SubsectionThe Assignment

• Read section chapter 5 section 1 of Strang.
• Read the following and complete the exercises below.

SubsectionLearning Goals

At some point, a student should be able to compute the determinant of a square matrix using the properties outlined here.

SubsectionDiscussion: The Determinant

Generally, the most interesting matrices to look at are the square ones. For square matrices, there is an important number called the determinant which helps us determine if the matrix is invertible or not.

Strang lists 10 important properties of determinants in this section, and verifies them for $2\times 2$ matrices. The verifications for general matrices aren't any harder, but they sure are longer, so I am glad he skipped them. Anyway, these properties are enough to get by when it is time to compute. In fact, clever use of these properties can save you a lot of time.

1. The determinant of the identity matrix is always one: $\det(I)=1$.
2. If we exchange two rows of a matrix, the determinant changes sign.
3. The determinant is linear in each row separately. (The fancy new word here is that $\det$ is a multilinear function.)
4. If two rows are equal, then the determinant is zero.
5. The row operation of “add a multiple of one row to another row” does not change the determinant of a matrix.
6. If a matrix has a row of zeros, then its determinant is zero.
7. If $A$ is triangular, then $\det(A)$ is the product of the diagonal entries of $A$.
8. A matrix is singular if its deterimant is zero. A matrix is invertible exactly when its determinant is non-zero.
9. The determinant is a multiplicative function at the level of matrices: $\det(AB) = \det(A)\det(B)$.
10. The determinant of a matrix and its transpose are the same.

That last property can be helpful in a variety of ways: it allows us to translate all of those statements about rows into statements about columns!

SubsubsectionInterpretation

Suppose that an $n\times n$ matrix $A$ is represented as a collection of its column vectors: \begin{equation*} A = \begin{pmatrix} | & | & & | \\ v_1 & v_2 & \dots & v_n \\ | & | & & | \end{pmatrix} . \end{equation*} Then the geometric significance of the determinant is this: The number $\det(A)$ represents the signed $n$-dimensional volume of the $n$-dimensional box in $\mathbb{R}^n$ with sides $v_1, v_2, \ldots, v_n$.

This takes a bit of getting used to, and the hardest part is the choice of signs. We choose a positive sign if the vectors $v_1, v_2, \ldots, v_n$ have the same orientation as the standard basis.

SubsectionSage and the Determinant

Sage has a built-in command for the determinant of a square matrix. It is just what you expect: A.determinant().

To be sure that this works properly, we can do it the old way, too:

Clearly, there are no row swaps used, so it is easy to see that $\det(A) = 4*(-5/2)*(21/2) = -105$.

SubsectionExercises

(Strang Ex 5.1.1) If a $4 \times 4$ matrix $A$ has $\det(A) = 1/2$, find
1. $\det(2A)$,
2. $\det(-A)$, and
3. $\det(A^2)$.
Explain which properties of determinants you need to make your deductions.
(Strang Ex 5.1.2) If a $3 \times 3$ matrix has $\det(A) = -1$, find
1. $\det(\frac{1}{2}A)$,
2. $\det(-A)$,
3. $\det(A^2)$, and
4. $\det(A^{-1})$
(Strang Ex 5.1.10)

Suppose that $A$ is a matrix with the property that the entries in each of its rows add to zero. Solve the equation $Ax = 0$ to prove that $\det(A)=0$.

Then suppose that instead the entries in each row add to one. Show that $\det(A-I)=0$. Does this mean that $\det(A) = 1$? Give an example or an argument.

(Strang Ex 5.1.12) The inverse of a $2 \times 2$ matrix seems to have determinant $1$ all the time: \begin{equation*} \det(A^{-1}) = \det \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{ad-bc}{ad-bc} = 1. \end{equation*} What is wrong with this argument? What should the value of $\det(A^{-1})$ be?
Suppose that $A = \left( \begin{smallmatrix} 4 & 1 \\ 2 & 3 \end{smallmatrix} \right)$ and that $\lambda$ is some number. Find $A^2$, $A^{-1}$ and $A-\lambda I$ and their determinants.
Which two numbers $\lambda$ lead to $\det(A-\lambda I) = 0$?