91Ó°ÊÓ

Represent the system of equations in matrix form as \( AX = B \), where \( A \) is the matrix of coefficients, \( X \) is the column matrix \([x, y, z]^T\), and \( B \) is the column matrix of constants. The given system is: \[\begin{bmatrix} 4 & -1 & 3 \3 & 1 & 1 \2 & -1 & 4\end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} -3 \ 0 \ 0 \end{bmatrix}\]

Calculate the determinant of the coefficient matrix \( A \) to determine if Cramer's rule can be applied. The determinant is:\[D = \begin{vmatrix} 4 & -1 & 3 \3 & 1 & 1 \2 & -1 & 4\end{vmatrix}\]Using row expansion, calculate:\[D = 4 \begin{vmatrix} 1 & 1 \ -1 & 4 \end{vmatrix} + 1 \begin{vmatrix} 3 & 1 \ 2 & 4 \end{vmatrix} + 3 \begin{vmatrix} 3 & 1 \ 2 & -1 \end{vmatrix}\]Perform the calculations for each 2x2 determinant, then solve for the total determinant:\[D = 4(1 \cdot 4 - 1 \cdot -1) - 1(3 \cdot 4 - 1 \cdot 2) + 3(3 \cdot -1 - 1 \cdot 2)\]\[D = 4(4 + 1) - 1(12 - 2) + 3(-3 - 2)\]\[D = 4 \cdot 5 - 1 \cdot 10 + 3 \cdot -5\]\[D = 20 - 10 - 15\]\[D = -5\]

With \( D eq 0 \), use Cramer's Rule. Find the determinants \( D_x, D_y, \) and \( D_z \):1. \( D_x \) (replace the first column of \( A \) with \( B \)):\[D_x = \begin{vmatrix} -3 & -1 & 3 \0 & 1 & 1 \0 & -1 & 4\end{vmatrix}\]Calculate:\[D_x = -3\left(1 \cdot 4 - (-1) \cdot 1\right) -1\left(0\cdot 4 - 0\cdot -1\right) + 3\left(0\cdot -1 - 1 \cdot 0\right)\]\[D_x = -3 \cdot 5 = -15\]2. \( D_y \) (replace the second column of \( A \) with \( B \)):\[D_y = \begin{vmatrix} 4 & -3 & 3 \3 & 0 & 1 \2 & 0 & 4\end{vmatrix}\]Calculate:\[D_y = 4(0 \cdot 4 - 0 \cdot 1) - (-3)(3 \cdot 4 - 1 \cdot 2) + 3(3 \cdot 0 - 0 \cdot 2)\]\[D_y = 0 + 3(12 - 2) = 30\]3. \( D_z \) (replace the third column of \( A \) with \( B \)):\[D_z = \begin{vmatrix} 4 & -1 & -3 \3 & 1 & 0 \2 & -1 & 0\end{vmatrix}\]Calculate:\[D_z = 4(1 \cdot 0 - (-1)\cdot 0) - (-1)(3\cdot 0 - 1\cdot 2) + (-3)(3\cdot -1 -1\cdot2)\]\[D_z = 0 - 2 + 3(3 \cdot 1 + 1\cdot 2) = 0 - 2 + 3(3 - 2)= 13\]

Use Cramer's Rule to find \( x, y, \) and \( z \):\[ x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D} \]Plug in the values:\[ x = \frac{-15}{-5} = 3, \quad y = \frac{30}{-5} = -6, \quad z = \frac{13}{-5} = -2.6 \]

Substitute \( x = 3, y = -6, z = -2.6 \) back into the original equations to check correctness.1. First equation: \( 4(3) - (-6) + 3(-2.6) = -3 \). Check if holds true.2. Second equation: \( 3(3) + (-6) + (-2.6) = 0 \). Check if holds true.3. Third equation: \( 2(3) - (-6) + 4(-2.6) = 0 \). Check if holds true.If these hold true, the solution is verified.