Matrices

First meeting

After this meeting students should:

• Understand how to use the Sympy library to carry out basic Matrix related tasks
• Know what they need to do to prepare for their fourth tutorial.

Explain to students that we will be solving the following problem:

The matrix $A$ is given by $A=\begin{pmatrix}1 & 1 & a\\ 2 & a & 1\\ a & 1 & 2 a\end{pmatrix}$

1. Find the determinant of $A$
2. Hence find the values of $a$ for which $A$ is singular.
3. For the following values of $a$, when possible obtain $A ^ {- 1}$
   and confirm the result by computing $AA^{-1}$:

   1. $a = 0$;
   2. $a = 1$;
   3. $a = 2$;
   4. $a = 3$.

Solution

>>> import sympy as sym
>>> a = sym.Symbol("a")
>>> A = sym.Matrix(((1, 1, a), (2, a, 1), (a, 1, 2 * a)))
>>> A
Matrix([
[1, 1,   a],
[2, a,   1],
[a, 1, 2*a]])

Now to calculate the determinant::

>>> determinant = A.det()
>>> determinant
-a**3 + 2*a**2 - a - 1

This can now be used to give us our first equation::

>>> sym.solveset(determinant, a)
{-(3*sqrt(93)/2 + 29/2)**(1/3)/3 - 1/(3*(3*sqrt(93)/2 + 29/2)**(1/3)) + 2/3, 1/(6*(3*sqrt(93)/2 + 29/2)**(1/3)) + (3*sqrt(93)/2 + 29/2)**(1/3)/6 + 2/3 + I*(-sqrt(3)/(6*(3*sqrt(93)/2 + 29/2)**(1/3)) + sqrt(3)*(3*sqrt(93)/2 + 29/2)**(1/3)/6), 1/(6*(3*sqrt(93)/2 + 29/2)**(1/3)) + (3*sqrt(93)/2 + 29/2)**(1/3)/6 + 2/3 + I*(-sqrt(3)*(3*sqrt(93)/2 + 29/2)**(1/3)/6 + sqrt(3)/(6*(3*sqrt(93)/2 + 29/2)**(1/3)))}

Now for each value of :math:a we can compute the inverse and confirm that inverse acts as required::

>>> A.subs({a: 0}).inv()
Matrix([
[ 1, 0, -1],
[ 0, 0,  1],
[-2, 1,  2]])
>>> A.subs({a: 0}).inv() @ A.subs({a: 0})
Matrix([
[1, 0, 0],
[0, 1, 0],
[0, 0, 1]])
>>> A.subs({a: 1}).inv()
Matrix([
[-1,  1,  0],
[ 3, -1, -1],
[-1,  0,  1]])
>>> A.subs({a: 1}).inv() @ A.subs({a: 1})
Matrix([
[1, 0, 0],
[0, 1, 0],
[0, 0, 1]])
>>> A.subs({a: 2}).inv()
Matrix([
[-7/3,  2/3,  1],
[   2,    0, -1],
[ 2/3, -1/3,  0]])
>>> A.subs({a: 2}).inv() @ A.subs({a: 2})
Matrix([
[1, 0, 0],
[0, 1, 0],
[0, 0, 1]])
>>> A.subs({a: 3}).inv()
Matrix([
[-17/13,  3/13,  8/13],
[  9/13,  3/13, -5/13],
[  7/13, -2/13, -1/13]])
>>> A.subs({a: 3}).inv() @ A.subs({a: 3})
Matrix([
[1, 0, 0],
[0, 1, 0],
[0, 0, 1]])

Come back: with time take any questions.

Point at resources.

Here is a video recording of a short review given in the 2020/2021 academic year: https://www.youtube.com/watch?v=rq_2ZYKq904