Example Question - elementary operations

Matrix Inversion Using Elementary Operations

<p>For the matrix \(A = \left[ \begin{array}{ccc} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array} \right]\), we want to find \(A^{-1}\) using elementary operations. We start by writing the augmented matrix \( [A | I] \) where \(I\) is the \(3 \times 3\) identity matrix:</p> <p>\( \left[ \begin{array}{ccc|ccc} 0 & 1 & 2 & 1 & 0 & 0 \\ 1 & 2 & 3 & 0 & 1 & 0 \\ 3 & 1 & 1 & 0 & 0 & 1 \end{array} \right] \)</p> <p>Step 1: Swap row 1 with a row having a non-zero entry in the first column. We swap row 1 and row 2:</p> <p>\( \left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 3 & 1 & 1 & 0 & 0 & 1 \end{array} \right] \)</p> <p>Step 2: Subtract 3 times row 1 from row 3:</p> <p>\( \left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & -5 & -8 & 0 & -3 & 1 \end{array} \right] \)</p> <p>Step 3: Add 5 times row 2 to row 3:</p> <p>\( \left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 2 & 5 & -3 & 1 \end{array} \right] \)</p> <p>Step 4: Divide row 3 by 2:</p> <p>\( \left[ \begin{array}{ccc|ccc} 1 & 2 & 3 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right] \)</p> <p>Step 5: Subtract 2 times row 3 from row 2, and subtract 3 times row 3 from row 1:</p> <p>\( \left[ \begin{array}{ccc|ccc} 1 & 2 & 0 & -\frac{15}{2} & \frac{11}{2} & -\frac{3}{2} \\ 0 & 1 & 0 & -4 & 3 & -1 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right] \)</p> <p>Step 6: Subtract 2 times row 2 from row 1:</p> <p>\( \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 & -4 & 3 & -1 \\ 0 & 0 & 1 & \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right] \)</p> <p>The matrix on the right side of the augmented matrix is now the inverse of \(A\):</p> <p>\( A^{-1} = \left[ \begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array} \right] \)</p>