91影视

We have a plane, let鈥檚 call it$蟺$

$\mathrm{蟺}\mathrm{...3}{\mathrm{x}}_1+4{\mathrm{x}}_2+5{\mathrm{x}}_3=0$

For it to be kernel of matrix$A$ for any vector$\stackrel{鈫�}{v}鈭�蟺$ equality

$A\stackrel{鈫�}{v}=0$

But, we know that normal vector of $蟺$, that is$\stackrel{鈫�}{n}$

$\stackrel{鈫�}{\mathrm{n}}=\left[\begin{array}{c}3\\ 4\\ 5\end{array}\right]鈫�\stackrel{鈫�}{\mathrm{n}}.\stackrel{鈫�}{\mathrm{v}}=0鈫�\left[\begin{array}{ccc}3& 4& 5\end{array}\right].\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right]=0$

So we鈥檒l use it as requested matrix$A$

$A=\left[\begin{array}{ccc}3& 4& 5\end{array}\right]$

We know that plane is spanned by any two non collinear vectors that lay on it.

These vectors need to be perpendicular to the normal vector we computed earlier.

But we also know that image (B)is spanned by it's column vectors and that they span our plane, so we need to find 2 non collinear vectors from that plane and put them as column vectors of $B$.

Let's say that those two vectors are

role="math" localid="1660111125995" $\stackrel{鈫�}{{\mathrm{v}}_1}=\left[\begin{array}{c}4\\ 鈭�3\\ 0\end{array}\right],\stackrel{鈫�}{{\mathrm{v}}_2}=\left[\begin{array}{c}0\\ 5\\ 鈭�4\end{array}\right]$

Those are clearly perpendicular to normal, we can see it as follows

Use it as a requested matrix

Requested matrix is

$\begin{array}{l}\left[\begin{array}{ccc}3& 4& 5\end{array}\right].\left[\begin{array}{c}4\\ 鈭�3\\ 0\end{array}\right]=0\\ \left[\begin{array}{ccc}3& 4& 5\end{array}\right].\left[\begin{array}{c}0\\ 5\\ 鈭�4\end{array}\right]=0\end{array}$

Use it as a requested matrix$A$

$\mathrm{A}=\left[\begin{array}{ccc}3& 4& 5\end{array}\right]$

Requested matrix $B$is

$\mathrm{B}=\left[\begin{array}{cc}4& 0\\ 鈭�3& 5\\ 0& 鈭�4\end{array}\right]$

Matrix$A$

$\mathrm{A}=\left[\begin{array}{ccc}3& 4& 5\end{array}\right]$

Matrix$B$

$B=\left[\begin{array}{cc}4& 0\\ 鈭�3& 5\\ 0& 鈭�4\end{array}\right]$