Show that T maps the whole of the 3-dimension space $R^3$ into a plane $∏$ giving its equation.

Question

I have a question about an exam pastpaper question. The question is as follows:

The linear transformation T is represented by the matrix T given as:

$T = \begin{bmatrix}1&3&-1\\2&4&-4\\-1&2&6\end{bmatrix}$

The question first begins by asking the condition for which

$T\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}a\\b\\c\end{bmatrix}$

has an infinite set of solutions.

The condition was found to be: $8a - 5b - 2c = 0$

However, it then carries on asking to show that T maps the whole of the 3-dimensional space $R^3$ into a plane $∏$ giving its equation, for which the answer turns out to be: $8x - 5y - 2z =0$.

How does the condition found earlier directly relate to the equation of $∏$?

Answer 1

Reducing by rows, we find

$$T=\begin{pmatrix} 1&3&-1\\ 2&4&-4\\ -1&2&6\end{pmatrix}\stackrel{R_2-2R_1,\,R_3+R_1}\longrightarrow \begin{pmatrix} 1&3&-1\\ 0&-2&-2\\ 0&5&5\end{pmatrix}\longrightarrow\begin{pmatrix} 1&3&-1\\ 0&-2&-2\\ 0&0&0\end{pmatrix}$$

and we can clearly see that $\;\text{rank}\,T=2\;$ from which we deduce that $\;\dim Im(T)=2\implies T\;$ maps the space $\;\Bbb R^3\;$ to a two-dimensional space, i.e. to a plane. In order to find out the form of the plane, we need to look at the matrix's columns and find a basis for it. For this, we shall take the transpose of $\;T\;$ and on this carry on a process similar to the above one:

$$T^t=\begin{pmatrix} 1&2&-1\\ 3&4&2\\ -1&-4&6\end{pmatrix}\stackrel{R_2-3R_1\\R_3+R_1}\longrightarrow\begin{pmatrix} 1&2&-1\\ 0&-2&5\\ 0&-2&5\end{pmatrix}\longrightarrow \begin{pmatrix} 1&2&-1\\ 0&-2&5\\ 0&0&0\end{pmatrix}$$

and we get then that

$$\text{Im}\,T=Span\left\{\,v_1:=(1,2,-1)^t ,\;v_2:=(0,-2,5)^t\,\right\}$$

and if we want the equation of the plane spanned by the above two linearly independent vectors we first evaluate its normal vector:

$$\text{Normal to the plane:}\;v_1\times v_2=\begin{vmatrix}i&j&k\\1&2&-1\\0&-2&5\end{vmatrix}=(8,\,-5,\,-2)^t$$

From which the cartesian equation for the plane is $\;8x-5y-2z=0\;$

Answer 2

For there to be any solution at all, $(a,b,c)^T$ must lie in the column space—the image—of $T$. Moreover, since $T$ is rank-deficient (which you can determine in various ways) whenever there is a solution to the equation, then there must be an infinite number of them, so the condition $8a-5b-2c=0$ for there to be an infinite number of solutions in fact characterizes the image of $T$. Writing this equation as $(8,-5,-2)\cdot(a,b,c)=0$, it says that the image of $T$ consists of all vectors orthogonal to $(8,-5,-2)^T$, which is exactly the plane with equation $8x-5y-2z=0$.