Let's solve the system step by step using **Gaussian elimination**. ### **Given System** \[ \begin{cases} x_1 + 4x_2 + 4x_3 = 28 \\ -x_1 - 5x_2 + 5x_3 = -4 \\ x_1 - 6x_2 + 6x_3 = 4 \\ \end{cases} \] --- ## **Step 1: Write the Augmented Matrix** \[ \begin{bmatrix} 1 & 4 & 4 & | & 28 \\ -1 & -5 & 5 & | & -4 \\ 1 & -6 & 6 & | & 4 \\ \end{bmatrix} \] --- ## **Step 2: Eliminate \(x_1\) from Row 2 and Row 3** ### Row 2: \( R_2 + R_1 \) \[ (-1 + 1)x_1 + (-5 + 4)x_2 + (5 + 4)x_3 = -4 + 28 \\ 0x_1 - 1x_2 + 9x_3 = 24 \\ \] So, Row 2 becomes: \( [0, -1, 9 \ | \ 24] \) ### Row 3: \( R_3 - R_1 \) \[ (1 - 1)x_1 + (-6 - 4)x_2 + (6 - 4)x_3 = 4 - 28 \\ 0x_1 - 10x_2 + 2x_3 = -24 \\ \] So, Row 3 becomes: \( [0, -10, 2 \ | \ -24] \) **New Matrix:** \[ \begin{bmatrix} 1 & 4 & 4 & | & 28 \\ 0 & -1 & 9 & | & 24 \\ 0 & -10 & 2 & | & -24 \\ \end{bmatrix} \] --- ## **Step 3: Eliminate \(x_2\) from Row 3** Row 3: \( R_3 - 10(R_2) \) \[ 0x_1 + (-10 - (-10))x_2 + (2 - 90)x_3 = -24 - 240 \\ 0x_1 + 0x_2 - 88x_3 = -264 \\ \] So, Row 3 becomes: \( [0, 0, -88 \ | \ -264] \) --- ## **Step 4: Solve for \(x_3\)** From Row 3: \[ -88x_3 = -264 \implies x_3 = \frac{-264}{-88} = 3 \] --- ## **Step 5: Back Substitute for \(x_2\)** From Row 2: \[ -1x_2 + 9x_3 = 24 \\ -1x_2 + 9(3) = 24 \\ -1x_2 + 27 = 24 \\ -1x_2 = 24 - 27 = -3 \\ x_2 = 3 \] --- ## **Step 6: Back Substitute for \(x_1\)** From Row 1: \[ x_1 + 4x_2 + 4x_3 = 28 \\ x_1 + 4(3) + 4(3) = 28 \\ x_1 + 12 + 12 = 28 \\ x_1 + 24 = 28 \\ x_1 = 28 - 24 = 4 \\ \] --- ## **Final Answer** \[ \boxed{ x_1 = 4,\quad x_2 = 3, \quad x_3 = 3 } \]