Form 3 Mathematics

Chapter 6: Matrices

6.1. Introduction
6.2. Order of a matrix
6.3. Addition and subtraction of matrices
6.4. Scalar multiplication and equal matrices
6.5. Multiplication of matrices
6.6. Determinant of 2 × 2 matrices
6.7. Inverse of 2 × 2 matrices
6.8. Identity matrices
6.9. Singular and non-singular matrices
6.10. Simultaneous linear equations in two variables
6.11. Summary
6.12. Further reading
6.13. Test 6

6.1. Introduction

In this chapter you are going to cover mathematical operations in singular and non-singular matrices. You are also going to use matrices as operators, that is, solving matrices using simultaneous equations.

Objectives

After going through this chapter, you should be able to

  • perform operations with matrices.
  • solve problems involving 2 × 2 singular and non-singular matrices.
  • find the determinant of a 2 × 2 matrix.
  • find the inverse of a 2 × 2 non-singular matrix.
  • solve simultaneous problems using the matrix method.

Key terms

Determinant – Determines whether the matrix is invertible or not.
Matrix – Is the rectangular arrangement of elements or objects.
Order of a matrix – Is the number of rows and columns of a given matrix.
Singular matrix – Is a matrix which has a determinant of zero, it has no inverse.

Time

You should not spend more than 10 hours in this chapter.

Study skills

The key to mastery of mathematics is practice. You need to work out as many problems as possible about matrices to have a better understanding of the subject.

6.2. Order of a matrix

The order of a matrix is defined by stating number of rows (horizontally arranged) and number of columns (vertically arranged). If there are 3 rows and 2 columns, as follows, it’s called a 3 × 2 matrix.
\( \begin{pmatrix} a & b \\ c & d \\ e & f \\ \end{pmatrix} \)

Example 1
Questions

  • State the order of the following matrices.
    1. \( A = \begin{pmatrix} 2 & 1 \\ 2 & 3 \\ 4 & 1 \\ \end{pmatrix} \)
    2. \( B = \begin{pmatrix} 1 \\ 4 \\ \end{pmatrix} \)
    3. \( C = \begin{pmatrix} 5 & 3 & 4 \\ 3 & 1 & 2 \\ 5 & 3 & 2 \\ \end{pmatrix} \)
    4. \( D = \begin{pmatrix} 1 & 3 & -7 \\ \end{pmatrix} \)

Answers

  1. To get the order of matrix A, you count the number of rows in the matrix first and then the number of columns. There are 3 rows and 2 columns. This is a 3 by 2 matrix because it has 3 rows and 2 columns, that is, 3 × 2 matrix.
  2. To get the order of matrix B, you count the number of rows in the matrix first and then the number of columns. There are 2 rows and 1 column. This is a 2 by 1 matrix because it has 2 rows and 1 column, that is, 2 × 1 matrix. It is called a column matrix because it has only one column.
  3. To get the order of matrix C, you count the number of rows in the matrix first and then the number of columns. There are 3 rows and 3 columns. This is a 3 by 3 matrix because it has 3 rows and 3 columns, that is, 3 × 2 matrix. It is called a square matrix of order 3 because it has equal number of rows and columns.
  4. To get the order of matrix D, you count the number of rows in the matrix first and then the number of columns. There is 1 row and 3 columns. This is a 1 by 3 matrix because it has 1 row and 3 columns, that is, 1 × 3 matrix. It is called a row matrix because it has only one row.

You may attempt the following exercise.

Exercise 6.1: Order of a matrix
Questions

  1. State the order of the following matrices.
    1. \( \begin{pmatrix} 0 & 2 & 5 \\ 1 & 3 & 4 \\ \end{pmatrix} \)
    2. \( \begin{pmatrix} x & -4 \\ 3 & 6 \\ 6 & y \\ 8 & 5 \\ \end{pmatrix} \)
    3. \( \begin{pmatrix} 4 & 5 & -9 & 0 \\ 3 & 4 & 5 & 7 \\ 5 & -2 & 4 & 1 \\ \end{pmatrix} \)
    4. \( \begin{pmatrix} 3 & 2 \\ 1 & 3 \\ \end{pmatrix} \)
  2. State the special name given to the following matrices.
    1. \( \begin{pmatrix} 3 & 3 & 7 \\ 2 & 6 & 9 \\ 5 & -8 & 0 \\ \end{pmatrix} \)
    2. \( \begin{pmatrix} 0 \\ 2 \\ 1 \\ \end{pmatrix} \)
    3. \( \begin{pmatrix} 1 & -8 & 6 & 3 \\ \end{pmatrix} \)

Answers

    1. 2 × 3
    2. 4 × 2
    3. 3 × 4
    4. 2 × 2
    1. Square matrix of order 3
    2. 3 by 1 column matrix
    3. 1 by 4 row matrix

6.3. Addition and subtraction of matrices

When you add or subtract matrices, you add or subtract corresponding elements of the given matrices.

Note: You can only add or subtract matrices which have the same order> If matrices are not of the same order it is impossible to add or subtract them.

Example 2
Questions

  • Given that \( A = \begin{pmatrix} 1 & 3 \\ -3 & 4 \\ \end{pmatrix} \), \( B = \begin{pmatrix} 5 & 4 \\ 7 & 2 \\ \end{pmatrix} \), \( C = \begin{pmatrix} -4 & 8 \\ 6 & 9 \\ \end{pmatrix} \), \( D = \begin{pmatrix} 5 & 3 & 1 \\ 5 & 2 & 7 \\ \end{pmatrix} \), find
    1. A + B
    2. B - C
    3. A + C - B
    4. A + D

Answers

  1. Matrix A and matrix B are of the same order, that is, (2 × 2). They can be added.
    \( A + B = \begin{pmatrix} 1 & 3 \\ -3 & 4 \\ \end{pmatrix} + \begin{pmatrix} 5 & 4 \\ 7 & 2 \\ \end{pmatrix} = \begin{pmatrix} 1 + 5 & 3 + 4 \\ -3 + 7 & 4 + 2 \\ \end{pmatrix} = \begin{pmatrix} 6 & 7 \\ 4 & 6 \\ \end{pmatrix} \)
    From this example, you can see that elements (1 and 5), (3 and 4), (-3 and 7) and (4 and 2) are corresponding values, that is why it was possible to add them.
  2. Matrix B and matrix C are of the same order, that is, (2 × 2). They can subtract.
    \( B - C = \begin{pmatrix} 5 & 4 \\ 7 & 2 \\ \end{pmatrix} - \begin{pmatrix} -4 & 8 \\ 6 & 9 \\ \end{pmatrix} = \begin{pmatrix} 5 - -4 & 4 - 8 \\ 7 - 6 & 2 - 9 \\ \end{pmatrix} = \begin{pmatrix} 9 & -4 \\ 1 & -7 \\ \end{pmatrix} \)
    From this example, you can see that elements (5 and -4), (4 and 8), (7 and 6) and (2 and 9) are corresponding values, that is why it was possible for them to subtract.
  3. Matrix A, matrix C and matrix B are of the same order, that is, (2 × 2). They can add and subtract.
    \( A + C - B = \begin{pmatrix} 1 & 3 \\ -3 & 4 \\ \end{pmatrix} + \begin{pmatrix} -4 & 8 \\ 6 & 9 \\ \end{pmatrix} - \begin{pmatrix} 5 & 4 \\ 7 & 2 \\ \end{pmatrix} = \begin{pmatrix} 1 + -4 - 5 & 3 + 8 - 4 \\ -3 + 6 - 7 & 4 + 9 - 2 \\ \end{pmatrix} = \begin{pmatrix} -8 & 7 \\ -4 & 11 \\ \end{pmatrix} \)
    From this example, you can see that elements (1 and -4 and 5), (3 and 8 and 4), (-3 and 6 and 7) and (4 and 9 and 2) are corresponding values, that is why it was possible to add and subtract them.
  4. If you look at the order of matrix A and matrix D, you can see that matrix A is a 2 × 2 matrix whilst matrix D is a 2 × 3 matrix. Since the 2 matrices are not of the same order you cannot proceed with addition. It is impossible to add.

You may attempt the following exercise.

Exercise 6.2: Addition and subtraction of matrices
Questions

  1. \( \begin{pmatrix} -1 & -7 \\ 3 & 8 \\ \end{pmatrix} + \begin{pmatrix} 2 & 8 \\ 3 & -6 \\ \end{pmatrix} \)
  2. \( \begin{pmatrix} 2 & 6 \\ 3 & 8 \\ -1 & 5 \\ \end{pmatrix} - \begin{pmatrix} -7 & 4 \\ 6 & 0 \\ -6 & -3 \\ \end{pmatrix} \)
  3. \( \begin{pmatrix} 1 & -6 \\ -2 & 4 \\ \end{pmatrix} + \begin{pmatrix} 0 \\ 2 \\ \end{pmatrix} \)
  4. \( \begin{pmatrix} 4 & -8 \\ 5 & 4 \\ \end{pmatrix} - \begin{pmatrix} -7 & -6 \\ 0 & 4 \\ \end{pmatrix} + \begin{pmatrix} 3 & -3 \\ 6 & 5 \\ \end{pmatrix} \)
  5. \( \begin{pmatrix} 0 & 6 \\ \end{pmatrix} + \begin{pmatrix} 4 \\ -5 \\ \end{pmatrix} \)

Answers

  1. \( \begin{pmatrix} 1 & 1 \\ 6 & 2 \\ \end{pmatrix} \)
  2. \( \begin{pmatrix} -9 & 2 \\ -3 & 8 \\ 5 & 8 \\ \end{pmatrix} \)
  3. Impossible
  4. \( \begin{pmatrix} 14 & -5 \\ 11 & -5 \\ \end{pmatrix} \)
  5. Impossible

6.4. Scalar multiplication and equal matrices

A scalar is a numerical matrix multiplier, which has an effect of multiplying every element in the given matrix, for example \( 3 \begin{pmatrix} x \\ y \\ \end{pmatrix} \) matrix. Here the scalar is 3. You must be aware that a scalar can be either positive or negative and can also be a fraction.

Example 3
Questions

  • Given that \( X = \begin{pmatrix} 3 & 2 \\ 4 & -5 \\ \end{pmatrix} \), \( Y = \begin{pmatrix} 4 \\ 2 \\ \end{pmatrix} \), and \( Z = \begin{pmatrix} 3 & 3 & 2 \\ \end{pmatrix} \), find
    1. \( 3X \)
    2. \( \dfrac{1}{2}Y \)
    3. \( -5Z \)

Answers

  1. \( 3X = 3 \begin{pmatrix} 3 & 2 \\ 4 & -5 \\ \end{pmatrix} = \begin{pmatrix} 9 & 6 \\ 12 & -15 \\ \end{pmatrix} \)
  2. \( \dfrac{1}{2}Y = \dfrac{1}{2} \begin{pmatrix} 4 \\ 2 \\ \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ \end{pmatrix} \)
  3. \( -5Z = -5 \begin{pmatrix} 3 & 3 & 2 \\ \end{pmatrix} = \begin{pmatrix} -15 & -15 & -10 \\ \end{pmatrix} \)

In the following example we are going to make use of scalar multiplication in solving matrices.

Example 4
Questions

  • Given that \( X = \begin{pmatrix} 3 & -3 \\ 4 & 5 \\ \end{pmatrix} \), \( Y = \begin{pmatrix} x & -6 \\ 8 & 10 \\ \end{pmatrix} \), and \( Z = \begin{pmatrix} 1 & 8 \\ 7 & 6 \\ \end{pmatrix} \), find
    1. \( 3X + 2Z \)
    2. x given that \( X = \dfrac{1}{2}Y \)
    3. the value of a and b given that \( \begin{pmatrix} a & 1 \\ -1 & 2 \\ \end{pmatrix} - \begin{pmatrix} -4 & 5 \\ 1 & 6 \\ \end{pmatrix} = \begin{pmatrix} 6 & -4 \\ -2 & b \\ \end{pmatrix} \)

Answers

  1. Multiply matrix X by 3 and matrix Z by 2 and then proceed with your addition 3X + 2Z.
    \( 3X + 2Z = 3 \begin{pmatrix} 3 & -3 \\ 4 & 5 \\ \end{pmatrix} + 2 \begin{pmatrix} 1 & 8 \\ 7 & 6 \\ \end{pmatrix} \\ = \begin{pmatrix} 9 & -9 \\ 12 & 15 \\ \end{pmatrix} + \begin{pmatrix} 2 & 16 \\ 14 & 12 \\ \end{pmatrix} = \begin{pmatrix} 11 & 7 \\ 26 & 27 \\ \end{pmatrix} \)
  2. \( X = \dfrac{1}{2}Y \). Multiply \( \dfrac{1}{2} \) by Y and proceed with your workings.
    \( \begin{pmatrix} 3 & -3 \\ 4 & 5 \\ \end{pmatrix} = \dfrac{1}{2} \begin{pmatrix} x & 8 \\ -6 & 10 \\ \end{pmatrix} \)

    \( \begin{pmatrix} 3 & -3 \\ 4 & 5 \\ \end{pmatrix} = \begin{pmatrix} \dfrac{x}{2} & -3 \\ 4 & 5 \\ \end{pmatrix} \)   (equate corresponding values)
    \( -3 = \dfrac{x}{2} \)
    \( x = 6 \)
  3. \( \begin{pmatrix} a & 1 \\ -1 & 2 \\ \end{pmatrix} - \begin{pmatrix} -4 & 5 \\ 1 & 6 \\ \end{pmatrix} = \begin{pmatrix} 6 & -4 \\ -2 & b \\ \end{pmatrix} \)

    \( \begin{pmatrix} a - -4 & 1 - 5 \\ -1 - 1 & 2 - 6 \\ \end{pmatrix} = \begin{pmatrix} 6 & -4 \\ -2 & b \\ \end{pmatrix} \)

    \( \begin{pmatrix} a + 4 & -4 \\ -2 & -4 \\ \end{pmatrix} = \begin{pmatrix} 6 & -4 \\ -2 & b \\ \end{pmatrix} \)   (equate corresponding values)

    \( a + 4 = 6 \)
    \( a = 2 \)

    \( b = -4 \)

Example 5
Questions

  • Find a 2 × 2 matrix M such that \( 3M + \begin{pmatrix} -3 & 0 \\ -2 & 5 \\ \end{pmatrix} = \begin{pmatrix} 6 & 3 \\ 4 & 2 \\ \end{pmatrix} \)

Answers

  • We first rearrange the matrices and we get
    \( 3M = \begin{pmatrix} 6 & 3 \\ 4 & 2 \\ \end{pmatrix} - \begin{pmatrix} -3 & 0 \\ -2 & 5 \\ \end{pmatrix} \)
    \( 3M = \begin{pmatrix} 9 & 3 \\ 6 & -3 \\ \end{pmatrix} \),

    since matrix M is multiplied by 3, then we divide the matrix by 3
    \( \dfrac{3M}{3} = \dfrac{1}{3} \begin{pmatrix} 9 & 3 \\ 6 & -3 \\ \end{pmatrix} \)
    \( M = \begin{pmatrix} 3 & 1 \\ 2 & -1 \\ \end{pmatrix} \)

You may attempt the following exercise.

Exercise 6.3: Scalar multiplication and equal matrices
Questions

  1. Solve the following equations, in each case X represents a matrix.
    1. \( 2X + \begin{pmatrix} 2 & 3 \\ 1 & 4 \\ \end{pmatrix} = \begin{pmatrix} 8 & 7 \\ 3 & 10 \\ \end{pmatrix} \)
    2. \( \dfrac{1}{3}X = \begin{pmatrix} 2 & 3 \\ 1 & 0 \\ \end{pmatrix} \)
    3. \( 4X = \begin{pmatrix} 12 & 4 & 8 \\ 0 & 16 & 12 \\ \end{pmatrix} \)
  2. Given that \( \begin{pmatrix} x & 4 & 2 \\ 3 & 0 & 1 \\ \end{pmatrix} + \begin{pmatrix} -4 & 7 & -5 \\ 3 & w & -3 \\ \end{pmatrix} = \begin{pmatrix} 2 & y & -3 \\ 6 & -4 & -2 \\ \end{pmatrix} \), find w, x and y.

Answers

    1. \( X = \begin{pmatrix} 3 & 2 \\ 1 & 3 \\ \end{pmatrix} \)
    2. \( X = \begin{pmatrix} 6 & 9 \\ 3 & 0 \\ \end{pmatrix} \)
    3. \( X = \begin{pmatrix} 3 & 1 & 2 \\ 0 & 4 & 3 \\ \end{pmatrix} \)
  2. \( w = −4, x = 6 y = 11 \)

6.5. Multiplication of matrices

Matrices can be multiplied only if the number of columns in the first matrix is the same as number of rows in the second matrix. Let matrix A be a matrix of order \( p × r \) and let matrix B be a matrix of order \( r × m \), then

  1. \( AB = p × r × r × m \) - possible, since the number of columns in the first matrix is the same as the number of rows in the second matrix, multiplication is possible and the order of the resultant matrix is p × m.
  2. \( BA = r × m × p × r \) - impossible, since the number of columns in the first matrix is not the same as the number of rows in the second matrix, multiplication is impossible.

So, AB ≠ BA. Consider the following examples.

Example 6
Questions

  • Find the product of the following matrices if possible.
    1. \( \begin{pmatrix} 2 & 3 \\ \end{pmatrix} \begin{pmatrix} -1 \\ 4 \\ \end{pmatrix} \)
    2. \( \begin{pmatrix} 4 \\ 2 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 3 \\ \end{pmatrix} \)
    3. \( \begin{pmatrix} 1 & 5 \\ 2 & 8 \\ \end{pmatrix} \begin{pmatrix} -2 \\ 3 \\ \end{pmatrix} \)

Answers

  1. \( \begin{pmatrix} 2 & 3 \\ \end{pmatrix} \begin{pmatrix} -1 \\ 4 \\ \end{pmatrix} \)
    The first matrix is a 1 × 2 matrix and the second matrix is a 2 × 1 matrix. If you want to find the product of the 2 matrices it is possible, 1 × 2 × 2 × 1.
    \( \begin{pmatrix} 2 & 3 \\ \end{pmatrix} \begin{pmatrix} -1 \\ 4 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 2 \times -1 & 3 \times 4 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} -2 + 12 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 10 \\ \end{pmatrix} \)
  2. \( \begin{pmatrix} 4 \\ 2 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 3 \\ \end{pmatrix} \)
    The first matrix is a 2 × 1 matrix and the second matrix is a 2 × 1 matrix. If you want to find the product of the 2 matrices it is impossible, 2 × 1 × 2 × 1.
  3. \( \begin{pmatrix} 1 & 5 \\ 2 & 8 \\ \end{pmatrix} \begin{pmatrix} -2 \\ 3 \\ \end{pmatrix} \)
    The first matrix is a 2 × 2 matrix and the second matrix is a 2 × 1 matrix. If you want to find the product of the 2 matrices it is possible, 2 × 2 × 2 × 1.
    \( \begin{pmatrix} 1 & 5 \\ 2 & 8 \\ \end{pmatrix} \begin{pmatrix} -2 \\ 3 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 1 \times -2 + 5 \times 3 \\ 2 \times -2 + 8 \times 3 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} -2 + 15 \\ -4 + 24 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 13 \\ 20 \\ \end{pmatrix} \)

Example 7
Questions

  • Given that \( A = \begin{pmatrix} 1 & 3 \\ -1 & 0 \\ \end{pmatrix} \) and \( B = \begin{pmatrix} 4 & 9 \\ -2 & 5 \\ \end{pmatrix} \), find
    1. \( AB \)
    2. \( BA \)
    3. \( B^2 \)

Answers

  1. \( AB = \begin{pmatrix} 1 & 3 \\ -1 & 0 \\ \end{pmatrix} \begin{pmatrix} 4 & 9 \\ -2 & 5 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 1 \times 4 + 3 \times −2 & 1 \times 9 + 3 \times 5 \\ −1 \times 4 + 0 \times −2 & −1 \times 9 + 0 \times 5 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 4 − 6 & 9 + 15 \\ −4 + 0 & −9 + 0 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} -2 & 24 \\ −4 & −9 \\ \end{pmatrix} \)
  2. \( BA = \begin{pmatrix} 4 & 9 \\ -2 & 5 \\ \end{pmatrix} \begin{pmatrix} 1 & 3 \\ -1 & 0 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 4 \times 1 + 9 \times −1 & 4 \times 3 + 9 \times 0 \\ −2 \times 1 + 5 \times −1 & −2 \times 3 + 5 \times 0 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 4 − 9 & 12 + 0 \\ −2 − 5 & −6 + 0 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} -5 & 12 \\ −7 & −6 \\ \end{pmatrix} \)
  3. \( B^2 = \begin{pmatrix} 4 & 9 \\ -2 & 5 \\ \end{pmatrix} \begin{pmatrix} 4 & 9 \\ -2 & 5 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 4 \times 4 + 9 \times −2 & 4 \times 9 + 9 \times 5 \\ −2 \times 4 + 5 \times −2 & −2 \times 9 + 5 \times 5 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} 16 − 18 & 36 + 45 \\ −8 − 10 & −18 + 25 \\ \end{pmatrix} \)
    \(= \begin{pmatrix} -2 & 81 \\ −18 & 7 \\ \end{pmatrix} \)

You may attempt the following exercise.

Exercise 6.4: Multiplication of matrices
Questions

  1. Find matrix M, such that \( M + \begin{pmatrix} 2 & -1 \\ 0 & 3 \\ \end{pmatrix} = \begin{pmatrix} 5 & 1 \\ 1 & 7 \\ \end{pmatrix} \)
  2. Given that \( A = \begin{pmatrix} 4 & 3 \\ 2 & 1 \\ \end{pmatrix} \), \(B = \begin{pmatrix} 0 & -2 \\ 1 & 6 \\ \end{pmatrix} \), find
    1. \( AB \)
    2. \( BA \)
    3. \( A^2 \)
    4. \( B^2 \)
  3. Find the value of p, q and r.
    \( \begin{pmatrix} 5 & -1 \\ p & 3 \\ \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -1 & 2 \\ \end{pmatrix} = \begin{pmatrix} q & r \\ 1 & 8 \\ \end{pmatrix} \)

Answers

  1. \( M = \begin{pmatrix} 3 & 2 \\ 1 & 4 \\ \end{pmatrix} \)
    1. \( AB = \begin{pmatrix} 3 & 10 \\ 1 & 2 \\ \end{pmatrix} \)
    2. \( BA = \begin{pmatrix} -4 & -2 \\ 16 & 9 \\ \end{pmatrix} \)
    3. \( A^2 = \begin{pmatrix} 22 & 15 \\ 10 & 7 \\ \end{pmatrix} \)
    4. \( B^2 = \begin{pmatrix} -2 & -12 \\ 6 & 34 \\ \end{pmatrix} \)
  3. \( p = 2 \), \( q = 11 \) and \( r = 3 \)

6.6. Determinant of 2 × 2 matrices

Given a 2 × 2 matrix \( M = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \), the determinant (det) is \( M = ad − bc \).

Note that the determinant can be denoted by |M|.

Example 8
Questions

  • Given that \( M = \begin{pmatrix} 2 & -4 \\ 3 & 6 \\ \end{pmatrix} \), find the determinant of M.

Answers

  • By using the formula M = ad − bc we substitute the values in the formula.
    a = 2, b = -4, c = 3 and d = 6.
    M = 2 × 6 – −4 × 3
    M = 12 − −12
    M = 24

Example 9
Questions

  1. Given that \( \begin{pmatrix} 1 & -4 \\ -8 & 4x \\ \end{pmatrix} \), has a determinant of 12, find the value of x.
  2. Given that \( A = \begin{pmatrix} 4 & -2 \\ y & y \\ \end{pmatrix} \), and |A| = −6, find y.

Answers

  1. Determinant = ad − bc
    12 = 1 × 4x − −4 × −8
    12 = 4x − 32
    44 = 4x
    x = 11
  2. |A|= ad − bc
    −6 = 4y + 2y
    −6 = 6y
    y = −1

You may attempt the following exercise.

Exercise 6.5: Determinant of 2 × 2 matrices
Questions

  1. Find the determinant of the following matrices.
    1. \( \begin{pmatrix} 3 & -1 \\ 5 & 4 \\ \end{pmatrix} \)
    2. \( \begin{pmatrix} -2 & -8 \\ 4 & 6 \\ \end{pmatrix} \)
  2. The determinant of a matrix \( \begin{pmatrix} a - 1 & 9 \\ 6 & 3 \\ \end{pmatrix} \) is 9. Find the value of a.
  3. The matrix \( \begin{pmatrix} x + 1 & 3 \\ x & 2 \\ \end{pmatrix} \) has a determinant of −2. Find the value of x.

Answers

    1. 17
    2. 20
  2. a = 19
  3. x = 4

6.7. Inverse of 2 × 2 matrices<

Given a 2 × 2 matrix \( M = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \), the determinant is \( M = ad − bc \).
The inverse is \( M^{-1} = \dfrac{1}{detM} \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \).
The inverse of matrix M may be written as \( M^{-1} \).

Note that elements a and d have exchanged positions while b and c are multiplied by −1.

Example 10
Questions

  • If \( M = \begin{pmatrix} 2 & -6 \\ -1 & 4 \\ \end{pmatrix} \)
    1. Find the value of the determinant of M.
    2. Hence write down the inverse of M.

Answers

  1. The determinant of M = ad − bc
    = 2 × 4 − −6 × −1 = 8 - 6 = 2
  2. The inverse of M, \( M^{-1} = \dfrac{1}{detM} \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \)
    \( M = \begin{pmatrix} 2 & -6 \\ -1 & 4 \\ \end{pmatrix} \)
    \( M^{-1} = \dfrac{1}{2} \begin{pmatrix} 4 & 6 \\ 1 & 2 \\ \end{pmatrix} \)
    2 and 4 have exchanged positions while -6 and −2 have been multiplied by −1.

Example 11
Questions

  • Given that the matrix \( A = \begin{pmatrix} 3 & 0 \\ 2 & -4 \\ \end{pmatrix} \), find its inverse.

Answers

  • The determinant of M = ad − bc
    = 3 × −4 − 2 × 0
    = -12 - 0
    = -12

    \( A^{-1} = \dfrac{1}{detA} \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \)
    \( A^{-1} = -\dfrac{1}{12} \begin{pmatrix} -4 & 0 \\ -2 & 3 \\ \end{pmatrix} \)

Example 12
Questions

  • Given that the matrix \( N = \begin{pmatrix} -1 & 1 \\ 0 & 3 \\ \end{pmatrix} \), find \( N^{-1} \).

Answers

  • |N| = ad − bc
    = -1 × 3 − 1 × 0
    = -3 - 0
    = -3

    \( N^{-1} = \dfrac{1}{detN} \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \)
    \( N^{-1} = -\dfrac{1}{3} \begin{pmatrix} 3 & -1 \\ 0 & -1 \\ \end{pmatrix} \)

You may attempt the following exercise.

Exercise 6.6: Inverse of 2 × 2 matrices
Questions

  • Find the inverse of the following matrices.
    1. \( \begin{pmatrix} 2 & 3 \\ 0 & 1 \\ \end{pmatrix} \)
    2. \( \begin{pmatrix} 1 & 9 \\ -1 & 0 \\ \end{pmatrix} \)
    3. \( \begin{pmatrix} 1 & -4 \\ 3 & 2 \\ \end{pmatrix} \)
    4. \( \begin{pmatrix} -1 & 4 \\ 1 & 2 \\ \end{pmatrix} \)
    5. \( \begin{pmatrix} 1 & -6 \\ -4 & 2 \\ \end{pmatrix} \)

Answers

  1. \( \dfrac{1}{2} \begin{pmatrix} 1 & -3 \\ 0 & 2 \\ \end{pmatrix} \)
  2. \( \dfrac{1}{9} \begin{pmatrix} 0 & -9 \\ 1 & 1 \\ \end{pmatrix} \)
  3. \( \dfrac{1}{14} \begin{pmatrix} 2 & 4 \\ -3 & 1 \\ \end{pmatrix} \)
  4. \( -\dfrac{1}{6} \begin{pmatrix} 2 & -4 \\ -1 & -1 \\ \end{pmatrix} \)
  5. \( -\dfrac{1}{22} \begin{pmatrix} 2 & 6 \\ 4 & 1 \\ \end{pmatrix} \)

6.8. Identity matrices

An identity matrix is denoted by I. An identity matrix, \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \).

Note: If you pre-multiply or post-multiply any 2 × 2 matrix by an identity matrix it remains the same. If you multiply a matrix by its inverse you get an identity matrix, \( MM^{-1} = I \).

Example 13
Questions

  • Given that \( Z \times \begin{pmatrix} 6 & 0 \\ 0 & 6 \\ \end{pmatrix} - Z = \begin{pmatrix} -5 & 0 \\ 5 & 10 \\ \end{pmatrix} \), find Z.

Answers

  • First make matrix \( \begin{pmatrix} 6 & 0 \\ 0 & 6 \\ \end{pmatrix} \) an identity matrix by factoring out 6.
    \( Z \times 6 \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} - Z = \begin{pmatrix} -5 & 0 \\ 5 & 10 \\ \end{pmatrix} \)   (rearrange the matrices)
    \( Z \times \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} - Z = \begin{pmatrix} -5 & 0 \\ 5 & 10 \\ \end{pmatrix} \)
    \( 6Z \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} - Z = \begin{pmatrix} -5 & 0 \\ 5 & 10 \\ \end{pmatrix} \)   (since 6Z is multiplying an identity matrix it remains the same)
    \( 6Z - Z = \begin{pmatrix} -5 & 0 \\ 5 & 10 \\ \end{pmatrix} \)
    \( 5Z = \begin{pmatrix} -5 & 0 \\ 5 & 10 \\ \end{pmatrix} \)
    \( Z = \begin{pmatrix} -1 & 0 \\ 1 & 2 \\ \end{pmatrix} \)

    • Example 14
    Questions

    • Find x and y such that \( \begin{pmatrix} 3 & 7 \\ x & y \\ \end{pmatrix} \begin{pmatrix} y & -7 \\ -2 & 3 \\ \end{pmatrix} = I \), where I is an identity matrix.

    Answers

    • \( \begin{pmatrix} 3 & 7 \\ x & y \\ \end{pmatrix} \begin{pmatrix} y & -7 \\ -2 & 3 \\ \end{pmatrix} = I \)
      \( \begin{pmatrix} 3 & 7 \\ x & y \\ \end{pmatrix} \begin{pmatrix} y & -7 \\ -2 & 3 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \)
      \( \begin{pmatrix} 3 \times y + 7 \times −2 & 3 \times −7 + 7 \times 3 \\ x \times y + y \times −2 & x \times −7 + y \times 3 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \)
      \( \begin{pmatrix} 3y − 14 & 0 \\ xy − 2y & −7x + 3y \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \)   (equate corresponding values)

      3y − 14 = 1
      3y = 1 + 14
      3y = 15
      y = 5

      −7x + 3y = 1
      −7x + 3 × 5 = 1
      −7x + 15 = 1
      −7x = 1 - 15
      −7x = −14
      x = 2

    You may attempt the following exercise.

    Exercise 6.7: Identity matrices
    Questions

    1. If M is a 2 × 2 matrix such that \( M \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix} 5 & 3 \\ 1 & -7 \\ \end{pmatrix} \), find matrix M.
    2. Given that \( H = \begin{pmatrix} 4 & 2 \\ 0 & 3 \\ \end{pmatrix} \) and \( J = \begin{pmatrix} 0.25 & m \\ 0 & \dfrac{1}{3} \\ \end{pmatrix} \), find the value of m which makes HJ an identity matrix.
    3. Find a 2 × 2 matrix Y such that \( \begin{pmatrix} \dfrac{1}{2} & 0 \\ 0 & \dfrac{1}{2} \\ \end{pmatrix}Y - \begin{pmatrix} 1 & 0 \\ 2 & 1 \\ \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ -3 & 1 \\ \end{pmatrix} \), find matrix M.
    4. Given that R is a 2 × 2 matrix such that \( R \begin{pmatrix} 2 & 0 \\ 0 & 2 \\ \end{pmatrix} + R = \begin{pmatrix} 3 & 6 \\ 9 & 12 \\ \end{pmatrix} \), find R.

    Answers

    1. \( M = \begin{pmatrix} 5 & 3 \\ 1 & -7 \\ \end{pmatrix} \)
    2. \( m = -\dfrac{1}{6} \)
    3. \( Y = \begin{pmatrix} 8 & 8 \\ -2 & 4 \\ \end{pmatrix} \)
    4. \( R = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix} \)

    6.9. Singular and non-singular matrices

    Consider the following matrix \( \begin{pmatrix} 3 & -2 \\ 6 & 4 \\ \end{pmatrix} \) . What is its determinant?
    By calculation you will get the determinant of this matrix being zero.

    If you proceed to find the inverse of this matrix you will get \( \dfrac{1}{0} \begin{pmatrix} 4 & 2 \\ 6 & 3 \\ \end{pmatrix} \) , but \( \dfrac{1}{0} \) does not exist so the inverse does not exist. Remember that any matrix which has no inverse and having a determinant of 0 is known as a singular matrix.

    Example 15
    Questions

    • Given that \( \begin{pmatrix} w & 6 \\ -6 & 12 \\ \end{pmatrix} \) has a determinant of zero, find w.

    Answers

    • \( \begin{pmatrix} w & 6 \\ -6 & 12 \\ \end{pmatrix} \)
      First find the determinant of the matrix and equate it to zero.
      Determinant = ad - bc
      12 × w − −6 × 6 = 0
      12w − −36 = 0
      12w + 36 = 0
      12w = -36
      w = -3

    Example 16
    Questions

    • Given that \( A = \begin{pmatrix} 2x + 1 & 3 \\ 1 & x \\ \end{pmatrix} \), find
      1. the determinant of A in terms of x.
      2. the values of x given that A is singular.

    Answers

    1. \( A = \begin{pmatrix} 2x + 1 & 3 \\ 1 & x \\ \end{pmatrix} \)
      Determinant = ad - bc
      \( = x \times (2x + 1) − 3 \times 1 \)
      \( = 2x^2 + x - 3 \)
    2. \( 2x^2 + x - 3 = 0 \), since the determinant of a singular matrix is 0.
      \( 2x^2 - 2x + 3x - 3 = 0 \)
      \( 2x(x − 1) + 3(x − 1) = 0 \)
      \( (x − 1)(2x + 3) = 0 \)
      \( x = 1 \) or \( x = -1\dfrac{1}{2} \)

    Example 17
    Questions

    • If \( A = \begin{pmatrix} -2 & p \\ p + 3 & -4p \\ \end{pmatrix} \) has no inverse, find two possible values of p.

    Answers

    • Determinant = ad - bc = 0
      \( −2 \times −4p − p \times (p + 3) = 0 \)
      \( 8p − p^2 - 3p = 0 \)
      \( −p^2 + 5p = 0 \)
      \( p^2 - 5p = 0 \)
      \( p(p - 5) = 0 \)
      \( p = 0 \) or \( p = 5 \)

    You may attempt the following exercise.

    Exercise 6.8: Singular and non-singular matrices
    Questions

    1. State why matrix \( \begin{pmatrix} 12 & -6 \\ -4 & 2 \\ \end{pmatrix} \) has no inverse.
    2. Find 2 possible values of m, that makes \( \begin{pmatrix} m + 3 & 4m - 3 \\ m & 2m + 2 \\ \end{pmatrix} \) a singular matrix.
    3. Given that \( A = \begin{pmatrix} 7 & 5x - 2 \\ 2 & 2x + 4 \\ \end{pmatrix} \) , find the value of x such that matrix A has no inverse.
    4. Find 2 possible values of x such that \( \begin{pmatrix} x & 3x - 1 \\ 2 & 3x - 1 \\ \end{pmatrix} \) has a determinant of zero.

    Answers

    1. It has no inverse because it has a determinant of zero.
    2. \( m = 6 \) or \( m = −\dfrac{1}{2} \)
    3. \( x = −8 \)
    4. \( x = \dfrac{1}{3} \) or \( x = 2 \)

    6.10. Simultaneous linear equations in two variables

    Simultaneous equations can be solved using the matrix method rather than elimination, graphical or substitution methods.

    Example 18
    Questions

    • Find the value of x and y.
      \( \begin{pmatrix} 4 & 1 \\ 3 & -1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} 9 \\ 5 \\ \end{pmatrix} \)

    Answers

    • \( \begin{pmatrix} 4 & 1 \\ 3 & -1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} 9 \\ 5 \\ \end{pmatrix} \)
      In order for us to proceed with the calculation we need to find the inverse of the matrix \( \begin{pmatrix} 4 & 1 \\ 3 & -1 \\ \end{pmatrix} \) . The inverse of the matrix is \( -\dfrac{1}{7} \begin{pmatrix} -1 & -1 \\ -3 & 4 \\ \end{pmatrix} \) . We then pre-multiply the matrix equation with the inverse.
      \( -\dfrac{1}{7} \begin{pmatrix} -1 & -1 \\ -3 & 4 \\ \end{pmatrix} \begin{pmatrix} 4 & 1 \\ 3 & -1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} = -\dfrac{1}{7} \begin{pmatrix} -1 & -1 \\ -3 & 4 \\ \end{pmatrix} \begin{pmatrix} 9 \\ 5 \\ \end{pmatrix} \)
      Remember that if you multiply a matrix by its inverse you get an identity matrix so, \( \begin{pmatrix} x \\ y \\ \end{pmatrix} = -\dfrac{1}{7} \begin{pmatrix} -1 & -1 \\ -3 & 4 \\ \end{pmatrix} \begin{pmatrix} 9 \\ 5 \\ \end{pmatrix} \)
      \( \begin{pmatrix} x \\ y \\ \end{pmatrix} = -\dfrac{1}{7} \begin{pmatrix} -1 \times 9 + -1 \times 5 \\ -3 \times 9 + 4 \times 5 \\ \end{pmatrix} \)
      \( \begin{pmatrix} x \\ y \\ \end{pmatrix} = -\dfrac{1}{7} \begin{pmatrix} -14 \\ -7 \\ \end{pmatrix} \)
      \( \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} -2 \\ -1 \\ \end{pmatrix} \)
      \( x = 2 \) and \( y = 1 \)

    Example 19
    Questions

    • Solve the following equations using the matrix method, 2x + 3y = 6 and 6x + 4y = 5.

    Answers

    • 2x + 3y = 6 and 6x + 4y = 5
      First arrange the equations in matrices.
      \( \begin{pmatrix} 2 & 3 \\ 6 & 4 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \\ \end{pmatrix} \)
      Solve the equation by finding the inverse of the matrix on the left.
      \( -\dfrac{1}{10} \begin{pmatrix} 4 & -3 \\ -6 & 2 \\ \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 6 & 4 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} = -\dfrac{1}{10} \begin{pmatrix} 4 & -3 \\ -6 & 2 \\ \end{pmatrix} \begin{pmatrix} 6 \\ 5 \\ \end{pmatrix} \)
      \( \begin{pmatrix} x \\ y \\ \end{pmatrix} = -\dfrac{1}{10} \begin{pmatrix} 4 \times 6 + -3 \times 5 \\ -6 \times 6 + 2 \times 5 \\ \end{pmatrix} \)
      \( \begin{pmatrix} x \\ y \\ \end{pmatrix} = -\dfrac{1}{10} \begin{pmatrix} 9 \\ -26 \\ \end{pmatrix} \)
      \( \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} -0.9 \\ 2.6 \\ \end{pmatrix} \)
      \( x = −0.9 \) and \( y = 2.6 \)

    You may attempt the following exercise.

    Exercise 6.9: Application of matrices
    Questions

    1. Solve the following equations using the matrix method, 2x + 3y = 11 and 3x − 5y = −12.
    2. Solve the following equations using the matrix method, 8x + 15y = 11 and 4x − y = −3.

    Answers

    1. \( x = 1 \) and \( y = 3 \)
    2. \( x = -\dfrac{1}{2} \) and \( y = 1 \)

    6.11. Summary

    The chapter showed that matrices are used as a way of storing information. We have seen that the order of a matrix is given by counting the number of rows followed by the number of columns of a given matrix. The chapter also looked at the determinant of a 2 x 2 matrix which is given by ad − bc, inverse of a 2 × 2 matrix which is given by \( \dfrac{1}{det} \begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \) , identity matrix which is denoted by I and represented as \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \) and that singular matrix has no inverse. We also looked at the application of matrices in solving linear simultaneous equations.

    6.12. Further reading

    1. Macrae, M. F., Madungwe, L. and Mutangadura, A. (2017). New General Mathematics Book 3. Pearson Capetown.
    2. Macrae, M. F., Madungwe, L. and Mutangadura, A. (2017). New General Mathematics Book 4. Pearson Capetown.
    3. Meyers, C., Graham, B., Dawe, L. (2004). Mathscape Working Mathematically. 9th Edition. MacMillan.
    4. Pimentel, R. and Wall, T. (2011). International Mathematics. Hodder UK.
    5. Rayner, D. (2005). Extended Mathematics. Oxford New York.

    6.13. Test 6

    Questions

    1. Given that \( P = \begin{pmatrix} 3 & 5 \\ 4 & x \\ \end{pmatrix} \) and \( Q = \begin{pmatrix} -2 \\ 3 \\ \end{pmatrix} \)
      1. Find PQ in terms of x.
      2. Find the value of x that makes |P| = 7.
      3. Hence write down \( P^{-1} \).
    2. Given that \( A = \begin{pmatrix} 3 & 2 \\ 4 & 3 \\ \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & 3 \\ 2 & 0 \\ \end{pmatrix} \) , find
      1. A − 2B
      2. AB
      3. A 2 × 2 matrix X such that AX = I, where I is an identity matrix.
    3. Given that \( R = \begin{pmatrix} -1 & x \\ 3 & 2x \\ \end{pmatrix} \) and \( S = \begin{pmatrix} 5 & -4 \\ -2 & 3 \\ \end{pmatrix} \) , find
      1. The value of x if the determinant of R is 20.
      2. The inverse of S.