Let $M = \left[ \begin{array} { c l } { - 4 } & { 3 } \\ { - 10 } & { 7 } \end{array} \right]$.  

1. Show that $Ma = a$ if $a = \left[ \begin{array} { l } { 3 } \\ { 5 } \end{array} \right]$.

   Step (a):

   $\left[ \begin{array} { c c } { - 4 } & { 3 } \\ { - 10 } & { 7 } \end{array} \right] \left[ \begin{array} { l } { 3 } \\ { 5 } \end{array} \right] = \left[ \begin{array} { c } { - 4 ( 3 ) + 3 ( 5 ) } \\ { - 10 ( 3 ) + 7 ( 5 ) } \end{array} \right]$

2. Find two other vectors b and c such that $Mb = b$ and $Mc = c$. Think before rushing to compute.

   Steps (b):

   In general, $\left[ \begin{array} { c c } { - 4 } & { 3 } \\ { - 10 } & { 7 } \end{array} \right] \left[ \begin{array} { l } { x } \\ { y } \end{array} \right] = \left[ \begin{array} { c } { - 4 ( x ) + 3 ( y ) } \\ { - 10 ( x ) + 7 ( y ) } \end{array} \right] = \left[ \begin{array} { l } { x } \\ { y } \end{array} \right]$

   Set up a system of equations and solve for $y$ in terms of $x$.

3. Find a vector d such that $Md = 2d$.

   Hint (c):

   For parts (c), (d), and (e), follow the steps in part (b).

4. Show that there is no vector $e ≠ 0$ such that $Me = 3e$.

5. Show that there is no vector $v ≠ 0$ such that $Mv = kv$ if $k ≠ 1, 2$.

   Step 1(e):

   $\left[ \begin{array} { c } { - 4 x + 3 y } \\ { - 10 x + 7 y } \end{array} \right] = \left[ \begin{array} { l } { k x } \\ { k y } \end{array} \right]$

   Step 2(e):

   Solving for $x$ in the top equation and $y$ in the bottom equation:

   $x = \frac { 3 y } { k + 4 } , y = \frac { - 10 x } { k - 7 }$

   Step 3(e):

   By substitution:

   $x = \frac { 3 ( \frac { - 10 x } { k - 7 } ) } { k + 4 }$

   Step 4(e):

   $x\left(k + 4\right)\left(k − 7\right) = −30x$

   Since we have assumed $x ≠ 0$, divide by $x$ and solve for $k$.