Solve the equation Ax=b by using the LU factorization given for A. Also, solve Ax=b by ordinary row reduction.
Solution:
The matrix L, U and vector b are given.
L=\begin{pmatrix}1&0&0\\ -1&1&0\\ 2&-4&1\end{pmatrix},\:U=\begin{pmatrix}3&-5&-4\\ 0&-4&-1\\ 0&0&-1\end{pmatrix},\:b=\begin{pmatrix}4\\ -19\\ 73\end{pmatrix}\\
To find the vector y, solve the system Ly=b.
\begin{pmatrix}1&0&0\\ -1&1&0\\ 2&-4&1\end{pmatrix}\begin{pmatrix}y_1\\ y_2\\ y_3\end{pmatrix}=\begin{pmatrix}4\\ -19\\ 73\end{pmatrix}\\
From the first row of the equation, we can write,
y_1=4\\
From the second row of the equation, we can write,
-y_1+y_2=-19\\
-4+y_2=-19\\
y_2=-19+4\\
y_2=-15\\
From the third row of the equation, we can write,
2y_1-4y_2+y_3=73\\
2(4)-4(-15)+y_3=73\\
8+60+y_3=73\\
68+y_3=73\\
y_3=73-68\\
y_3=5\\
The vector y is,
y=\begin{pmatrix}y_1\\ y_2\\ y_3\end{pmatrix}=\begin{pmatrix}4\\ -15\\ 5\end{pmatrix}\\
To find the vector x, solve the system Ux=y.
\begin{pmatrix}3&-5&-4\\ 0&-4&-1\\ 0&0&-1\end{pmatrix}=\begin{pmatrix}x_1\\ x_2\\ x_3\end{pmatrix}=\begin{pmatrix}4\\ -15\\ 5\end{pmatrix}\\
From the third row of the equation, we can write,
-x_3=5\\
x_3=-5\\
From the second row of the equation, we can write,
-4x_2-x_3=-15\\
-4x_2+5=-15\\
-4x_2=-15-5\\
-4x_2=-20\\
x_2=20/4\\
x_2=5\\
From the first row of the equation, we can write,
3x_1-5x_2-4x_3=4\\
3x_1-5(5)-4(-5)=4\\
3x_1-25+20=4\\
3x_1-5=4\\
3x_1=4+5\\
3x_1=9\\
x_1=9/3\\
x_1=3\\
The vector x is,
Final answer:
{\color{Red}y=\begin{pmatrix}4\\ -15\\ 5\end{pmatrix} }\\
{\color{Red}x=\begin{pmatrix}3\\ 5\\ -5\end{pmatrix} }\\
