Answer:
Explanation:
Let's denote the matrix \( D \) as follows:
\[ D = \begin{bmatrix} 2 & -6 & 1 \\ 4 & 2 & 5 \\ -11 & 46 & -37 \end{bmatrix} \]
Also, let \( D_2 \) be the matrix obtained by replacing the second column of \( D \) with the column vector \( \begin{bmatrix} 4 \\ -6 \\ 2 \end{bmatrix} \):
\[ D_2 = \begin{bmatrix} 2 & 4 & 1 \\ 4 & -6 & 5 \\ -11 & 2 & -37 \end{bmatrix} \]
Now, let's calculate the determinant \( D \) (denoted as \( \text{det}(D) \)) and \( D_2 \) (denoted as \( \text{det}(D_2) \)).
\[ \text{det}(D) = 2 \cdot (2 \cdot (-37) - 5 \cdot 46) - (-6) \cdot (4 \cdot (-37) - 5 \cdot (-11)) + 1 \cdot (4 \cdot 46 - 2 \cdot (-11)) \]
Simplify the above expression to find \( \text{det}(D) \).
\[ \text{det}(D_2) = 2 \cdot (-6 \cdot (-37) - 5 \cdot 2) - 4 \cdot (4 \cdot (-37) - 5 \cdot (-11)) + 1 \cdot (4 \cdot 2 - (-6) \cdot (-11)) \]
Simplify the above expression to find \( \text{det}(D_2) \).
If \( \text{det}(D) \\eq 0 \), then the system of equations represented by \( D \) has a unique solution, and the solution can be found by using Cramer's rule:
\[ x = \frac{\text{det}(D_1)}{\text{det}(D)} \]
\[ y = \frac{\text{det}(D_2)}{\text{det}(D)} \]
\[ z = \frac{\text{det}(D_3)}{\text{det}(D)} \]
where \( D_1 \), \( D_2 \), and \( D_3 \) are matrices obtained by replacing the first, second, and third columns of \( D \) with the column vector on the right side, respectively.
If \( \text{det}(D) = 0 \), then the system may have no solution or infinitely many solutions.
Perform the calculations, and if you find \( \text{det}(D) \\eq 0 \), you can proceed to find the values of \( x \), \( y \), and \( z \). If \( \text{det}(D) = 0 \), then the system may have no unique solution.