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Introduction to Matrices Simplified Revision Notes

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2.1.1 Introduction to Matrices

What are Matrices?

Matrices are rectangular arrays of numbers arranged in rows and columns. They are a powerful tool in mathematics used to solve systems of equations, perform transformations in geometry, and more. Matrices are fundamental in many areas of maths, especially in linear algebra.

Definition of a Matrix

A matrix is typically written as:

A=(a11a12a1na21a22a2nam1am2amn)A = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{pmatrix}
  • The numbers aija_{ij} are called the elements or entries of the matrix.
  • The subscript ii refers to the row, and jj refers to the column, so aija_{ij} is the element in row ii and column jj .
  • A matrix with mm rows and nn columns is called an mm × nn matrix (read "mm by nn").

Types of Matrices

  1. Row Matrix: A matrix with only one row (e.g., 1×n1 \times n ).
(123)\begin{pmatrix} 1 & 2 & 3 \end{pmatrix}
  1. Column Matrix: A matrix with only one column (e.g., m×1m \times 1 ).
(123)\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}
  1. Square Matrix: A matrix with the same number of rows and columns ( n×n n \times n ).
(1234)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}
  1. Zero Matrix: A matrix where all entries are zero.
(0000)\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}

Matrix Notation

  • Capital letters like A, B, and C are used to denote matrices.
  • Each element in the matrix is denoted by aija_{ij}, where ii is the row number and jj is the column number.
infoNote

For example, in the matrix A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}

  • a11=1a_{11} = 1
  • a12=2a_{12} = 2
  • a21=3a_{21} = 3
  • a22=4a_{22} = 4

Operations on Matrices

Matrix Addition:

Matrices can only be added if they have the same dimensions. Add corresponding elements.

lightbulbExample

Example:

A=(1234),B=(5678)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}A+B=(1+52+63+74+8)=(681012)A + B = \begin{pmatrix} 1+5 & 2+6 \\ 3+7 & 4+8 \end{pmatrix} = \begin{pmatrix} 6 & 8 \\ 10 & 12 \end{pmatrix}

Matrix Subtraction:

Like addition, subtraction is performed element by element.

lightbulbExample

Example:

AB=(15263748)=(4444)A - B = \begin{pmatrix} 1-5 & 2-6 \\ 3-7 & 4-8 \end{pmatrix} = \begin{pmatrix} -4 & -4 \\ -4 & -4 \end{pmatrix}

Scalar Multiplication:

Multiply every element of the matrix by the scalar (a single number).

lightbulbExample

Example:

2×(1234)=(2×12×22×32×4)=(2468)2 \times \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times 3 & 2 \times 4 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix}

Square Matrices Multiplication:

To multiply square matrices, we multiply certain elements of the matrices together. We move from left to right in the first matrix and from top to bottom in the second.

lightbulbExample

Example**:** Multiply these Matrices

(2413)(7826)\begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 7 & 8 \\ 2 & 6 \end{pmatrix}

Step 1: Start at the top left of the first matrix and the top left of the second.

Multiply these numbers together and add the product of the two numbers obtained while moving 11 right on matrix 11 and 11 down on matrix 22.

(2413)(7826)=(2×7+4×2)\begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 7 & 8 \\ 2 & 6 \end{pmatrix} = \begin{pmatrix} 2 \times 7 + 4 \times 2 & & & \\ \\ \end{pmatrix}

We started at the top of the first matrix and the left of the second, so this entry goes top left.


Step 2: Now start at the bottom of the first and left of the second.

(2413)(7826)\begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 7 & 8 \\ 2 & 6 \end{pmatrix}

Move from left to right in the first and top to bottom in the second.

(1×7+3×2)\begin{pmatrix} \\ 1 \times 7 + 3 \times 2 \\ \end{pmatrix}

Step 3: Repeat this method starting at the other two possible start points

(2413)(7826)=(2×7+4×22×8+4×61×7+3×21×8+3×2)\begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 7 & 8 \\ 2 & 6 \end{pmatrix}=\begin{pmatrix} 2 \times 7 + 4 \times 2 & 2 \times 8 + 4 \times 6\\ 1 \times 7 + 3 \times 2 & 1 \times 8 + 3 \times 2 \end{pmatrix}=(22401326)=\begin{pmatrix} 22 & 40\\ 13 & 26 \end{pmatrix}

Understanding Basic Matrix Operations is Crucial

A matrix is an arrangement of numbers in rows and columns.

You can perform addition, subtraction, and scalar multiplication on matrices, but addition and subtraction require matrices to be of the same dimension.

Matrix dimensions are important for determining whether operations like addition and multiplication are possible.

Worked Examples

lightbulbExample

Example Given the matrices:

A=(2563),B=(1203)\mathbf{A} = \begin{pmatrix} 2 & 5 \\ 6 & -3 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} -1 & 2 \\ 0 & 3 \end{pmatrix}

Calculate A×B\mathbf{A} \times \mathbf{B}


Step 1: Multiply the first row of A\mathbf{A} by the first column of B\mathbf{B} to find the top-left element.

(2563)(10)=(2×1)+(5×0)\begin{pmatrix} 2 & 5 \\ 6 & -3 \end{pmatrix} \begin{pmatrix} -1 \\ 0 \end{pmatrix} = (2 \times -1) + (5 \times 0) =2+0=2= -2 + 0 = -2

Step 2: Multiply the first row of A\mathbf{A} by the second column of B\mathbf{B} to find the top-right element.

(2563)(23)=(2×2)+(5×3)\begin{pmatrix} 2 & 5 \\ 6 & -3 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = (2 \times 2) + (5 \times 3)=4+15=19 = 4 + 15 = 19

Step 3: Multiply the second row of A\mathbf{A} by the first column of B\mathbf{B} to find the bottom-left element.

(63)(10)=(6×1)+(3×0)\begin{pmatrix} 6 & -3 \end{pmatrix} \begin{pmatrix} -1 \\ 0 \end{pmatrix} = (6 \times -1) + (-3 \times 0)=6+0=6 = -6 + 0 = -6

Step 4: Multiply the second row of A\mathbf{A} by the second column of B\mathbf{B} to find the bottom-right element.

(63)(23)=(6×2)+(3×3)\begin{pmatrix} 6 & -3 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = (6 \times 2) + (-3 \times 3) =129=3= 12 - 9 = 3

Step 5: Combine the elements to form the resulting matrix:

A×B=(21963)\mathbf{A} \times \mathbf{B} = \begin{pmatrix} -2 & 19 \\ -6 & 3 \end{pmatrix}
lightbulbExample

Example Given the matrices:

A=(263114712),B=(2752372610)\mathbf{A} = \begin{pmatrix} 2 & 6 & 3 \\ 1 & 1 & 4 \\ 7 & 1 & 2 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 2 & 7 & 5 \\ 2 & -3 & 7 \\ 2 & 6 & 10 \end{pmatrix}

Calculate A×B\mathbf{A} \times \mathbf{B}


Step 1: Multiply the first row of A\mathbf{A} by the first column of B\mathbf{B} to find the top-left element.

(263)(222)=(2×2)+(6×2)+(3×2)\begin{pmatrix} 2 & 6 & 3 \end{pmatrix} \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} = (2 \times 2) + (6 \times 2) + (3 \times 2) =4+12+6=22= 4 + 12 + 6 = 22

Step 2: Multiply the first row of A\mathbf{A} by the second column of B\mathbf{B} to find the top-middle element.

(263)(736)=(2×7)+(6×3)+(3×6)\begin{pmatrix} 2 & 6 & 3 \end{pmatrix} \begin{pmatrix} 7 \\ -3 \\ 6 \end{pmatrix} = (2 \times 7) + (6 \times -3) + (3 \times 6) =1418+18=14= 14 - 18 + 18 = 14

Step 3: Multiply the first row of A\mathbf{A} by the third column of B\mathbf{B} to find the top-right element.

(263)(5710)=(2×5)+(6×7)+(3×10)\begin{pmatrix} 2 & 6 & 3 \end{pmatrix} \begin{pmatrix} 5 \\ 7 \\ 10 \end{pmatrix} = (2 \times 5) + (6 \times 7) + (3 \times 10) =10+42+30=82= 10 + 42 + 30 = 82

Step 4: Multiply the second row of A\mathbf{A} by the first column of B\mathbf{B} to find the middle-left element.

(114)(222)=(1×2)+(1×2)+(4×2)\begin{pmatrix} 1 & 1 & 4 \end{pmatrix} \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} = (1 \times 2) + (1 \times 2) + (4 \times 2) =2+2+8=12= 2 + 2 + 8 = 12

Step 5: Multiply the second row of A\mathbf{A} by the second column of B\mathbf{B} to find the middle-middle element.

(114)(736)=(1×7)+(1×3)+(4×6)\begin{pmatrix} 1 & 1 & 4 \end{pmatrix} \begin{pmatrix} 7 \\ -3 \\ 6 \end{pmatrix} = (1 \times 7) + (1 \times -3) + (4 \times 6) =73+24=28= 7 - 3 + 24 = 28

Step 6: Multiply the second row of A\mathbf{A} by the third column of B\mathbf{B} to find the middle-right element.

(114)(5710)=(1×5)+(1×7)+(4×10)\begin{pmatrix} 1 & 1 & 4 \end{pmatrix} \begin{pmatrix} 5 \\ 7 \\ 10 \end{pmatrix} = (1 \times 5) + (1 \times 7) + (4 \times 10) =5+7+40=52= 5 + 7 + 40 = 52

Step 7: Multiply the third row of A\mathbf{A} by the first column of B\mathbf{B} to find the bottom-left element.

(712)(222)=(7×2)+(1×2)+(2×2)\begin{pmatrix} 7 & 1 & 2 \end{pmatrix} \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} = (7 \times 2) + (1 \times 2) + (2 \times 2) =14+2+4=20= 14 + 2 + 4 = 20

Step 8: Multiply the third row of A\mathbf{A} by the second column of B\mathbf{B} to find the bottom-middle element.

(712)(736)=(7×7)+(1×3)+(2×6)\begin{pmatrix} 7 & 1 & 2 \end{pmatrix} \begin{pmatrix} 7 \\ -3 \\ 6 \end{pmatrix} = (7 \times 7) + (1 \times -3) + (2 \times 6) =493+12=58= 49 - 3 + 12 = 58

Step 9: Multiply the third row of A\mathbf{A} by the third column of B\mathbf{B} to find the bottom-right element.

(712)(5710)=(7×5)+(1×7)+(2×10)\begin{pmatrix} 7 & 1 & 2 \end{pmatrix} \begin{pmatrix} 5 \\ 7 \\ 10 \end{pmatrix} = (7 \times 5) + (1 \times 7) + (2 \times 10) =35+7+20=62= 35 + 7 + 20 = 62

Step 10: Combine all the calculated elements to form the resulting matrix.

A×B=(2514102162852225862)\mathbf{A} \times \mathbf{B} = \begin{pmatrix} 25 & 14 & 102 \\ 16 & 28 & 52 \\ 22 & 58 & 62 \end{pmatrix}
lightbulbExample

Example Using the Calculator for Matrix Multiplications Multiply the following matrices:

(2317)(6304)\begin{pmatrix} 2 & 3 \\ 1 & 7 \end{pmatrix} \begin{pmatrix} 6 & -3 \\ 0 & 4 \end{pmatrix}

Calculator Steps:

Non-Commutativity of Matrices

Two objects, AA and BB, are said to be commutative (or "they commute") if:

A×B=B×AA \times B = B \times A

For example, real and complex numbers are commutative under multiplication. However, matrices are not commutative in general.

lightbulbExample

Example: Let:

A=(1324),B=(7211)A = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 7 & -2 \\ 1 & 1 \end{pmatrix}

Step 1: Calculate ABAB

AB=(1324)(7211)AB = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 7 & -2 \\ 1 & 1 \end{pmatrix}

Multiply the rows of AA by the columns of BB:

AB=((1×7)+(3×1)(1×2)+(3×1)(2×7)+(4×1)(2×2)+(4×1))AB = \begin{pmatrix} (1 \times 7) + (3 \times 1) & (1 \times -2) + (3 \times 1) \\ (2 \times 7) + (4 \times 1) & (2 \times -2) + (4 \times 1) \end{pmatrix} =(101180)= \begin{pmatrix} 10 & 1 \\ 18 & 0 \end{pmatrix}

Step 2: Calculate BABA

BA=(7211)(1324)BA = \begin{pmatrix} 7 & -2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}

Multiply the rows of BB by the columns of AA:

BA=((7×1)+(2×2)(7×3)+(2×4)(1×1)+(1×2)(1×3)+(1×4))BA = \begin{pmatrix} (7 \times 1) + (-2 \times 2) & (7 \times 3) + (-2 \times 4) \\ (1 \times 1) + (1 \times 2) & (1 \times 3) + (1 \times 4) \end{pmatrix} =(31337)= \begin{pmatrix} 3 & 13 \\ 3 & 7 \end{pmatrix}

Conclusion:

AB=(101180),BA=(31337)AB = \begin{pmatrix} 10 & 1 \\ 18 & 0 \end{pmatrix}, \quad BA = \begin{pmatrix} 3 & 13 \\ 3 & 7 \end{pmatrix}

Clearly,

ABBAAB \neq BA

This shows that matrices do not generally commute under multiplication.

While there are special cases where specific pairs of matrices do commute (i.e., AB=BAAB = BA), this is the exception rather than the rule, as demonstrated in this example.

Multiplying Non-Square Matrices

The same rules for multiplying square matrices apply to non-square matrices.

lightbulbExample

Example: Let:

A=(241134),B=(2711),C=(36)A = \begin{pmatrix} 2 & 4 \\ 1 & 1 \\ 3 & -4 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 7 \\ 1 & 1 \end{pmatrix}, \quad C = \begin{pmatrix} 3 & 6 \end{pmatrix}

Matrix Dimensions:

  • AA is 3×23 \times 2
  • BB is 2×22 \times 2
  • CC is 1×21 \times 2 An m×nm \times n matrix can only be multiplied by an n×kn \times k matrix because the inner dimensions must match.

Checking Multiplications:

Multiplying ABAB:

A is 3×2,B is 2×2A \text{ is } 3 \times 2, \quad B \text{ is } 2 \times 2

The inner dimensions (2,22, 2) match, so multiplication is possible.

The resulting matrix will have dimensions of the outer dimensions:

3×23 \times 2

Multiplying BABA:

B is 2×2,A is 3×2B \text{ is } 2 \times 2, \quad A \text{ is } 3 \times 2

The inner dimensions (2,32, 3) do not match, so multiplication is not possible.

lightbulbExample

Example: Let:

A=(621134),B=(11)A = \begin{pmatrix} 6 & 2 \\ 1 & 1 \\ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} -1 \\ 1 \end{pmatrix}

Matrix Dimensions:

  • AA is 3×23 \times 2
  • BB is 2×12 \times 1 Here, the inner dimensions (2,22, 2) match, so multiplication is possible.

The resulting matrix will have dimensions of the outer dimensions:

3×13 \times 1
infoNote

Calculating ABAB:

AB=(621134)(11)AB = \begin{pmatrix} 6 & 2 \\ 1 & 1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix}

Multiply the rows of AA by the single column of BB:

AB=((6×1)+(2×1)(1×1)+(1×1)(3×1)+(4×1))AB = \begin{pmatrix} (6 \times -1) + (2 \times 1) \\ (1 \times -1) + (1 \times 1) \\ (3 \times -1) + (4 \times 1) \end{pmatrix} =(6+21+13+4)=(401)= \begin{pmatrix} -6 + 2 \\ -1 + 1 \\ -3 + 4 \end{pmatrix} = \begin{pmatrix} -4 \\ 0 \\ 1 \end{pmatrix}
infoNote

Key Takeaways:

  • To multiply an m×nm \times n matrix by an n×kn \times k matrix, the inner dimensions nn must match.
  • The result will be an m×km \times k matrix.
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