Lab7Solutions

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% MATH1019 Linear Algebra and Statistics for Engineers

% Laboratory Session 7

% Solutions

% From row operations to rank and solving linear systems

% Exercises

AI =eye(3)

AI =

1 0 0

0 1 0

0 0 1

AI([1 3],:) = AI([3 1],:) % swaps row1 and row3

AI =

0 0 1

0 1 0

1 0 0

P=AI

% The matrix P is called a permutation matrix.

P =

0 0 1

0 1 0

1 0 0

% Show the P^2 is the identity matrix.

PSquared = P*P

PSquared =

1 0 0

0 1 0

0 0 1

M = [1,2,3,4;5,6,7,8;9,10,11,12]

M =

1 2 3 4

5 6 7 8

9 10 11 12

% Let's now do some other row operations, performing them on the matrix M.

M0=M; % save a copy of M as the next commands modify M, and will use M0 in a later exercise

M(2,:) = M(2,:)-5*M(1,:) % row2 <- row2-5*row1

M =

1 2 3 4

0 -4 -8 -12

9 10 11 12

M(3,:) = M(3,:)-9*M(1,:) % row3 <- row3 -9*row1

M =

1 2 3 4

0 -4 -8 -12

0 -8 -16 -24

M(3,:) = M(3,:)-2*M(2,:) % row3 <- row3-2row2

M =

1 2 3 4

0 -4 -8 -12

0 0 0 0

% The rank of a matrix is the number of nonzero rows in the matrix when it’s in row echelon form.

rank(M)

ans = 2

% 2 This agrees with what you have found with your row operations.

% 2.

% Define the matrices

M1= [1,2,3,4;5,6,7,8;9,10,11,11]

M2= [1,2,3,4;5,6,7,8;9,10,12,12]

M1 =

1 2 3 4

5 6 7 8

9 10 11 11

M2 =

1 2 3 4

5 6 7 8

9 10 12 12

% What are their ranks?

rank(M1)

ans =3

rank(M2)

ans = 3

% 3.

% The system of linear equations Ax=b can be solved by reducing the augmented matrix A|b to row echelon form.

% Consider next systems defined as follows

% 3.

% The system of linear equations Ax=b can be solved by reducing the augmented matrix A|b to row echelon form.

% Consider next systems defined as follows

A0=M0(:,[1:3]); b0=M0(:,4)

b0 =

4

8

12

A1=M1(:,[1:3]); b1=M1(:,4)

b1 =

4

8

11

A2=M2(:,[1:3]); b2=M2(:,4)

b2 =

4

8

12

% Use the rank command to decide which of the systems have solutions.

rank(M0)

rank(A1)

% ans = 2

rank(A2)

% ans = 3

% A2 is full rank

% Solving using Gaussian Elimination and row operations

M2(2,:) = M2(2,:)-5*M2(1,:)

M2 =

1 2 3 4

0 -4 -8 -12

9 10 12 12

M2(3,:) = M2(3,:) - 9*M2(1,:)

M2 =

1 2 3 4

0 -4 -8 -12

0 -8 -15 -24

M2(3,:) = M2(3,:) - 2*M2(2,:)

M2 =

1 2 3 4

0 -4 -8 -12

0 0 1 0

% By hand backward substitution could then be used to solve for x

% Solving using the command x=A\b

x2= A2\b2

x2 =

-2.0000

3.0000

-0.0000

% 4.

% Start with Gaussian elimination

A=[8,3;4,6]; A0=A; b=[-2;8]; A(2,:)= A(2,:)-(1/2)*A(1,:)

A =

8.0000 3.0000

0 4.5000

% This gives a row echelon form (an upper triangular matrix).

% Gaussian elimination in matlab is lu factorization:

[L,U,P] = lu(A0)

L =

1.0000 0

0.5000 1.0000

U =

8.0000 3.0000

0 4.5000

P =

1 0

0 1

% Look at the upper triangular factor, U it’s the same as you found with the row operation above.

% If one solves Ly=b and then Ux=y the x so found solves Ax=b

y=L\b

y =

-2

9

x=U\y

x =
-1

2

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