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.
Best Answer
$$ A = \left[\begin{array}{rrr} 1 & -1 & 3\\ 5 & 6 & -4\\ 7 & 4 & 2\\ \end{array}\right] $$ 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 .