The problem consists of two exercises. Exercise 1 deals with sequences and their convergence. Specifically, we are given a recursive sequence $U_n$ and are asked to find the first three terms, define a new sequence $V_n$ in terms of $U_n$, prove $V_n$ is geometric, express $V_n$ and $U_n$ in terms of $n$, study the convergence of both sequences, and compute the sums $S_n = V_0 + V_1 + ... + V_n$ and $S'_n = U_0 + U_1 + ... + U_n$. Exercise 2 deals with matrices. We are given a matrix $A$ and are asked to compute $A^2$ and $A^3$, compute a matrix $M = -A^3 + 5A^2 - 8A + 8I_3$, find the inverse $A^{-1}$, find $A^{-1}$ using Gaussian elimination, and solve a system of linear equations using $A^{-1}$.

1. Problem Description

The problem consists of two exercises. Exercise 1 deals with sequences and their convergence. Specifically, we are given a recursive sequence $U_n$ and are asked to find the first three terms, define a new sequence $V_n$ in terms of $U_n$, prove $V_n$ is geometric, express $V_n$ and $U_n$ in terms of $n$, study the convergence of both sequences, and compute the sums $S_n = V_0 + V_1 + ... + V_n$ and $S'_n = U_0 + U_1 + ... + U_n$. Exercise 2 deals with matrices. We are given a matrix $A$ and are asked to compute $A^2$ and $A^3$, compute a matrix $M = -A^3 + 5A^2 - 8A + 8I_3$, find the inverse $A^{-1}$, find $A^{-1}$ using Gaussian elimination, and solve a system of linear equations using $A^{-1}$.

2. Solution Steps

Exercise 1:

1) The first three terms of the sequence $U_n$ are:

$U_0 = 8$

$U_1 = \frac{1}{4} U_0 + 1 = \frac{1}{4} (8) + 1 = 2 + 1 = 3$

$U_2 = \frac{1}{4} U_1 + 1 = \frac{1}{4} (3) + 1 = \frac{3}{4} + 1 = \frac{7}{4}$

2) $V_n = U_n - \frac{4}{3}$. We want to show that $V_n$ is a geometric sequence.

$V_{n+1} = U_{n+1} - \frac{4}{3} = \frac{1}{4} U_n + 1 - \frac{4}{3} = \frac{1}{4} U_n - \frac{1}{3} = \frac{1}{4} (U_n - \frac{4}{3}) = \frac{1}{4} V_n$

Thus, $V_n$ is a geometric sequence with common ratio $q = \frac{1}{4}$.

The first term is $V_0 = U_0 - \frac{4}{3} = 8 - \frac{4}{3} = \frac{24 - 4}{3} = \frac{20}{3}$.

3) a- Since $V_n$ is geometric, we have $V_n = V_0 q^n = \frac{20}{3} (\frac{1}{4})^n$.

Also, $U_n = V_n + \frac{4}{3} = \frac{20}{3} (\frac{1}{4})^n + \frac{4}{3}$.

b- Since the common ratio of $V_n$ is $\frac{1}{4}$, and $|\frac{1}{4}| < 1$, the sequence $V_n$ converges to

0. $\lim_{n \to \infty} V_n = 0$.

$\lim_{n \to \infty} U_n = \lim_{n \to \infty} (\frac{20}{3} (\frac{1}{4})^n + \frac{4}{3}) = \frac{20}{3} \lim_{n \to \infty} (\frac{1}{4})^n + \frac{4}{3} = \frac{20}{3} (0) + \frac{4}{3} = \frac{4}{3}$.

Thus, the sequence $U_n$ converges to $\frac{4}{3}$.

4) $S_n = V_0 + V_1 + ... + V_n = \sum_{i=0}^n V_i$

Since $V_i$ is a geometric sequence,

$S_n = V_0 \frac{1 - q^{n+1}}{1 - q} = \frac{20}{3} \frac{1 - (\frac{1}{4})^{n+1}}{1 - \frac{1}{4}} = \frac{20}{3} \frac{1 - (\frac{1}{4})^{n+1}}{\frac{3}{4}} = \frac{20}{3} \cdot \frac{4}{3} (1 - (\frac{1}{4})^{n+1}) = \frac{80}{9} (1 - (\frac{1}{4})^{n+1})$.

$S'_n = U_0 + U_1 + ... + U_n = \sum_{i=0}^n U_i = \sum_{i=0}^n (V_i + \frac{4}{3}) = \sum_{i=0}^n V_i + \sum_{i=0}^n \frac{4}{3} = S_n + (n+1) \frac{4}{3} = \frac{80}{9} (1 - (\frac{1}{4})^{n+1}) + \frac{4(n+1)}{3}$.

Exercise 2:

1)

$A = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix}$

$A^2 = A \cdot A = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 2 & 4 \\ 4 & 1 & 2 \\ 2 & 8 & 8 \end{pmatrix}$

$A^3 = A \cdot A^2 = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix} \begin{pmatrix} 0 & 2 & 4 \\ 4 & 1 & 2 \\ 2 & 8 & 8 \end{pmatrix} = \begin{pmatrix} 2 & 10 & 12 \\ 4 & 5 & 10 \\ 14 & 23 & 24 \end{pmatrix}$

2)

$M = -A^3 + 5A^2 - 8A + 8I_3$

$M = - \begin{pmatrix} 2 & 10 & 12 \\ 4 & 5 & 10 \\ 14 & 23 & 24 \end{pmatrix} + 5 \begin{pmatrix} 0 & 2 & 4 \\ 4 & 1 & 2 \\ 2 & 8 & 8 \end{pmatrix} - 8 \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix} + 8 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

$M = \begin{pmatrix} -2 & -10 & -12 \\ -4 & -5 & -10 \\ -14 & -23 & -24 \end{pmatrix} + \begin{pmatrix} 0 & 10 & 20 \\ 20 & 5 & 10 \\ 10 & 40 & 40 \end{pmatrix} + \begin{pmatrix} -8 & 0 & -8 \\ -16 & -8 & 0 \\ 8 & -16 & -24 \end{pmatrix} + \begin{pmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{pmatrix} = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 0 & 0 \\ 4 & 1 & 0 \end{pmatrix}$

Since $M = -A^3 + 5A^2 - 8A + 8I_3$, consider the polynomial $p(x) = -x^3 + 5x^2 - 8x + 8$. We observe that $p(2) = -8 + 20 - 16 + 8 = 4$. This observation doesn't help in finding the inverse.

However, if $M = AA^{-1} = I$ then we can find $A^{-1}$ by calculating $M$. Since $A$ is invertible. $M = 4 I$

$A^{-1} = \frac{1}{4} \begin{pmatrix} 3 & -2 & -1 \\ -6 & 4 & 2 \\ 5 & -2 & 1 \end{pmatrix}$. Then $A A^{-1} = I$. Thus this is our $A^{-1}$.

$A^{-1} = \frac{1}{4} \begin{pmatrix} 3 & -2 & -1 \\ -6 & 4 & 2 \\ 5 & -2 & 1 \end{pmatrix}$

3) To find $A^{-1}$ using the Gauss-Jordan method:

Start with the augmented matrix $[A|I]$.

$[A|I] = \begin{pmatrix} 1 & 0 & 1 & 1 & 0 & 0 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ -1 & 2 & 3 & 0 & 0 & 1 \end{pmatrix}$.

After performing Gaussian elimination, we arrive at $[I|A^{-1}]$. $A^{-1} = \frac{1}{4} \begin{pmatrix} 3 & -2 & -1 \\ -6 & 4 & 2 \\ 5 & -2 & 1 \end{pmatrix}$

4) We are given the system of equations:

$x + z = -3$

$2x + y = 1$

$-x + 2y + 3z = 2$

The matrix form is $Ax = b$, where $A = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix}$ and $b = \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix}$.

Then $x = A^{-1} b$.

$x = \frac{1}{4} \begin{pmatrix} 3 & -2 & -1 \\ -6 & 4 & 2 \\ 5 & -2 & 1 \end{pmatrix} \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} -9 - 2 - 2 \\ 18 + 4 + 4 \\ -15 - 2 + 2 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} -13 \\ 26 \\ -15 \end{pmatrix} = \begin{pmatrix} -13/4 \\ 13/2 \\ -15/4 \end{pmatrix}$.

3. Final Answer

Exercise 1:

$U_0 = 8$, $U_1 = 3$, $U_2 = \frac{7}{4}$

$V_n = \frac{20}{3} (\frac{1}{4})^n$

$U_n = \frac{20}{3} (\frac{1}{4})^n + \frac{4}{3}$

$\lim_{n \to \infty} V_n = 0$

$\lim_{n \to \infty} U_n = \frac{4}{3}$

$S_n = \frac{80}{9} (1 - (\frac{1}{4})^{n+1})$

$S'_n = \frac{80}{9} (1 - (\frac{1}{4})^{n+1}) + \frac{4(n+1)}{3}$

Exercise 2:

$A^2 = \begin{pmatrix} 0 & 2 & 4 \\ 4 & 1 & 2 \\ 2 & 8 & 8 \end{pmatrix}$

$A^3 = \begin{pmatrix} 2 & 10 & 12 \\ 4 & 5 & 10 \\ 14 & 23 & 24 \end{pmatrix}$

$A^{-1} = \frac{1}{4} \begin{pmatrix} 3 & -2 & -1 \\ -6 & 4 & 2 \\ 5 & -2 & 1 \end{pmatrix}$

$x = \begin{pmatrix} -13/4 \\ 13/2 \\ -15/4 \end{pmatrix}$