Question

Ax = b: A = ( −2 −1 2 −2 2 3 −4 1 3 ) b = ( −1 −1 4 ) x = ( x1 x2 x3 ) (a) (6 pts) Use the elimination method to find the matrices L and U such that A = LU. (b) (3 pts) Solve the intermediate system Ly = b for the intermediate variable y. (c) (3 pts) Solve the system Ux = y and confirm your solution from Problem 2 above

Answer 1

`\underbrace{\begin{bmatrix}-2&-1&2\\-2&2&3\\-4&1&3\end{bmatrix}}_A\underbrace{\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}}_x=\underbrace{\begin{bmatrix}-1\\-1\\4\end{bmatrix}}_b`

Multiply
`A` on the left side with the following elimination matrix
`E_1`:

`\underbrace{\begin{bmatrix}1&0&0\\-1&1&0\\-2&0&1\end{bmatrix}}_(E_1)A=\begin{bmatrix}-2&-1&2\\0&3&1\\0&3&-1\end{bmatrix}`

Multiply
`E_1A` on the left by another elimination matrix
`E_2`:

`\underbrace{\begin{bmatrix}1&0&0\\0&1&0\\0&-1&1\end{bmatrix}}_(E_2)(E_1A)=\begin{bmatrix}-2&-1&2\\0&3&1\\0&0&-2\end{bmatrix}`

`\implies\boxed{U=\begin{bmatrix}-2&-1&2\\0&3&1\\0&0&-2\end{bmatrix}}`

Multiply on the left by the inverse of
`E_2E_1`:

`(E_2E_1)^(-1)(E_2E_1)A=(E_2E_1)^(-1)U`

`A=\underbrace{({E_1}^(-1){E_2}^(-1))}_LU`

We have

`{E_1}^(-1)=\begin{bmatrix}1&0&0\\1&1&0\\2&0&1\end{bmatrix}`

`{E_2}^(-1)=\begin{bmatrix}1&0&0\\0&1&0\\0&1&1\end{bmatrix}`

`\implies\boxed{L=\begin{bmatrix}1&0&0\\1&1&0\\3&1&1\end{bmatrix}}`