<s>[INST] Finding kernel and range of a linear transformation We are given:

Find $\ker(T)$, and $\textrm{rng}(T)$, where $T$ is the linear transformation given by
$$T:\mathbb{R^3} \rightarrow \mathbb{R^3}$$ 
with standard matrix 
$$ A = \left[\begin{array}{rrr}
    1 & -1 & 3\\
    5 & 6 & -4\\
    7 & 4 & 2\\
    \end{array}\right]\textrm{.}
$$

The kernel can be found in a $2 \times 2$ matrix as follows:
$$ L = \left[\begin{array}{rrr}
    a & b\\
    c & d\\
    \end{array}\right] = (a+d) + (b+c)t
$$
Then to find the kernel of $L$ we set
$$(a+d) + (b+c)t = 0$$
$$d = -a$$
$$c = -b$$
so that the kernel of $L$ is the set of all matrices of the form
$$ A = \left[\begin{array}{rrr}
    a & b\\
    -b & -a\\
    \end{array}\right]
$$
but I do not know how to apply that to this problem.
 [/INST] $$
A = \left[\begin{array}{rrr}
    1 & -1 & 3\\
    5 & 6 & -4\\
    7 & 4 & 2\\
    \end{array}\right]
$$
Consider a linear map represented as a $m × n$ matrix $A$ .
The kernel of this linear map is the set of solutions to the equation $Ax = 0$
$$
ker(A)=\{x \in R^n|Ax=0\}
$$
$$
det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0
$$
Since $det(A)=0$ , $x\ne0$ and $0$ is a vector here. 
$$
\left[\begin{array}{rrr}
    1 & -1 & 3\\
    5 & 6 & -4\\
    7 & 4 & 2\\
    \end{array}\right]
\left[\begin{array}{r}
   a\\b\\c
   \end{array}\right]
=\left[\begin{array}{r}
   0\\0\\0
   \end{array}\right]
$$
In row-reduced form,
$$
A = \left[\begin{array}{rrr}
    1 & 0 & \frac{14}{11}\\
    0 & 1 & \frac{-19}{11}\\
    0 & 0 & 0\\
    \end{array}\right]
$$
$$x=\frac{-14}{11}z$$
$$y=\frac{19}{11}z$$
$$
\left[\begin{array}{r}
   a\\b\\c
   \end{array}\right]
=\left[\begin{array}{r}
   -14\\19\\11
   \end{array}\right]z
$$
Similarly for $2×2$ matrix .
</s>