1. The problem provides several matrices and a quadratic form $q(x,y,z) = 3x^2 + 4xy - y^2 + 8xz - 6yz + z^2$. We will analyze each given matrix and the quadratic form step-by-step. 2. First matrix: $$\begin{bmatrix}4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \end{bmatrix}$$ This is a $3 \times 3$ matrix. We can find its determinant or use it for other operations if needed. 3. Quadratic form $q(x,y,z)$: $$q(x,y,z) = 3x^2 + 4xy - y^2 + 8xz - 6yz + z^2$$ This can be represented in matrix form as $q(\mathbf{v}) = \mathbf{v}^T Q \mathbf{v}$ where $\mathbf{v} = \begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and $Q$ is a symmetric matrix. 4. Construct matrix $Q$ for $q$: The coefficients correspond to: - $3x^2$ means $Q_{11} = 3$ - $4xy$ means $Q_{12} + Q_{21} = 4$, so $Q_{12} = Q_{21} = 2$ - $-y^2$ means $Q_{22} = -1$ - $8xz$ means $Q_{13} + Q_{31} = 8$, so $Q_{13} = Q_{31} = 4$ - $-6yz$ means $Q_{23} + Q_{32} = -6$, so $Q_{23} = Q_{32} = -3$ - $z^2$ means $Q_{33} = 1$ Thus, $$Q = \begin{bmatrix}3 & 2 & 4 \\ 2 & -1 & -3 \\ 4 & -3 & 1 \end{bmatrix}$$ 5. Next matrix $M$: $$M = \begin{bmatrix}1 & 2 \\ 3 & 4 \end{bmatrix}$$ This is a $2 \times 2$ matrix, repeated twice in the input. 6. Matrix $A$ is a $6 \times 6$ upper-triangular matrix: $$A = \begin{bmatrix}2 & 2 & 1 & 8 & 6 & -9 \\ 0 & 1 & 2 & 4 & 7 & 5 \\ 0 & 0 & -1 & -6 & 9 & -2 \\ 0 & 0 & 0 & 2 & -1 & -1 \\ 0 & 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & -1 & 1 & 2 \end{bmatrix}$$ 7. Since $A$ is upper-triangular, its determinant is the product of diagonal entries: $$\det(A) = 2 \times 1 \times (-1) \times 2 \times 0 \times 2 = 0$$ Because one diagonal entry is zero, determinant is zero. 8. Summary: - Matrix $Q$ represents the quadratic form $q(x,y,z)$. - Matrix $M$ is a $2 \times 2$ matrix. - Matrix $A$ is upper-triangular with zero determinant. Final answers: - Quadratic form matrix $Q$: $$\begin{bmatrix}3 & 2 & 4 \\ 2 & -1 & -3 \\ 4 & -3 & 1 \end{bmatrix}$$ - Determinant of $A$ is $0$.