Elementary Row Operations on Augmented Matrix [1 1 1 : 1 0 0] [2 3 -1 : 0 1 0] [-4 7 4 : 0 0 1]

Math Problem Statement

perform elementary row operations, A:In = [1 1 1 : 1 0 0 0] [2 3 -1 : 0 1 0] [-4 7 4 : 0 0 1]

We are tasked with performing elementary row operations on the augmented matrix $A : I_n$ , where:

1 & 1 & 1 & : & 1 & 0 & 0 \\ 2 & 3 & -1 & : & 0 & 1 & 0 \\ -4 & 7 & 4 & : & 0 & 0 & 1 \end{bmatrix}$$ Our goal is to reduce the left-hand side to the identity matrix using elementary row operations. ### Step 1: Make the first column's elements below the pivot (in row 1, column 1) equal to zero. - We will first use row 1 to eliminate the $$2$$ in row 2 and the $$-4$$ in row 3. **Row operation for row 2 (R2):** $$R2 \rightarrow R2 - 2 \cdot R1$$ \[ R2 = [2, 3, -1, 0, 1, 0] - 2 \times [1, 1, 1, 1, 0, 0] \] $$R2 = [0, 1, -3, -2, 1, 0]$$ **Row operation for row 3 (R3):** $$R3 \rightarrow R3 + 4 \cdot R1$$ \[ R3 = [-4, 7, 4, 0, 0, 1] + 4 \times [1, 1, 1, 1, 0, 0] \] $$R3 = [0, 11, 8, 4, 0, 1]$$ The updated matrix is: $$\begin{bmatrix} 1 & 1 & 1 & : & 1 & 0 & 0 \\ 0 & 1 & -3 & : & -2 & 1 & 0 \\ 0 & 11 & 8 & : & 4 & 0 & 1 \end{bmatrix}$$ ### Step 2: Eliminate the element in row 3, column 2. **Row operation for row 3 (R3):** $$R3 \rightarrow R3 - 11 \cdot R2$$ \[ R3 = [0, 11, 8, 4, 0, 1] - 11 \times [0, 1, -3, -2, 1, 0] \] $$R3 = [0, 0, 41, 26, -11, 1]$$ The updated matrix is: $$\begin{bmatrix} 1 & 1 & 1 & : & 1 & 0 & 0 \\ 0 & 1 & -3 & : & -2 & 1 & 0 \\ 0 & 0 & 41 & : & 26 & -11 & 1 \end{bmatrix}$$ ### Step 3: Make the third row’s pivot (row 3, column 3) equal to 1. **Row operation for row 3 (R3):** \( R3 \rightarrow \frac{R3}{41} \

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Linear Algebra
Matrix Operations
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