= Solution
The first-column largest <pivot element> is four, so eliminate with multipliers $1/2,0,1/4$ below it. The augmented <matrix> becomes
$$
\begin{pmatrix}4&1&2&6&|&4\\0&1/2&2&-1&|&2\\0&2&4&4&|&4\\0&1&2&7/2&|&7/2\end{pmatrix}.
$$
<Partial pivoting> in the second column swaps rows two and three because two is largest in modulus. The next elimination multipliers are $1/4$ and $1/2$. This gives
$$
\begin{pmatrix}4&1&2&6&|&4\\0&2&4&4&|&4\\0&0&1&-2&|&1\\0&0&0&3/2&|&3/2\end{pmatrix}.
$$
The third <pivot element> is already largest and the final multiplier below it is zero. <Back substitution> gives $z=1$, $y=3$, $x=-6$, $w=-1/2$. Therefore
$$
\boxed{(w,x,y,z)=(-1/2,-6,3,1).}
$$
The row swap must also reorder the previously stored first-column multipliers in $L$. The factors are
$$
\boxed{P=\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix},\quad
L=\begin{pmatrix}1&0&0&0\\0&1&0&0\\1/2&1/4&1&0\\1/4&1/2&0&1\end{pmatrix},}
$$
$$
\boxed{U=\begin{pmatrix}4&1&2&6\\0&2&4&4\\0&0&1&-2\\0&0&0&3/2\end{pmatrix}.}
$$
Multiplying these factors verifies $PA=LU$. This is the required elimination with <partial pivoting>, not an alternative solution method.
Back to article page
