IB Mathematics AI SL Definition of a matrix Study Notes - New Syllabus

MATRICES

♦ A matrix is simply a rectangular array of numbers (called elements). Some typical examples:

\(
\begin{pmatrix}
1 & 2 & 7 \\
-3 & 0 & 5
\end{pmatrix}, \quad
\begin{pmatrix}
4 & 6 \\
2 & 11
\end{pmatrix}, \quad
\begin{pmatrix}
4 & -8 \\
2.5 & 4 \\
0 & 1/2 \\
1 & 3.1
\end{pmatrix}
\)

A matrix is usually denoted by a capital letter, say:

\(
A = \begin{pmatrix}
1 & 2 & 7 \\
-3 & 0 & 5
\end{pmatrix}, \quad
B = \begin{pmatrix}
4 & 6 \\
2 & 11
\end{pmatrix}, \quad
C = \begin{pmatrix}
4 & -8 \\
2.5 & 4 \\
0 & 1/2 \\
1 & 3.1
\end{pmatrix}
\)

The first matrix above has 2 rows and 3 columns. We say that \( A \) is a \( 2 \times 3 \) (“two by three”) matrix, or otherwise that the order of \( A \) is \( 2 \times 3 \).

Likewise, \( B \) is a \( 2 \times 2 \) matrix while \( C \) is a \( 4 \times 2 \) matrix.

The general form of a \( 2 \times 3 \) matrix is:

\(
A = \begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{pmatrix}
\)

(notice that \( a_{23} \), for example, is the element in row 2 and column 3).

♦ SQUARE MATRICES
No of rows = No of columns. The order of a square matrix is \( n \times n \), e.g., \( 2 \times 2 \), \( 3 \times 3 \), \( 4 \times 4 \), etc. (in this case, we may also say: “a square matrix of order \( n \)”).

For example,

\(
\begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix}, \quad
\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{pmatrix}
\)

are square matrices of order 2 and 3, respectively.

We also say that the elements \( a_{11}, a_{22}, a_{33}, \ldots \) (indicated above) form the main diagonal of the square matrix.

♦ ROW MATRICES
Matrices of order \( 1 \times n \). For example:

\(
A = \begin{pmatrix} 2 & 7 & 3 \end{pmatrix} \quad \text{is a } 1 \times 3 \text{ row matrix.}
\)

\(
B = \begin{pmatrix} 4 & 2 & 6 & -1 & 0 & 6 \end{pmatrix} \quad \text{is a } 1 \times 6 \text{ row matrix.}
\)

♦ COLUMN MATRICES (or VECTORS)
Matrices of order \( m \times 1 \). For example:

\(
A = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \quad \text{is a } 3 \times 1 \text{ vector.}
\)

\(
B = \begin{pmatrix} -1 \\ 3 \end{pmatrix} \quad \text{is a } 2 \times 1 \text{ vector.}
\)

Notice: Matrices of order \( 1 \times 1 \) are also defined, for example \( C = (5) \).

♦ THE ZERO MATRIX O
All elements are 0. The \( 2 \times 3 \) zero matrix is:

\(
O = \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\)

The \( 2 \times 2 \) zero matrix is:

\(
O = \begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}, \quad \text{and so on!}
\)

♦ EQUAL MATRICES: \( A = B \)
\( A = B \) if:
\( A \) and \( B \) have the same order.
 The corresponding elements are equal.

Let \( A \) and \( B \) have the same order. We define some new matrices:

♦ THE SUM \( A + B \)
We simply add the corresponding elements.

♦ THE DIFFERENCE \( A – B \)
We simply subtract the corresponding elements.

♦ THE SCALAR PRODUCT \( nA \) (\( n \) is a scalar, i.e., a number)
We simply multiply each element of \( A \) by \( n \).

♦ THE PRODUCT OF TWO MATRICES: \( AB \)

\(
\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 5 & 1 \end{pmatrix} = ?
\)

It is NOT \( \begin{pmatrix} 2 & 6 \\ 15 & 4 \end{pmatrix} \) as someone would expect!! Here, we do not multiply the corresponding elements.

Multiply a row-matrix by a column-matrix:

\(
\begin{pmatrix} a & b & c \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = (ax + by + cz)
\)

Notice that we multiply the corresponding elements and add the results.

\(
\begin{pmatrix} 2 & 3 & 5 \end{pmatrix} \begin{pmatrix} 3 \\ 4 \\ 1 \end{pmatrix} = (2 \cdot 3 + 3 \cdot 4 + 5 \cdot 1) = (23)
\)

Now we are ready to multiply two matrices \( A \) and \( B \) in general. First of all, the orders of \( A \) and \( B \) must be as follows:

\(
\begin{array}{c|c}
A & B \\
\hline
m \times k & k \times n
\end{array}
\)

That is, the number of columns of \( A \) = number of rows of \( B \).

The order of the new matrix \( AB \) will be \( m \times n \).

\(
\begin{array}{|c|c|c|}
\hline
\text{Order of } A & \text{Order of } B & \text{Order of } AB \\
\hline
3 \times 5 & 5 \times 8 & 3 \times 8 \\
\hline
3 \times 2 & 2 \times 1 & 3 \times 1 \\
\hline
2 \times 2 & 2 \times 2 & 2 \times 2 \\
\hline
2 \times 3 & 2 \times 3 & \text{Not defined} \\
\hline
\end{array}
\)

The multiplication takes place as follows:
 We multiply rows of \( A \) by columns of \( B \).
 (Row \( i \)) \( \times \) (Column \( j \)) will give the element \( a_{ij} \).

Let:

\(
A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}, \quad
B = \begin{pmatrix} 2 & 5 \\ 3 & 3 \\ 1 & 2 \end{pmatrix}
\)

Notice that the order of \( AB \) is expected to be \( 2 \times 2 \).

For the first row of \( AB \): multiply row 1 of \( A \) by each column of \( B \) separately:

\(
AB = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \begin{pmatrix} 2 & 5 \\ 3 & 3 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 1 & 1 \cdot 5 + 2 \cdot 3 + 3 \cdot 2 \\ – & – \end{pmatrix} = \begin{pmatrix} 11 & 17 \\ – & – \end{pmatrix}
\)

For the second row of \( AB \): multiply row 2 of \( A \) by each column of \( B \) separately:

\(
AB = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \begin{pmatrix} 2 & 5 \\ 3 & 3 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 11 & 17 \\ 4 \cdot 2 + 5 \cdot 3 + 6 \cdot 1 & 4 \cdot 5 + 5 \cdot 3 + 6 \cdot 2 \end{pmatrix} = \begin{pmatrix} 11 & 17 \\ 29 & 47 \end{pmatrix}
\)

♦ THE IDENTITY MATRIX \( I \)
1 in the main diagonal, 0 elsewhere. Namely:

\(
I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad
I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad
I = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \quad \text{etc.}
\)

♦ PROPERTIES OF MATRIX OPERATIONS
The following properties hold:

For addition of matrices 
\( A + B = B + A \) Commutative law 
\( (A + B) + C = A + (B + C) \) Associative law 
 \( A + O = A = O + A \)  \( O \) is the identity for addition 

 For multiplication of matrices 
\( (AB)C = A(BC) \) | Associative law 
 \( AI = A = IA \) | \( I \) is the identity for multiplication 

Remember that the commutative law does not hold for multiplication, i.e., \( AB \neq BA \) in general.

For addition and multiplication of matrices together 
 \( (A + B)C = AC + BC \) Distributive laws 
 \( C(A + B) = CA + CB \) 

 For multiplication by scalars \( m, n \in \mathbb{R} \) 
\( m(A + B) = mA + mB \)  Distributive laws 
 \( (m + n)A = mA + nA \) 
 \( m(nA) = (mn)A \) 

THE DETERMINANT detA – THE INVERSE \( A^{-1} \)

♦ 2×2 DETERMINANT

Let \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \). The determinant of \( A \) is the number:

\(
\text{det}A = ad – bc
\)

It is also denoted by:

\(
\text{det} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad – bc \quad \text{or} \quad \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad – bc
\)

♦ 3×3 DETERMINANT

Let:

\(
A = \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix}
\)

In exams, the \( 3 \times 3 \) determinant, \( \text{det}A \), will be asked only by GDC.

Although it is not in the syllabus, just for information we mention the definition of the determinant of \( A \). It is the number:

\(
\text{det}A = a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2 – a_1b_3c_2 – a_2b_1c_3 – a_3b_2c_1
\)

A more elegant way to estimate \( \text{det}A \) is:

\(
\text{det}A = a_1 \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} – a_2 \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix}
\)

 We multiply the elements of the first row, \( a_1, a_2, a_3 \), by three smaller determinants respectively.
For \( a_1 \), the corresponding \( 2 \times 2 \) determinant can be obtained if we eliminate the row and the column of \( a_1 \).
 Similarly for \( a_2 \) and \( a_3 \).
 We alternate the signs (\( + – + \)).

Consider the following special cases:

\(
A = \begin{pmatrix} a & x & y \\ 0 & b & z \\ 0 & 0 & c \end{pmatrix}, \quad
B = \begin{pmatrix} a & 0 & 0 \\ x & b & 0 \\ y & z & c \end{pmatrix}, \quad
C = \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}
\)

Matrices like \( A \) are called upper-triangular (0’s below main diagonal).
Matrices like \( B \) are called lower-triangular (0’s above main diagonal).
Matrices like \( C \) are called diagonal (0’s below and above main diagonal).

For the determinant of such a matrix, we just multiply the elements of the main diagonal:

\(
\text{det}A = abc, \quad \text{det}B = abc, \quad \text{det}C = abc
\)

For example:

\(
\begin{vmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{vmatrix} = 2 \cdot 3 \cdot 4 = 24, \quad
\begin{vmatrix} 2 & \sqrt{2} & 5 \\ 0 & 3 & -8 \\ 0 & 0 & 4 \end{vmatrix} = 2 \cdot 3 \cdot 4 = 24
\)

♦ THE INVERSE \( A^{-1} \) OF A 2×2 MATRIX

Let \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \). Given that \( \text{det}A \neq 0 \) (and only then), we define the inverse of \( A \) as follows:

\(
A^{-1} = \frac{1}{\text{det}A} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
\)

♦ NUMBERS  MATRICES 

 The inverse of number \( a \) is \( a^{-1} \). \( aa^{-1} = 1 \) and \( a^{-1}a = 1 \). | The inverse of a matrix \( A \) is \( A^{-1} \). \( AA^{-1} = I \) and \( A^{-1}A = I \). 
 Is any number invertible? NO. Only if \( a \neq 0 \) (if \( a = 0 \), \( a^{-1} \) is not defined). | Is any Matrix invertible? NO. Only if \( \text{det}A \neq 0 \) (if \( \text{det}A = 0 \), \( A^{-1} \) is not defined). 
 If \( a \) is invertible, then \( a^{-1} = \frac{1}{a} \). | If \( A \) is invertible, then \( A^{-1} = \frac{1}{\text{det}A} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \). 

NOTICE: If \( AB = I \) or \( BA = I \), we know that \( A \) is invertible and \( B \) is the inverse of \( A \), that is:

\(
A^{-1} = B
\)

(\( B \) is also invertible and \( B^{-1} = A \)).

♦ THE INVERSE \( A^{-1} \) OF A 3×3 MATRIX

The explicit formula for \( A^{-1} \) is out of our scope! It is enough to know that:
 \( A^{-1} \) exists only if \( \text{det}A \neq 0 \).
 \( AA^{-1} = I \) and \( A^{-1}A = I \).
 If \( AB = I \) or \( BA = I \), then \( B \) is the inverse of \( A \).
 \( A^{-1} \) can be found by GDC.