Exercises 3.2 - 3.4

Chapter 3. Systems of Linear Equations

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

\[A = \begin{bmatrix} 1 & 2 & -3 & 0\\ 1 & 1 & -2 & 1\\ 2 & 1 & 3 & 4\\ -3 & 2 & 2 & -8 \end{bmatrix}.\]

1. Using the syntax A^k which produces the \(k\)-th power of a square matrix and the command eye for the identity matrix, compute the above matrix polynomial.

2. Using the command polyvalm, compute the above matrix polynomial.

3. 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

\[A = \begin{bmatrix} 3 & 3 & -4 & -3 \\ 0 & 6 & 1 & 1\\ 5 & 4 & 2 & 1 \\ 2 & 3 & 3 & 2 \end{bmatrix} \quad \mathrm{and} \quad \mathbf{b} = \begin{bmatrix} -2 \\ 3 \\ 5 \\ 1 \end{bmatrix}.\]

1. Use the MATLAB command inv or the syntax A^(-1) to find the inverse of \(A\).

2. Display the output matrix as a rational form, NOT decimally. You may use the command format.

3. 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

\[\mathbf{v}=\Sigma_{j=1}^{25} c_{j}\mathbf{v}_{j}\]

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 \(\textrm{span}\{\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{3}}\}\).