Answer to Question #249535 in Linear Algebra for kofi

Question

Consider the following system of linear algebraic equations 𝐴𝑥=𝑏:

[1 1 3 [ X = [2

5 3 1 Y 3

2 3 1] Z] -1]

Solve the system using (you can compare your solution with MATLAB)

1. Cramer’s Rule

2. Gauss elimination without pivoting

3. Gauss-Jordan elimination

4. LU factorization

Answer 1

$A=\begin{pmatrix} 1 & 1&3 \\ 5&3&1\\ 2&3&1 \end{pmatrix}, x=\begin{pmatrix} x \\ y\\z \end{pmatrix},b=\begin{pmatrix} 2 \\ 3\\-1 \end{pmatrix}$

1. Cramer’s Rule

$\Delta=\begin{vmatrix} 1 & 1&3 \\ 5&3&1\\ 2&3&1 \end{vmatrix}=3+2+45-18-5-3=24\\ \Delta_x=\begin{vmatrix} 2 & 1&3 \\ 3&3&1\\ -1&3&1 \end{vmatrix}=6-1+27+9-3-6=32\\ \Delta_y=\begin{vmatrix} 1 & 2&3 \\ 5&3&1\\ 2&-1&1 \end{vmatrix}=3+4-15-18-10+1=-35$

$\Delta_z=\begin{vmatrix} 1 & 1&2 \\ 5&3&3\\ 2&3&-1 \end{vmatrix}=-3+6+30-12+5-9=17$

$x=\frac{\Delta _1}{\Delta}=\frac{32}{24}=\frac{4}{3}\\ y=\frac{\Delta _2}{\Delta}=\frac{-35}{24}=-\frac{35}{24}\\ z=\frac{\Delta _3}{\Delta}=\frac{17}{24}$

2. Gauss elimination without pivoting

$\begin{pmatrix} 1 & 1&3 &|&2\\ 5&3&1&|&3\\ 2&3&1&|&-1 \end{pmatrix}\to$

2r+1r(-5)

3r+1r(-2)

$\begin{pmatrix} 1 & 1&3 &|&2\\ 0&-2&-14&|&-7\\ 0&1&-5&|&-5 \end{pmatrix}\to$

2r$\leftrightarrow$ 3r

$\begin{pmatrix} 1 & 1&3 &|&2\\ 0&1&-5&|&-5\\ 0&0&-24&|&-17\\ \end{pmatrix}$

$x+y+3z=2\\ -2y-14z=-7\\ -24z=-17\mapsto z=\frac{17}{24}$

$-2y-14\cdot\frac{17}{24}=-7\mapsto x_2=-\frac{35}{24}\\ x-\frac{35}{24}+3\cdot\frac{17}{24}=2\mapsto x=\frac{3}{4}$

3. Gauss-Jordan elimination

$\begin{pmatrix} 1 & 1&3 &|&2\\ 5&3&1&|&3\\ 2&3&1&|&-1 \end{pmatrix}\to$

2r+1r(-5)

3r+1r(-2)

$\begin{pmatrix} 1 & 1&3 &|&2\\ 0&-2&-14&|&-7\\ 0&1&-5&|&-5 \end{pmatrix}\to$

2r$\leftrightarrow$ 3r

$\begin{pmatrix} 1 & 1&3 &|&2\\ 0&1&-5&|&-5\\ 0&-2&-14&|&-7 \end{pmatrix}\to$

3r+2r(2)

$\begin{pmatrix} 1 & 1&3 &|&2\\ 0&1&-5&|&-5\\ 0&0&-24&|&-17\\ \end{pmatrix}\to$

3r($\frac{-1}{24}$ )

$\begin{pmatrix} 1 & 1&3 &|&2\\ 0&1&-5&|&-5\\ 0&0&1&|&\frac{17}{24} \end{pmatrix}\to$

2r+3r(5)

1r+3r(-3)

$\begin{pmatrix} 1 & 1&0 &|&-\frac{3}{24}\\ 0&1&0&|&-\frac{35}{24}\\ 0&0&1&|&\frac{17}{24} \end{pmatrix}\to$

1r+2r(-1)

$\begin{pmatrix} 1 & 0&0 &|&\frac{32}{24}\\ 0&1&0&|&-\frac{35}{24}\\ 0&0&1&|&\frac{17}{24} \end{pmatrix}$

$x=\frac{32}{24}=\frac{4}{3}, y=-\frac{35}{24}, z=\frac{17}{24}$

4. LU factorization

$L=\begin{pmatrix} 1 & 0&0 \\ 5 & 1&0\\ 2&-\frac{1}{2}&1 \end{pmatrix}$

5: 2:

$\begin{pmatrix} 1 & 1&3 \\ 0&-2&-14\\ 0&1&-5 \end{pmatrix}$

$-\frac{1}{2}$ :

$\begin{pmatrix} 1 & 1&3 \\ 0&-2&-14\\ 0&0&-12 \end{pmatrix}=U$

$LY=B\\$

$\begin{pmatrix} 1 & 0&0 \\ 5 & 1&0\\ 2&-\frac{1}{2}&1 \end{pmatrix}\begin{pmatrix} y_1\\ y_2\\ y_3 \end{pmatrix}=\begin{pmatrix} 2\\ 3\\ -1 \end{pmatrix}$

$y_1=2\\ 5y_1+y_2=3\mapsto y_2=-7\\ 2y_1-\frac{1}{2}y_2+y_3=-1\mapsto y_3=-\frac{17}{2}$

$UX=Y\\$

$\begin{pmatrix} 1 & 1&3 \\ 0&-2&-14\\ 0&0&-12 \end{pmatrix}\begin{pmatrix} x \\ y\\z \end{pmatrix}=\begin{pmatrix} 2 \\ -7\\-\frac{17}{2} \end{pmatrix}$

$x+y+3z=2\mapsto x=-\frac{32}{24} \\ -2y-14z=-7\mapsto y=-\frac{35}{24}\\ -12z=-\frac{17}{2}\mapsto z=\frac{17}{24}$