Given Information
We are given that A is a $3 \times 3$ matrix with eigenvalues as 3 and 4/5 and 3/5. The corresponding eigenvectors are \[ \mathbf { v } _ { 1 } = \left[ \begin{array} { c } { 1 } \\ { 0 } \\ { - 3 } \end{array} \right] , \mathbf { v } _ { 2 } = \left[ \begin{array} { c } { 2 } \\ { 1 } \\ { - 5 } \end{array} \right] , \text { and } \mathbf { v } _ { 3 } = \left[ \begin{array} { c } { - 3 } \\ { - 3 } \\ { 7 } \end{array} \right] \] We have to find the solution of the equation $\mathbf { x } _ { k + 1 } = A \mathbf { x } _ { k }$ for given $x_0$: \[ \mathbf { x } _ { 0 } = \left[ \begin{array} { r } { - 2 } \\ { - 5 } \\ { 3 } \end{array} \right] \]

Step-1:
Write $x_0$ as combination of vectors $v_1$, $v_2$ and $v_3$ \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&{ - 2}\\ 0&1&{ - 3}&{ - 5}\\ { - 3}&{ - 5}&7&3 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&{ - 2}\\ 0&1&{ - 3}&{ - 5}\\ 0&1&{ - 2}&{ - 3} \end{array}} \right]::\left\{ \begin{array}{l} {R_3} = {R_3} + 3{R_1}\\ {R_2} \Leftrightarrow {R_3} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&{ - 2}\\ 0&1&{ - 3}&{ - 5}\\ 0&0&1&2 \end{array}} \right]::\left\{ {{R_3} = {R_3} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 3}&{ - 2}\\ 0&1&0&1\\ 0&0&1&2 \end{array}} \right]::\left\{ {{R_2} = {R_2} + 3{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{l}} 1&2&0&4\\ 0&1&0&1\\ 0&0&1&2 \end{array}} \right]::\left\{ {{R_1} = {R_1} + 3{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{l}} 1&0&0&2\\ 0&1&0&1\\ 0&0&1&2 \end{array}} \right]::\left\{ {{R_1} = {R_1} - 2{R_2}} \right\} \end{array}\] Therefore, \[ \mathbf { x } _ { 0 } = 2 \mathbf { v } _ { 1 } + \mathbf { v } _ { 2 } + 2 \mathbf { v } _ { 3 } \]

Step-2: Compute $Av_1$, $Av_2$ and $Av_3$
Since $v_1$, $v_2$ and $v_3$ are eigenvalues of A, hence they satisfy the equations \[ [ A - 3 I ] \mathbf { v } _ { 1 } = \mathbf { 0 } , \left[ A - \dfrac { 4 } { 5 } I \right] \mathbf { v } _ { 2 } = \mathbf { 0 } , \text { and } \left[ A - \dfrac { 3 } { 5 } I \right] \mathbf { v } _ { 3 } = \mathbf { 0 } \] Thus, \[ A \mathbf { v } _ { 1 } = 3 \mathbf { v } _ { 1 } , A \mathbf { v } _ { 2 } = \dfrac { 4 } { 5 } \mathbf { v } _ { 2 } , \text { and } A \mathbf { v } _ { 3 } = \dfrac { 4 } { 5 } \mathbf { v } _ { 3 } \]

Step-3:
Multiply $x_0$ by A: \[\begin{array}{l} A{{\bf{x}}_0} = 2A{{\bf{v}}_1} + A{{\bf{v}}_2} + 2A{{\bf{v}}_3}\\ {{\bf{x}}_1} = 2\cdot3\cdot{{\bf{v}}_1} + \left( {\dfrac{4}{5}} \right)\cdot{{\bf{v}}_2} + 2\cdot\left( {\dfrac{3}{5}} \right){{\bf{v}}_3}\\ {{\bf{x}}_1} = 2\cdot3\cdot{{\bf{v}}_1} + \left( {\dfrac{4}{5}} \right)\cdot{{\bf{v}}_2} + 2\cdot\left( {\dfrac{3}{5}} \right){{\bf{v}}_3} \end{array}\] The required vector $x_1$ is:

\[ \mathbf { x } _ { k } = 2 \cdot 3 ^ { k } \mathbf { v } _ { 1 } + \left( \dfrac { 4 } { 5 } \right) ^ { k } \mathbf { v } _ { 2 } + 2 \cdot \left( \dfrac { 3 } { 5 } \right) ^ { k } \mathbf { v } _ { 3 } \] As k reaches infinite: \[\begin{array}{l} {{\bf{x}}_k} = 2\cdot{3^k}{{\bf{v}}_1} + {\left( {\dfrac{4}{5}} \right)^k}{{\bf{v}}_2} + 2\cdot{\left( {\dfrac{3}{5}} \right)^k}{{\bf{v}}_3}\\ {{\bf{x}}_k} = 2\cdot{3^k}{{\bf{v}}_1} + 0 + 0\\ {{\bf{x}}_k} = 2\cdot{3^k}\left[ {\begin{array}{*{20}{c}} 1\\ 0\\ { - 3} \end{array}} \right] \end{array}\] Therefore, when k reaches infinite,

\[{{\bf{x}}_k} = 2\cdot{3^k}\left[ {\begin{array}{*{20}{c}} 1\\ 0\\ { - 3} \end{array}} \right]\]