{though}\mathrm{B}=\mathrm{rref}\left(\mathrm{A}\right). \end{gathered}$$