EXERCISE $4$

Enter the matrix A and the vector $b$ in MATLAB:

$A=\left[\begin{array}{cccc}-8& 7& -2& 1\\ -5& 2& -2& 9\\ 8& 8& -1& -4\\ 5& 2& 2& 7\end{array}\right],b=\left[\begin{array}{c}48\\ 81\\ -75\\ 35\end{array}\right]$

The exact solution to the system $Ax=b$ is the vector $x=(-6,1,7,7{)}^T.$

$($a$)$ Enter $[L,U,P]=lu(A)$ to find the $LU$ factorization of the matrix $PA.$ Verify that

$PA=LU$ by computing $PA-LU.$

$($b$)$ Use the $LU$ factorization you found in part $($a$)$ to solve the system $Ax=b.$ Call the

computed solution $x_{l}u($don$\text{'}$t forget that you will need $P*b).$

$($c$)$ Enter the vector x and compare your solution $x-1u$ from part $($b$)$ with the exact solution

x by computing norm $(x_{l}u-x)$

NOTE: the norm function gives the magnitude of the vector, that is$,$ for a vector $a=$

$(a_1,a_2,$dots,$a_n{)}^T,$ the norm of $a$ is defined as: norm$(a)=\sqrt[\phantom{2}]{a_1^2+a_2^2+\text{d}\text{o}\text{t}\text{s}+a_n^2}.$ Thus

the command norm$(x_{l}u-x)$ measures how close the computed solution $x_{l}u$ is to the

exact solution x $.$ You should expect a very small number.