**Solution to Linear Algebra Hoffman & Kunze Chapter 1.3 Exercise 1.3.3**

Solution: The system $AX=2X$ is

$$\left[\begin{array}{ccc}6&-4&0\\4&-2&0\\-1&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=2\left[\begin{array}{c}x\\y\\z\end{array}\right]$$which is the same as

\begin{alignat*}{1}

6x-4y&=2x\\

4x-2y&=2y\\

-x+3z&=2z

\end{alignat*}which is equivalent to

\begin{alignat*}{1}

4x-4y&=0\\

4x-4y&=0\\

-x+z&=0

\end{alignat*}The matrix of coefficients is

$$\left[\begin{array}{ccc}4&-4&0\\4&-4&0\\-1&0&1\end{array}\right]$$which row-reduces to

$$\left[\begin{array}{ccc}1&0&-1\\0&1&-1\\0&0&0\end{array}\right]$$Thus the solutions are all elements of $F^3$ of the form $(x,x,x)$ where $x\in F$.

The system $AX=3X$ is

$$\left[\begin{array}{ccc}6&-4&0\\4&-2&0\\-1&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=3\left[\begin{array}{c}x\\y\\z\end{array}\right]$$which is the same as

\begin{alignat*}{1}

6x-4y &=3x\\

4x-2y&=3y\\

-x+3z&=3z

\end{alignat*}

which is equivalent to

\begin{alignat*}{1}

3x-4y&=0\\

x-2y&=0\\

-x&=0

\end{alignat*}The matrix of coefficients is

$$\left[\begin{array}{ccc}3&-4&0\\1&-2&0\\-1&0&0\end{array}\right]$$which row-reduces to$$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right]$$Thus the solutions are all elements of $F^3$ of the form $(0,0,z)$ where $z\in F$.