Question - Solving a System of Linear Equations Using Augmented Matrix Method
Solution:
The given image shows a system of linear equations represented in an augmented matrix form, which we aim to solve. To find the solution, we must reduce this matrix to its reduced row echelon form (RREF).
Here's the step-by-step solution:
Starting matrix:
\[ \begin{bmatrix} 3 & -4 & | & 1 & 0 & 0 \\ 2 & 4 & | & 0 & 1 & 0 \\ -4 & 2 & | & 0 & 0 & 1 \\ \end{bmatrix} \]Step 1: Make the first element of the first row a 1 (if it's not already) by dividing the entire first row by 3.
\[ \begin{bmatrix} 1 & -\frac{4}{3} & | & \frac{1}{3} & 0 & 0 \\ 2 & 4 & | & 0 & 1 & 0 \\ -4 & 2 & | & 0 & 0 & 1 \\ \end{bmatrix} \]Step 2: Eliminate the first element of the second and third row to make them 0 by adding (-2) times the first row to the second row and adding 4 times the first row to the third row.
\[ \begin{bmatrix} 1 & -\frac{4}{3} & | & \frac{1}{3} & 0 & 0 \\ 0 & \frac{14}{3} & | & -\frac{2}{3} & 1 & 0 \\ 0 & \frac{2}{3} & | & \frac{4}{3} & 0 & 1 \\ \end{bmatrix} \]Step 3: Make the second element of the second row a 1 by dividing the entire second row by $\frac{14}{3}$.
\[ \begin{bmatrix} 1 & -\frac{4}{3} & | & \frac{1}{3} & 0 & 0 \\ 0 & 1 & | & -\frac{1}{7} & \frac{3}{14} & 0 \\ 0 & \frac{2}{3} & | & \frac{4}{3} & 0 & 1 \\ \end{bmatrix} \]Step 4: Eliminate the second element of the first and third row to make them 0 by adding $\frac{4}{3}$ times the second row to the first row and adding $-\frac{2}{3}$ times the second row to the third row.
\[ \begin{bmatrix} 1 & 0 & | & \frac{5}{21} & \frac{4}{21} & 0 \\ 0 & 1 & | & -\frac{1}{7} & \frac{3}{14} & 0 \\ 0 & 0 & | & \frac{8}{7} & -\frac{1}{7} & 1 \\ \end{bmatrix} \]This matrix is in RREF. The solutions to the system of equations can be read directly from the matrix:
\[ x = \frac{5}{21}, y = -\frac{1}{7}, z = \frac{8}{7} \]CamTutor
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