Past exam of the mathematical methods I course of the University of Oxford / 2013 / 9 / c / Solution

The coefficient matrix and right-hand side areFor an LU decomposition with unit diagonal in the lower factor, subtract row 1 from row 2, add twice row 1 to row 3, and then add the new row 2 to row 3. The elimination multipliers are $l_{21}=1$, $l_{31}=-2$, and $l_{32}=-1$. ThusMultiplication verifies $LU=A$. Forward substitution in $L\mathbf{y}=\mathbf{b}$ gives $y_1=4$, $y_2=-4$, and $y_3=-2$. Back substitution then gives $x_3=-1$, $x_2=5/3$, and $x_1=1/6$:Every pivot is nonzero, so no row permutation is needed and the linear system has a unique solution.