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\( B=\left[\begin{array}{ccc}1 & 2 & 5 \\ 3 & 4 & 1 \\ 1 & 3 & -2\end{array}\right] \) Determine the inverse of \( B \) using elementary row operations (ERO)

Ask by Thornton Bright. in Ghana
Mar 19,2025

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To find the inverse of matrix \( B \) using elementary row operations, augment \( B \) with the identity matrix and perform row operations to transform \( B \) into the identity matrix. The resulting inverse matrix is: \[ B^{-1} = \begin{bmatrix} -\frac{1}{28} & \frac{5}{28} & \frac{5}{14} \\ -\frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\ \frac{1}{28} & -\frac{1}{28} & \frac{1}{14} \end{bmatrix} \]

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To find the inverse of matrix \( B \) using elementary row operations, we augment the matrix \( B \) with the identity matrix: \[ \left[\begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 3 & 4 & 1 & 0 & 1 & 0 \\ 1 & 3 & -2 & 0 & 0 & 1 \end{array}\right] \] We will perform row operations to transform the left side into the identity matrix. 1. **Row Operation**: \( R_2 \leftarrow R_2 - 3R_1 \) \[ \left[\begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & -2 & -14 & -3 & 1 & 0 \\ 1 & 3 & -2 & 0 & 0 & 1 \end{array}\right] \] 2. **Row Operation**: \( R_3 \leftarrow R_3 - R_1 \) \[ \left[\begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & -2 & -14 & -3 & 1 & 0 \\ 0 & 1 & -7 & -1 & 0 & 1 \end{array}\right] \] 3. **Row Operation**: \( R_2 \leftarrow -\frac{1}{2}R_2 \) \[ \left[\begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 1 & 7 & \frac{3}{2} & -\frac{1}{2} & 0 \\ 0 & 1 & -7 & -1 & 0 & 1 \end{array}\right] \] 4. **Row Operation**: \( R_3 \leftarrow R_3 - R_2 \) \[ \left[\begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 1 & 7 & \frac{3}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & -14 & -\frac{5}{2} & \frac{1}{2} & 1 \end{array}\right] \] 5. **Row Operation**: \( R_3 \leftarrow -\frac{1}{14} R_3 \) \[ \left[\begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 1 & 7 & \frac{3}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & \frac{5}{28} & -\frac{1}{28} & -\frac{1}{14} \end{array}\right] \] 6. **Row Operation**: \( R_2 \leftarrow R_2 - 7R_3 \) \[ \left[\begin{array}{ccc|ccc} 1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 1 & 0 & -2 & 1 & \frac{1}{2} \\ 0 & 0 & 1 & \frac{5}{28} & -\frac{1}{28} & -\frac{1}{14} \end{array}\right] \] 7. **Row Operation**: \( R_1 \leftarrow R_1 - 5R_3 \) \[ \left[\begin{array}{ccc|ccc} 1 & 2 & 0 & -\frac{9}{28} & \frac{5}{28} & \frac{5}{14} \\ 0 & 1 & 0 & -2 & 1 & \frac{1}{2} \\ 0 & 0 & 1 & \frac{5}{28} & -\frac{1}{28} & -\frac{1}{14} \end{array}\right] \] 8. **Row Operation**: \( R_1 \leftarrow R_1 - 2R_2 \) \[ \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{15}{28

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