1) Consider the linear system \( \left\{\begin{array}{c}x+y+3 z=3 \\ -x+y+z=-1 \\ 2 x+3 y+8 z=4\end{array}\right. \) i- Solve the linear system. ii- Find \( \left|2 A^{\top} A^{2}\right| \).

**Step 1. Solve the linear system** The system is given by \[ \begin{cases} x + y + 3z = 3, \\ -x + y + z = -1, \\ 2x + 3y + 8z = 4. \end{cases} \] 1. Subtract the second equation from the first: \[ (x+y+3z) - (-x+y+z) = 3 - (-1). \] This simplifies to \[ 2x + 2z = 4 \quad\Longrightarrow\quad x + z = 2. \] Hence, \[ x = 2 - z. \] 2. Substitute \(x = 2-z\) into the second equation: \[ - (2-z) + y + z = -1. \] Expanding the left-hand side gives \[ -2 + z + y + z = -1 \quad\Longrightarrow\quad y + 2z = 1. \] Thus, \[ y = 1 - 2z. \] 3. Substitute \(x = 2-z\) and \(y = 1-2z\) into the first equation: \[ (2-z) + (1-2z) + 3z = 3. \] Simplify: \[ 2 - z + 1 - 2z + 3z = 3 \quad\Longrightarrow\quad 3 + 0z = 3. \] This equation holds for all \(z\) in the first equation. 4. Now substitute \(x = 2-z\) and \(y = 1-2z\) into the third equation: \[ 2(2-z) + 3(1-2z) + 8z = 4. \] Compute each term: \[ 4 - 2z + 3 - 6z + 8z = 4. \] Combine like terms: \[ 7 + ( -2z -6z + 8z) = 7 + 0z = 7. \] So the third equation becomes \[ 7 = 4, \] which is a contradiction. Thus, the system of equations is inconsistent and has **no solution**. --- **Step 2. Find \(\left|2 A^{\top} A^{2}\right|\)** Let \[ A = \begin{pmatrix} 1 & 1 & 3 \\ -1 & 1 & 1 \\ 2 & 3 & 8 \end{pmatrix}. \] We first note some properties of determinants: - For any square matrix \(B\) and scalar \(c\) in an \(n \times n\) matrix, \[ |cB| = c^n |B|. \] - The determinant of a transpose is the same as the determinant of the original matrix: \[ |A^{\top}| = |A|. \] - For powers of matrices, \[ |A^{2}| = |A|^2. \] Now, consider \[ \left|2 A^{\top} A^{2}\right|. \] Since the matrix \(2A^{\top}A^{2}\) is a \(3 \times 3\) matrix, factor out the scalar \(2\): \[ \left|2 A^{\top} A^{2}\right| = 2^{3}\left|A^{\top}A^{2}\right| = 8\,|A^{\top}||A^{2}|. \] Using the multiplicative property of determinants: \[ |A^{\top}| = |A| \quad \text{and} \quad |A^{2}| = |A|^2. \] Thus, \[ \left|2 A^{\top} A^{2}\right| = 8\,|A|\,|A|^{2} = 8\,|A|^3. \] Next, compute \(|A|\) for \[ A = \begin{pmatrix} 1 & 1 & 3 \\ -1 & 1 & 1 \\ 2 & 3 & 8 \end{pmatrix}. \] Using expansion for the determinant, \[ |A| = 1\begin{vmatrix}1 & 1 \\ 3 & 8\end{vmatrix} - 1\begin{vmatrix}-1 & 1 \\ 2 & 8\end{vmatrix} + 3\begin{vmatrix}-1 & 1 \\ 2 & 3\end{vmatrix}. \] Calculate each \(2 \times 2\) determinant: - \(\begin{vmatrix}1 & 1 \\ 3 & 8\end{vmatrix} = 1\cdot8 - 1\cdot3 = 8-3 = 5.\) - \(\begin{vmatrix}-1 & 1 \\ 2 & 8\end{vmatrix} = (-1)(8) - (1)(2) = -8-2 = -10.\) - \(\begin{vmatrix}-1 & 1 \\ 2 & 3\end{vmatrix} = (-1)(3) - (1)(2) = -3-2 = -5.\) Now substitute back: \[ |A| = 1\cdot5 - 1\cdot(-10) + 3\cdot(-5) = 5 + 10 - 15 = 0. \] Finally, \[ \left|2 A^{\top} A^{2}\right| = 8\,|A|^3 = 8\cdot0^3 = 0. \] --- **Final Answers:** i) The system has no solution. ii) \(\left|2 A^{\top} A^{2}\right| = 0.\)