Question

Use the following system of equations to solve problems x1 3x2 2x3 4 6x1 4x2 7x3 10 5x1 8x2 6x3 14 6) (4 points) Use Doolittle's Decomposition without any pivoting to solve the system above, what would the "d" vector be? a. [7:-40; -24] b. [8;-43; -22] c. [9-45; -20] d. [10:-48; -18] None of the above e.

NOTE: Plz solve step by step method so i can learn the process. Thanks

Answer 1

Given System of equations

$x_{1}+3x_{2}+2x_{3}=4$

$6x_{1}+4x_{2}+7x_{3}=10$

$5x_{1}+8x_{2}+6x_{3}=14$

Writing the equations in the matrix form AX=B

$\begin{bmatrix} 1 &3 &2 \\ 6&4 &7 \\ 5& 8 &6 \end{bmatrix}.\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}=\begin{bmatrix} 4\\ 10\\ 14 \end{bmatrix}$

So in Doolittle's Decomposition we have to decompose the matrix A into two matrices L and U such that

A=LU where

1 L=1L21 10 し31 し32 1J
and $U=\begin{bmatrix} U_{11} &U_{12} &U_{13} \\ 0&U_{22} &U_{23} \\ 0& 0 &U_{33} \end{bmatrix}$

i.e, $\begin{bmatrix} 1 &3 &2 \\ 6&4 &7 \\ 5& 8 &6 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ L_{21}&1 &0 \\ L_{31}& L_{32} &1 \end{bmatrix}.\begin{bmatrix} U_{11} &U_{12} &U_{13} \\ 0&U_{22} &U_{23} \\ 0& 0 &U_{33} \end{bmatrix}$

equating corresponding elements on both the sides

$U_{11}=1$

$U_{12}+0*U_{22}=3\Rightarrow U_{12}=3$

$U_{13}+0*U_{23}+0*U_{33}=1\Rightarrow U_{13}=2$

$L_{21}U_{11}=6\Rightarrow L_{21}=6$

$L_{21}U_{12}+1*U_{22}=4\Rightarrow U_{22}=-14$

$L_{21}U_{13}+1*U_{23}=7\Rightarrow U_{23}=-5$

$L_{31}U_{11}=5\Rightarrow L_{31}=5$

32022 8

$L_{31}U_{13}+L_{32}U_{23}+1*U_{33}=6\Rightarrow U_{33}=\frac{-3}{2}$

putting the values into the matrices

$\begin{bmatrix} 1 &3 &2 \\ 6&4 &7 \\ 5& 8 &6 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 6&1 &0 \\ 5& \frac{1}{2} &1 \end{bmatrix}.\begin{bmatrix} 1 &3 &2 \\ 0&-14 &-5 \\ 0& 0 &\frac{-3}{2} \end{bmatrix}$

Solving the System of Linear equations

$Ax=B$

But A =L U

$LUx=B$

Let $Ux=d$

$\Rightarrow Ld=b$

$\begin{bmatrix} 1 &0 &0 \\ 6& 1 &0 \\ 5 &\frac{1}{2} &1 \end{bmatrix}\begin{bmatrix} d_{1}\\ d_{2}\\ d_{3} \end{bmatrix}=\begin{bmatrix} 4\\ 10\\ 14 \end{bmatrix}$

$\Rightarrow$

$d_{1}=4$

$6d_{1}+d_{2}=10\Rightarrow d_{2}=-14$

$5d_{1}+\frac{1}{2}d_{2}+d_{3}=14\Rightarrow d_{3}=1$

so d vector is $d=\begin{bmatrix} 4\\ -14\\ 1 \end{bmatrix}$

Now Ux=d

$\begin{bmatrix} 1 &3 &2 \\ 0&-14 &-5 \\ 0& 0 &\frac{-3}{2} \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}=\begin{bmatrix} 4\\ -14\\ 1 \end{bmatrix}$

$\Rightarrow$

$x_{3}=\frac{-2}{3}$

$-14x_{2}-5x_{3}=-14\Rightarrow x_{2}=\frac{26}{21}$

$x_{1}+3x_{2}+2x_{3}=4\Rightarrow x_{1}=\frac{34}{21}$

Solution for the given set of equations is

$x=\begin{bmatrix} 34/21\\ 26/21\\ -2/3 \end{bmatrix}$

for the given question

option E (None of the Above ) is the right answer.