Solve for the unknown variables, using Cramer's rule \( \begin{array}{l}2 x+z=4+y \\ 3 x+6 y=21-9 z \\ -2 z+3 y-x=-5\end{array} \)
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To solve the system of equations using Cramer's rule, we first need to rewrite the given equations in a standard linear form \(Ax = b\). Here's how we rearrange them: 1. \(2x - y + z = 4\) 2. \(3x - 6y + 9z = 21\) (divide by 3 to simplify to \(x - 2y + 3z = 7\)) 3. \(-x + 3y - 2z = -5\) (multiply by -1 to simplify to \(x - 3y + 2z = 5\)) Now the system can be written in matrix form \(Ax = b\): \[ \begin{bmatrix} 2 & -1 & 1 \\ 1 & -2 & 3 \\ 1 & 3 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ 7 \\ 5 \end{bmatrix} \] Next, we compute the determinant of matrix \(A\): \[ D = \begin{vmatrix} 2 & -1 & 1 \\ 1 & -2 & 3 \\ 1 & 3 & -2 \end{vmatrix} = 2 \begin{vmatrix} -2 & 3 \\ 3 & -2 \end{vmatrix} - (-1) \begin{vmatrix} 1 & 3 \\ 1 & -2 \end{vmatrix} + 1 \begin{vmatrix} 1 & -2 \\ 1 & 3 \end{vmatrix} \] Calculating each of these 2x2 determinants: 1. \(\begin{vmatrix} -2 & 3 \\ 3 & -2 \end{vmatrix} = (-2)(-2) - (3)(3) = 4 - 9 = -5\) 2. \(\begin{vmatrix} 1 & 3 \\ 1 & -2 \end{vmatrix} = (1)(-2) - (3)(1) = -2 - 3 = -5\) 3. \(\begin{vmatrix} 1 & -2 \\ 1 & 3 \end{vmatrix} = (1)(3) - (-2)(1) = 3 + 2 = 5\) Now substituting back: \[ D = 2(-5) + 5 + 5 = -10 + 5 + 5 = 0 \] Since the determinant \(D = 0\), the system of equations does not have a unique solution. This means it is either inconsistent (no solutions) or has infinitely many solutions (dependent equations). For further steps, we would need to analyze the rank of the coefficient matrix \(A\) and the augmented matrix \([A|b]\) to draw a proper conclusion.
