Calculadora de decomposição LU

Encontre a decomposição LU de $\left[\begin{array}{ccc}2 & 7 & 1\\3 & -2 & 0\\1 & 5 & 3\end{array}\right]$.

Comece com a matriz de identidade $L = \left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]$.

Subtraia a linha $1$ multiplicada por $\frac{3}{2}$ da linha $2$: $R_{2} = R_{2} - \frac{3 R_{1}}{2}$.

$\left[\begin{array}{ccc}2 & 7 & 1\\0 & - \frac{25}{2} & - \frac{3}{2}\\1 & 5 & 3\end{array}\right]$

Escreva o coeficiente $\frac{3}{2}$ na matriz $L$ na linha $2$, coluna $1$:

$L = \left[\begin{array}{ccc}1 & 0 & 0\\\frac{3}{2} & 1 & 0\\0 & 0 & 1\end{array}\right]$

Subtraia a linha $1$ multiplicada por $\frac{1}{2}$ da linha $3$: $R_{3} = R_{3} - \frac{R_{1}}{2}$.

$\left[\begin{array}{ccc}2 & 7 & 1\\0 & - \frac{25}{2} & - \frac{3}{2}\\0 & \frac{3}{2} & \frac{5}{2}\end{array}\right]$

Escreva o coeficiente $\frac{1}{2}$ na matriz $L$ na linha $3$, coluna $1$:

$L = \left[\begin{array}{ccc}1 & 0 & 0\\\frac{3}{2} & 1 & 0\\\frac{1}{2} & 0 & 1\end{array}\right]$

Adicione a linha $2$ multiplicada por $\frac{3}{25}$ à linha $3$: $R_{3} = R_{3} + \frac{3 R_{2}}{25}$.

$\left[\begin{array}{ccc}2 & 7 & 1\\0 & - \frac{25}{2} & - \frac{3}{2}\\0 & 0 & \frac{58}{25}\end{array}\right]$

Escreva o coeficiente $- \frac{3}{25}$ na matriz $L$ na linha $3$, coluna $2$:

$L = \left[\begin{array}{ccc}1 & 0 & 0\\\frac{3}{2} & 1 & 0\\\frac{1}{2} & - \frac{3}{25} & 1\end{array}\right]$

A matriz obtida é a matriz $U$.

$L = \left[\begin{array}{ccc}1 & 0 & 0\\\frac{3}{2} & 1 & 0\\\frac{1}{2} & - \frac{3}{25} & 1\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0\\1.5 & 1 & 0\\0.5 & -0.12 & 1\end{array}\right]$A

$U = \left[\begin{array}{ccc}2 & 7 & 1\\0 & - \frac{25}{2} & - \frac{3}{2}\\0 & 0 & \frac{58}{25}\end{array}\right] = \left[\begin{array}{ccc}2 & 7 & 1\\0 & -12.5 & -1.5\\0 & 0 & 2.32\end{array}\right]$A