Exercises 2.1 - 3.4
Chapter 2. Systems of Linear Equations
2.1 Introduction to Systems of Linear Equatios
No MATLAB problems in this section.
2.2 Solving Linear Systems by Row Reduction
Exercise 2.1. (Reduced Row Echelon Form with Pivot Columns and Ranks)
In MATLAB, there are several useful commands for matrices such as rref
command which produces the reduced row echelon form together with the
pivot columns, and rank command which gives the number of the leading
\(1\)’s without finding its row echelon form. Find the reduced row echelon form, the pivot columns, and the rank of the matrix \(A\), where
% Construct the matrix A.
A=[2 -3 1 0 4; 1 1 2 2 0; 3 0 -1 4 5; 1 6 5 6 -4];
% Display the format of each entry as a rational form
format rat;
% Find the reduced row echelon form
% and the pivot columns of the matrix A.
[rref_A, pivotcols] = rref(A);
% Find the rank of the matrix A.
rank_A = rank(A);
disp('The reduced row echelon form is'); disp(rref_A);
disp('The pivot columns are'); disp(pivotcols);
disp('The number of the leading 1 is'); disp(rank_A);
MATLAB results.
The reduced row echelon form is
1 0 0 17/13 3/2
0 1 0 11/13 -1/2
0 0 1 -1/13 -1/2
0 0 0 0 0
The pivot columns are
1 2 3
The number of the leading 1 is
3
Exercise 2.2. (Linear Combinations)
Use the MATLAB command rref to express the
vector \(\mathbf{b}=(-21, \hspace{1mm}-60, \hspace{1mm}-3, \hspace{1mm}108, \hspace{1mm}84)\) as a linear combination of \(\mathbf{v_{1}}\), \(\mathbf{v_{2}}\), and \(\mathbf{v_{3}}\) where \(\mathbf{v_{1}}=(1, \hspace{1mm} -1, \hspace{1mm}3, \hspace{1mm}11, \hspace{1mm}20)\), \(\mathbf{v_{2}}=(10, \hspace{1mm}5, \hspace{1mm}15, \hspace{1mm}20, \hspace{1mm}11)\), and \(\mathbf{v_{3}}=(3, \hspace{1mm}3, \hspace{1mm}4, \hspace{1mm}4, \hspace{1mm}9)\).
% Construct b as a column vector.
b = [-21 -60 -3 108 84]';
% Set v1, v2, v3 as column vectors.
v1 = [1 -1 3 11 20]';
v2 = [10 5 15 20 11]';
v3 = [3 3 4 4 9]';
% Set a matrix A with column vectors v1, v2 and v3.
A = [v1 v2 v3];
% Augmented matrix [A | b].
augA = [A b];
% Reduced row echelon form of augA.
rref_augA = rref(augA);
% Solution vector from rref_augA.
x = rref_augA(1:3, 4);
% Display the result as an integer form.
format rat;
disp('b is a linear combination of x(1)*v1+x(2)*v2+x(3)*v3, where');
disp('x(1) ='); disp(x(1)); disp('x(2) ='); disp(x(2));
disp('x(3) ='); disp(x(3));
MATLAB results.
b is a linear combination of x(1)*v1+x(2)*v2+x(3)*v3, where
x(1) =
12
x(2) =
3
x(3) =
-21
Chapter 3. Matrices and Matrix Algebra
3.1 Operations on Matrices
No MATLAB problems in this section.
3.2 Inverses; Algebraic Properties of Matrices
Exercise 3.1.
In this problem, we compute \(A^{5} - 3A^{3} + 7A - 4I\) for the matrix \(A\), where
\[A = \begin{bmatrix} 1 & 2 & -3 & 0\\ 1 & 1 & -2 & 1\\ 2 & 1 & 3 & 4\\ -3 & 2 & 2 & -8 \end{bmatrix}.\]-
Using the syntax
A^kwhich produces the \(k\)-th power of a square matrix and the commandeyefor the identity matrix, compute the above matrix polynomial. -
Using the command
polyvalm, compute the above matrix polynomial. -
Tell what happens if you type the syntax
A.^k.
Solution.
% Construct the matrix A.
A = [1 2 -3 0; 1 1 -2 1; 2 1 3 4; -3 2 2 -8];
% (a)
result_a = A^5 + (-3)*A^3 + 7*A + (-4)*eye(4);
% Display the matrix polynomial.
disp('The result of the matrix polynomial is');
disp(result_a)
% (b)
% Coefficient of the matrix polynomial.
coeff_poly = [1 0 -3 0 7 -4];
% Evaluate the matrix polynomial of coefficient
% with coeff_poly vector with the input matrix A.
result_b = polyvalm(coeff_poly, A);
% Display the matrix polynomial.
disp('The result of the matrix polynomial is');
disp(result_b);
% (c)
disp('The result of A.^2 is'); disp(A.^2);
disp('The result of A.^3 is'); disp(A.^3);
disp('The result of A.^4 is'); disp(A.^4);
MATLAB results.
The result of the matrix polynomial is
874 -1272 -39 3021
2580 -2306 -723 7536
5191 -4121 -2444 14563
-16852 12539 5649 -46917
The result of the matrix polynomial is
874 -1272 -39 3021
2580 -2306 -723 7536
5191 -4121 -2444 14563
-16852 12539 5649 -46917
The result of A.^2 is
1 4 9 0
1 1 4 1
4 1 9 16
9 4 4 64
The result of A.^3 is
1 8 -27 0
1 1 -8 1
8 1 27 64
-27 8 8 -512
The result of A.^4 is
1 16 81 0
1 1 16 1
16 1 81 256
81 16 16 4096
From the results, we can see that the syntax A.^k produces
the entrywise \(k\)-th powers of the matrix \(A\).
3.3 Elementary Matrices; A Method for Finding \(A^{-1}\)
Exercise 3.2.
In this problem, we solve the linear system \(A \mathbf{x} = \mathbf{b}\) by using matrix inversion, where
\[A = \begin{bmatrix} 3 & 3 & -4 & -3 \\ 0 & 6 & 1 & 1\\ 5 & 4 & 2 & 1 \\ 2 & 3 & 3 & 2 \end{bmatrix} \quad \mathrm{and} \quad \mathbf{b} = \begin{bmatrix} -2 \\ 3 \\ 5 \\ 1 \end{bmatrix}.\]-
Use the MATLAB command
invor the syntaxA^(-1)to find the inverse of \(A\). -
Display the output matrix as a rational form, NOT decimally. You may use the command
format. -
Using the result of (a), compute the solution of the linear system \(A \mathbf{x} = \mathbf{b}\) by taking \(\mathbf{x} = A^{-1} \mathbf{b}\).
Solution.
% Construct the matrix A and the right-hand-side vector b.
A = [3 3 -4 -3; 0 6 1 1; 5 4 2 1; 2 3 3 2];
b = [-2 3 5 1]';
% (a)
% Use the command inv.
Inv_A1 = inv(A);
% Use the syntax A^(-1).
Inv_A2 = A^(-1);
% (b)
format rat;
disp('The result of the command inv is'); disp(Inv_A1);
disp('The result of the syntax A^(-1) is'); disp(Inv_A2);
% (c)
% Since A is invertible, the solution to Ax=b is x=A^(-1)*b.
x = Inv_A1 * b;
disp('The solution to Ax=b is x = A^(-1)*b'); disp(x');
MATLAB results.
The result of the command inv is
-7 5 12 -19
3 -2 -5 8
41 -30 -69 111
-59 43 99 -159
The result of the syntax A^(-1) is
-7 5 12 -19
3 -2 -5 8
41 -30 -69 111
-59 43 99 -159
The solution to Ax=b is x = A^(-1)*b
70 -29 -406 583
3.4 Subspaces and Linear Independence
Exercise 3.3. (Sigma notation)
Compute the linear combination
\[\mathbf{v}=\sum_{j=1}^{25} c_{j}\mathbf{v}_{j}\]for \(c_{j}=1/j\) and \(\mathbf{v}_{j}=(\sin j, \cos j).\)
Solution.
v=zeros(1,2);
for i=1:25
v=v+(1/i)*[sin(i), cos(i)];
end
disp(v);
MATLAB results.
1.0322 0.0553
Exercise 3.4.
Let \(\mathbf{v_{1}}=(4, 3, 2, 1)\), \(\mathbf{v_{2}}=(5, 1, 2, 4)\), \(\mathbf{v_{3}}=(7, 1, 5, 3)\), \(\mathbf{x}=(16, 5, 9, 8)\), and \(\mathbf{y}=(3, 1, 2, 7)\). Determine whether \(\mathbf{x}\) and \(\mathbf{y}\) lie in \(\textrm{span}\{\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{3}}\}\).
Solution.
% Construct v1, v2, v3, x, y
v1=[4 3 2 1]'; v2=[5 1 2 4]'; v3=[7 1 5 3]';
x=[16 5 9 8]'; y=[3 1 2 7]';
% Augmented matrices [v1|v2|v3|x] and [v1|v2|v3|y]
X=[v1 v2 v3 x];
Y=[v1 v2 v3 y];
disp('Reduced row echelon form of [v1 v2 v3 x] is');
disp(rref(X));
disp('Reduced row echelon form of [v1 v2 v3 y] is');
disp(rref(Y));
MATLAB results.
Reduced row echelon form of [v1 v2 v3 x] is
1 0 0 1
0 1 0 1
0 0 1 1
0 0 0 0
Reduced row echelon form of [v1 v2 v3 y] is
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
Therefore, \(\mathbf{x}\) lies in \(\textrm{span}\{\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{3}}\}\) and \(\mathbf{y}\) does not lie in \(\textrm{span}\{\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{3}}\}\).
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