To solve the given system of equations using Gauss Elimination, we first express the system in augmented matrix form and then apply Gaussian elimination to transform the matrix into row-echelon form, followed by back substitution.

Given system of equations:

10x - 2y + 3z = 23 \tag{1} 2x + 10y - 5z = -33 \tag{2} 3x - 4y + 10z = 41 \tag{3}

Step 1: Write the augmented matrix

The augmented matrix corresponding to this system is:

10 & -2 & 3 & | & 23 \\ 2 & 10 & -5 & | & -33 \\ 3 & -4 & 10 & | & 41 \end{bmatrix}$$ ### Step 2: Perform Gaussian Elimination We will use row operations to reduce this matrix to upper triangular form. #### Row 1: Normalize the first pivot by dividing row 1 by 10 $$R_1 = \frac{1}{10} R_1 \quad \text{(Divide row 1 by 10)}$$ $$\begin{bmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 2 & 10 & -5 & | & -33 \\ 3 & -4 & 10 & | & 41 \end{bmatrix}$$ #### Row 2: Eliminate the first element in row 2 by subtracting 2 times row 1 from row 2 $$R_2 = R_2 - 2R_1$$ $$\begin{bmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 0 & \frac{52}{5} & -\frac{31}{5} & | & -\frac{299}{5} \\ 3 & -4 & 10 & | & 41 \end{bmatrix}$$ #### Row 3: Eliminate the first element in row 3 by subtracting 3 times row 1 from row 3 $$R_3 = R_3 - 3R_1$$ $$\begin{bmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 0 & \frac{52}{5} & -\frac{31}{5} & | & -\frac{299}{5} \\ 0 & -\frac{11}{5} & \frac{41}{10} & | & \frac{87}{10} \end{bmatrix}$$ ### Step 3: Normalize the second pivot by multiplying row 2 by $$\frac{5}{52}$$ $$R_2 = \frac{5}{52}R_2$$ $$\begin{bmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 0 & 1 & -\frac{31}{52} & | & -\frac{299}{52} \\ 0 & -\frac{11}{5} & \frac{41}{10} & | & \frac{87}{10} \end{bmatrix}$$ #### Row 3: Eliminate the second element in row 3 by adding $$\frac{11}{5}$$ times row 2 to row 3 $$R_3 = R_3 + \frac{11}{5}R_2$$ $$\begin{bmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 0 & 1 & -\frac{31}{52} & | & -\frac{299}{52} \\ 0 & 0 & \frac{579}{104} & | & \frac{2997}{104} \end{bmatrix}$$ ### Step 4: Normalize the third pivot by multiplying row 3 by $$\frac{104}{579}$$ $$R_3 = \frac{104}{579} R_3$$ $$\begin{bmatrix} 1 & -\frac{1}{5} & \frac{3}{10} & | & \frac{23}{10} \\ 0 & 1 & -\frac{31}{52} & | & -\frac{299}{52} \\ 0 & 0 & 1 & | & \frac{2997}{579} \end{bmatrix}$$ ### Step 5: Perform back substitution - From row 3: $$z = \frac{2997}{579} = 5.175$$ - From row 2: $$y = -\frac{299}{52} + \frac{31}{52} \times 5.175$$ - From row 1: $$x = \dots$$ I will now finalize these