Laplaceov razvoj

Laplaceovim razvojem izračunajte determinantu matrice

$\displaystyle A=\begin{bmatrix} 1 & 5 & -1 & 1 \\ 2 & 0 & 1 & -1 \\ 0 & 1 & 2 & 3 \\ 1 & 0 & 0 & -1 \end{bmatrix}.$

Rješenje. Laplaceovim razvojem [M1, poglavlje 2.9.3] po četvrtom retku slijedi

$\displaystyle \det A=\begin{vmatrix} 1 & 5 & -1 & 1 \\ 2 & 0 & 1 & -1 \\ 0 & ... ...-1)^{4+4}\begin{vmatrix} 1 & 5 & -1 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \end{vmatrix}.$

Sada izračunajmo dobivene determinante trećeg reda. Laplaceovim razvojem po drugom retku dobivamo

$\displaystyle \begin{vmatrix}5 & -1 & 1 \\ 0 & 1 & -1 \\ 1 & 2 & 3 \end{vmatrix}$	$\displaystyle =1\cdot(-1)^{2+2}\begin{vmatrix}5 & 1\\ 1 & 3 \end{vmatrix}+(-1)\cdot(-1)^{2+3}\begin{vmatrix}5 & -1\\ 1 & 2 \end{vmatrix}$
	$\displaystyle =(5\cdot 3-1\cdot1)+\left[5\cdot2-(-1)\cdot1\right]=14+11=25.$

Razvojem po prvom stupcu slijedi

$\displaystyle \begin{vmatrix}1 & 5 & -1 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \end{vmatrix}$	$\displaystyle =1\cdot(-1)^{1+1}\begin{vmatrix}0 & 1\\ 1 & 2 \end{vmatrix}+2\cdot(-1)^{2+1}\begin{vmatrix}5 & -1\\ 1 & 2 \end{vmatrix}$
	$\displaystyle =(0\cdot 2-1\cdot1)-2\left[5\cdot2-(-1)\cdot1\right]=-1-22=-23.$

Dakle,

$\displaystyle \det A=-1\cdot25-1\cdot(-23)=-2.$