Differential Equations - Systems: Matrices & Vectors - i

This section is intended to be a catch all for many of the basic concepts that are used occasionally in working with systems of differential equations. There will not be a lot of details in this section, nor will we be working large numbers of examples. Also, in many cases we will not be looking at the general case since we won’t need the general cases in our differential equations work.

Let’s start with some of the basic notation for matrices. An $n\times m$ (this is often called the size or dimension of the matrix) matrix is a matrix with $n$ rows and $m$ columns and the entry in the $i^{\text{th}}$ row and $j^{\text{th}}$ column is denoted by $a_{ij}$. A short hand method of writing a general $n\times m$matrix is the following.

$$A={\left(\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1m}\\ a_{21}& a_{22}& \cdots & a_{2m}\\ ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nm}\end{array}\right)}_{n\times m}={\left(a_{ij}\right)}_{n\times m}$$

The size or dimension of a matrix is subscripted as shown if required. If it’s not required or clear from the problem the subscripted size is often dropped from the matrix.

Special Matrices

There are a few “special” matrices out there that we may use on occasion. The first special matrix is the square matrix. A square matrix is any matrix whose size (or dimension) is $n\times n$. In other words, it has the same number of rows as columns. In a square matrix the diagonal that starts in the upper left and ends in the lower right is often called the main diagonal.

The next two special matrices that we want to look at are the zero matrix and the identity matrix. The zero matrix, denoted $0_{n\times m}$, is a matrix all of whose entries are zeroes. The identity matrix is a square $n\times n$ matrix, denoted $I_n$, whose main diagonals are all 1’s and all the other elements are zero. Here are the general zero and identity matrices.

$$0_{n\times m}={\left(\begin{array}{cccc}0& 0& \cdots & 0\\ ⋮& ⋮& & ⋮\\ 0& 0& \cdots & 0\end{array}\right)}_{n\times m}{I}_n={\left(\begin{array}{cccc}1& 0& \cdots & 0\\ 0& 1& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 1\end{array}\right)}_{n\times n}$$

In matrix arithmetic these two matrices will act in matrix work like zero and one act in the real number system.

The last two special matrices that we’ll look at here are the column matrix and the row matrix. These are matrices that consist of a single column or a single row. In general, they are,

$$x={\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)}_{n\times 1}y={\left(\begin{array}{cccc}{y}_1& y_2& \cdots & y_m\end{array}\right)}_{1\times m}$$

We will often refer to these as vectors.

Arithmetic

We next need to take a look at arithmetic involving matrices. We’ll start with addition and subtraction of two matrices. So, suppose that we have two $n\times m$ matrices, $A$ and $B$. The sum (or difference) of these two matrices is then,

$$A_{n\times m}\pm B_{n\times m}={\left(a_{ij}\right)}_{n\times m}\pm {\left(b_{ij}\right)}_{n\times m}={\left(a_{ij}\pm b_{ij}\right)}_{n\times m}$$

The sum or difference of two matrices of the same size is a new matrix of identical size whose entries are the sum or difference of the corresponding entries from the original two matrices. Note that we can’t add or subtract entries with different sizes.

Next, let’s look at scalar multiplication. In scalar multiplication we are going to multiply a matrix $A$ by a constant (sometimes called a scalar) $\alpha$. In this case we get a new matrix whose entries have all been multiplied by the constant, $\alpha$.

$$\alpha A_{n\times m}=\alpha {\left(a_{ij}\right)}_{n\times m}={\left(\alpha a_{ij}\right)}_{n\times m}$$

Example 1 Given the following two matrices,$$A=\left(\begin{array}{cc}3& -2\\ -9& 1\end{array}\right)B=\left(\begin{array}{cc}-4& 1\\ 0& -5\end{array}\right)$$compute $A-5B$.

There isn’t much to do here other than the work.
$$\begin{array}{cc}A-5B& =\left(\begin{array}{cc}3& -2\\ -9& 1\end{array}\right)-5\left(\begin{array}{cc}-4& 1\\ 0& -5\end{array}\right)\\ & =\left(\begin{array}{cc}3& -2\\ -9& 1\end{array}\right)-\left(\begin{array}{cc}-20& 5\\ 0& -25\end{array}\right)\\ & =\left(\begin{array}{cc}23& -7\\ -9& 26\end{array}\right)\end{array}$$We first multiplied all the entries of $B$ by 5 then subtracted corresponding entries to get the entries in the new matrix.

The final matrix operation that we’ll take a look at is matrix multiplication. Here we will start with two matrices, $A_{n\times p}$ and $B_{p\times m}$. Note that $A$must have the same number of columns as $B$ has rows. If this isn’t true, then we can’t perform the multiplication. If it is true, then we can perform the following multiplication.

$$A_{n\times p}{B}_{p\times m}={\left(c_{ij}\right)}_{n\times m}$$

The new matrix will have size $n\times m$ and the entry in the $i^{\text{th}}$ row and $j^{\text{th}}$ column, $c_{ij}$, is found by multiplying row $i$ of matrix $A$ by column $j$ of matrix $B$. This doesn’t always make sense in words so let’s look at an example.

Example 2 Given$$A={\left(\begin{array}{ccc}2& -1& 0\\ -3& 6& 1\end{array}\right)}_{2\times 3}B={\left(\begin{array}{cccc}1& 0& -1& 2\\ -4& 3& 1& 0\\ 0& 3& 0& -2\end{array}\right)}_{3\times 4}$$compute $AB$.

The new matrix will have size $2\times 4$. The entry in row 1 and column 1 of the new matrix will be found by multiplying row 1 of $A$ by column 1 of $B$. This means that we multiply corresponding entries from the row of $A$ and the column of $B$ and then add the results up. Here are a couple of the entries computed all the way out.
$$\begin{array}{cc}{c}_{11}& =\left(2\right)\left(1\right)+\left(-1\right)\left(-4\right)+\left(0\right)\left(0\right)=6\\ c_{13}& =\left(2\right)\left(-1\right)+\left(-1\right)\left(1\right)+\left(0\right)\left(0\right)=-3\\ c_{24}& =\left(-3\right)\left(2\right)+\left(6\right)\left(0\right)+\left(1\right)\left(-2\right)=-8\end{array}$$Here’s the complete solution.
$$C=\left(\begin{array}{cccc}6& -3& -3& 4\\ -27& 21& 9& -8\end{array}\right)$$

In this last example notice that we could not have done the product BA since the number of columns of $B$ does not match the number of row of $A$. It is important to note that just because we can compute $AB$ doesn’t mean that we can compute $BA$. Likewise, even if we can compute both $AB$ and $BA$ they may or may not be the same matrix.

Determinant

The next topic that we need to take a look at is the determinant of a matrix. The determinant is actually a function that takes a square matrix and converts it into a number. The actual formula for the function is somewhat complex and definitely beyond the scope of this review.

The main method for computing determinants of any square matrix is called the method of cofactors. Since we are going to be dealing almost exclusively with $2\times 2$ matrices and the occasional $3\times 3$ matrix we won’t go into the method here. We can give simple formulas for each of these cases. The standard notation for the determinant of the matrix $A$ is.

$$det\left(A\right)=\left|A\right|$$

Here are the formulas for the determinant of $2\times 2$ and $3\times 3$ matrices.

$$\left|\begin{array}{cc}a& c\\ b& d\end{array}\right|=ad-cb$$$$\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|=a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|$$

Example 3 Find the determinant of each of the following matrices.
$$A=\left(\begin{array}{cc}-9& -18\\ 2& 4\end{array}\right)B=\left(\begin{array}{ccc}2& 3& 1\\ -1& -6& 7\\ 4& 5& -1\end{array}\right)$$

For the $2\times 2$ there isn’t much to do other than to plug it into the formula.
$$det\left(A\right)=\left|\begin{array}{cc}-9& -18\\ 2& 4\end{array}\right|=\left(-9\right)\left(4\right)-\left(-18\right)\left(2\right)=0$$For the $3\times 3$ we could plug it into the formula, however unlike the $2\times 2$ case this is not an easy formula to remember. There is an easier way to get the same result. A quicker way of getting the same result is to do the following. First write down the matrix and tack a copy of the first two columns onto the end as follows.
$$det\left(B\right)=\left|\begin{array}{ccc}2& 3& 1\\ -1& -6& 7\\ 4& 5& -1\end{array}\right|\begin{array}{cc}2& 3\\ -1& -6\\ 4& 5\end{array}$$Now, notice that there are three diagonals that run from left to right and three diagonals that run from right to left. What we do is multiply the entries on each diagonal up and the if the diagonal runs from left to right we add them up and if the diagonal runs from right to left we subtract them.
Here is the work for this matrix.
$$\begin{array}{cc}det\left(B\right)& =\left|\begin{array}{ccc}2& 3& 1\\ -1& -6& 7\\ 4& 5& -1\end{array}\right|\begin{array}{cc}2& 3\\ -1& -6\\ 4& 5\end{array}\\ & =\left(2\right)\left(-6\right)\left(-1\right)+\left(3\right)\left(7\right)\left(4\right)+\left(1\right)\left(-1\right)\left(5\right)-\\ & \left(3\right)\left(-1\right)\left(-1\right)-\left(2\right)\left(7\right)\left(5\right)-\left(1\right)\left(-6\right)\left(4\right)\\ & =42\end{array}$$

You can either use the formula or the short cut to get the determinant of a $3\times 3$.

If the determinant of a matrix is zero we call that matrix singular and if the determinant of a matrix isn’t zero we call the matrix nonsingular. The $2\times 2$ matrix in the above example was singular while the $3\times 3$ matrix is nonsingular.

Matrix Inverse

Next, we need to take a look at the inverse of a matrix. Given a square matrix, $A$, of size n x $n$ if we can find another matrix of the same size, $B$ such that,

$$AB=BA=I_n$$

then we call $B$ the inverse of $A$ and denote it by $B=A^{-1}$.

Computing the inverse of a matrix, $A$, is fairly simple. First, we form a new matrix,

$$\left(A{I}_n\right)$$

and then use the row operations from the previous section and try to convert this matrix into the form,

$$\left(I_{n}B\right)$$

If we can then $B$ is the inverse of $A$. If we can’t then there is no inverse of the matrix $A$.in which the inverse doesn’t exist.