Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 5B / Solution

Eliminate the first entries of rows two, three and four using multipliers $a^3,a^2,a$. All remaining subdiagonal entries are then already zero. The unit-lower-triangular LU decomposition isMultiplication verifies $LU=A$. Since $\gamma \ne 0$, all pivots are nonzero. Forward substitution in $Ly=b$ givesBackward substitution in $Ux=y$ gives $x_4=0$, $x_3=-a^2$, $x_2=0$, and $x_1=\gamma +a^4=1$. ThusDirect substitution yields $Ax=(\gamma ,0,0,a\gamma {)}^T$ as a check.