Step-by-Step Solutions to Exercise 8e Matrix Products | NEW General Mathematics for Senior Secondary School 3 (SS3)

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Step-by-Step Solutions to Exercise 8e Matrix Products | NEW General Mathematics for Senior Secondary School 3 (SS3)

Introduction:

Are you struggling with matrix multiplication or trying to find clear solutions for Exercise 8e from the NEW General Mathematics for Senior Secondary School 3 (SS3)? Look no further! In this detailed guide, we provide step-by-step solutions to each question from Exercise 8e to help you master matrix products with ease. This article is designed for students looking for simple and clear explanations to excel in their exams.

Let’s break down the solutions to all the matrix product questions in Exercise 8e.

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Question 1:

Multiply the following matrices: \[ \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} \times \begin{pmatrix} 2 \\ 3 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row of the first matrix by the column matrix: \[ (3 \times 2) + (2 \times 3) = 6 + 6 = 12 \]
2. Multiply the second row of the first matrix by the column matrix: \[ (1 \times 2) + (4 \times 3) = 2 + 12 = 14 \] So, the result is: \[ \begin{pmatrix} 12 \\ 14 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 2:

Multiply the matrices: \[ \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \times \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the first column: \[ (4 \times 3) + (1 \times 2) = 12 + 2 = 14 \]
2. Multiply the first row by the second column: \[ (4 \times 1) + (1 \times 1) = 4 + 1 = 5 \] 3. Multiply the second row by the first column: \[ (3 \times 3) + (2 \times 2) = 9 + 4 = 13 \] 4. Multiply the second row by the second column: \[ (3 \times 1) + (2 \times 1) = 3 + 2 = 5 \] So, the result is: \[ \begin{pmatrix} 14 & 5 \\ 13 & 5 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 3:

Multiply the matrices: \[ \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} \times \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the first column: \[ (3 \times 4) + (1 \times 3) = 12 + 3 = 15 \]
2. Multiply the first row by the second column: \[ (3 \times 1) + (1 \times 2) = 3 + 2 = 5 \] 3. Multiply the second row by the first column: \[ (2 \times 4) + (1 \times 3) = 8 + 3 = 11 \] 4. Multiply the second row by the second column: \[ (2 \times 1) + (1 \times 2) = 2 + 2 = 4 \] So, the result is: \[ \begin{pmatrix} 15 & 5 \\ 11 & 4 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 4:

Multiply the matrices: \[ \begin{pmatrix} 11 & 2 \\ 5 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & -2 \\ -5 & 11 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the first column: \[ (11 \times 1) + (2 \times -5) = 11 – 10 = 1 \]
2. Multiply the first row by the second column: \[ (11 \times -2) + (2 \times 11) = -22 + 22 = 0 \] 3. Multiply the second row by the first column: \[ (5 \times 1) + (1 \times -5) = 5 – 5 = 0 \] 4. Multiply the second row by the second column: \[ (5 \times -2) + (1 \times 11) = -10 + 11 = 1 \] So, the result is: \[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 5:

Multiply the matrices: \[ \begin{pmatrix} 2 & -1 \\ -3 & 6 \end{pmatrix} \times \begin{pmatrix} 3 & 1 \\ 3 & 2 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the first column: \[ (2 \times 3) + (-1 \times 3) = 6 – 3 = 3 \]
2. Multiply the first row by the second column: \[ (2 \times 1) + (-1 \times 2) = 2 – 2 = 0 \] 3. Multiply the second row by the first column: \[ (-3 \times 3) + (6 \times 3) = -9 + 18 = 9 \] 4. Multiply the second row by the second column: \[ (-3 \times 1) + (6 \times 2) = -3 + 12 = 9 \] So, the result is: \[ \begin{pmatrix} 3 & 0 \\ 9 & 9 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 6:

Multiply the matrices: \[ \begin{pmatrix} 2 & 3 & 1 \\ 4 & 2 & 1 \\ 3 & 1 & 1 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the column matrix: \[ (2 \times 2) + (3 \times 1) + (1 \times 3) = 4 + 3 + 3 = 10 \]
2. Multiply the second row by the column matrix: \[ (4 \times 2) + (2 \times 1) + (1 \times 3) = 8 + 2 + 3 = 13 \] 3. Multiply the third row by the column matrix: \[ (3 \times 2) + (1 \times 1) + (1 \times 3) = 6 + 1 + 3 = 10 \] So, the result is: \[ \begin{pmatrix} 10 \\ 13 \\ 10 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 7:

Multiply the matrices: \[ \begin{pmatrix} 2 & 2 \\ 0 & -4 \end{pmatrix} \times \begin{pmatrix} 2 & 2 \\ 0 & -4 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the first column: \[ (2 \times 2) + (2 \times 0) = 4 + 0 = 4 \]
2. Multiply the first row by the second column: \[ (2 \times 2) + (2 \times -4) = 4 – 8 = -4 \] 3. Multiply the second row by the first column: \[ (0 \times 2) + (-4 \times 0) = 0 + 0 = 0 \] 4. Multiply the second row by the second column: \[ (0 \times 2) + (-4 \times -4) = 0 + 16 = 16 \] So, the result is: \[ \begin{pmatrix} 4 & -4 \\ 0 & 16 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 8:

Multiply the matrices: \[ \begin{pmatrix} 1 & 2 & -1 \\ 3 & 1 & -2 \end{pmatrix} \times \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the column matrix: \[ (1 \times 2) + (2 \times -2) + (-1 \times 1) = 2 – 4 – 1 = -3 \]
2. Multiply the second row by the column matrix: \[ (3 \times 2) + (1 \times -2) + (-2 \times 1) = 6 – 2 – 2 = 2 \] So, the result for Question 8 is: \[ \begin{pmatrix} -3 \\ 2 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 9:

Multiply the matrices: \[ \begin{pmatrix} 0 & 1 & 2 \\ -2 & 1 & 3 \end{pmatrix} \times \begin{pmatrix} 1 & 2 & -1 \\ 3 & 1 & -2 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the first column: \[ (0 \times 1) + (1 \times 3) + (2 \times 1) = 0 + 3 + 2 = 5 \]
2. Multiply the first row by the second column: \[ (0 \times 2) + (1 \times 1) + (2 \times -2) = 0 + 1 – 4 = -3 \] 3. Multiply the first row by the third column: \[ (0 \times -1) + (1 \times -2) + (2 \times -1) = 0 – 2 – 2 = -4 \] 4. Multiply the second row by the first column: \[ (-2 \times 1) + (1 \times 3) + (3 \times 1) = -2 + 3 + 3 = 4 \] 5. Multiply the second row by the second column: \[ (-2 \times 2) + (1 \times 1) + (3 \times -2) = -4 + 1 – 6 = -9 \] 6. Multiply the second row by the third column: \[ (-2 \times -1) + (1 \times -2) + (3 \times -1) = 2 – 2 – 3 = -3 \] So, the result is: \[ \begin{pmatrix} 5 & -3 & -4 \\ 4 & -9 & -3 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 10:

Multiply the matrices: \[ \begin{pmatrix} 3 & 5 \\ -2 & 4 \end{pmatrix} \times \begin{pmatrix} -1 & 0 \\ 4 & 6 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the first column: \[ (3 \times -1) + (5 \times 4) = -3 + 20 = 17 \]
2. Multiply the first row by the second column: \[ (3 \times 0) + (5 \times 6) = 0 + 30 = 30 \] 3. Multiply the second row by the first column: \[ (-2 \times -1) + (4 \times 4) = 2 + 16 = 18 \] 4. Multiply the second row by the second column: \[ (-2 \times 0) + (4 \times 6) = 0 + 24 = 24 \] So, the result is: \[ \begin{pmatrix} 17 & 30 \\ 18 & 24 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 11:

Find the square of the matrix: \[ \begin{pmatrix} -2 & 1 \\ -3 & 4 \end{pmatrix}^2 \] Step-by-step solution:
To square the matrix, multiply the matrix by itself: \[ \begin{pmatrix} -2 & 1 \\ -3 & 4 \end{pmatrix} \times \begin{pmatrix} -2 & 1 \\ -3 & 4 \end{pmatrix} \]

1. Multiply the first row by the first column: \[ (-2 \times -2) + (1 \times -3) = 4 – 3 = 1 \]
2. Multiply the first row by the second column: \[ (-2 \times 1) + (1 \times 4) = -2 + 4 = 2 \] 3. Multiply the second row by the first column: \[ (-3 \times -2) + (4 \times -3) = 6 – 12 = -6 \] 4. Multiply the second row by the second column: \[ (-3 \times 1) + (4 \times 4) = -3 + 16 = 13 \] So, the result is: \[ \begin{pmatrix} 1 & 2 \\ -6 & 13 \end{pmatrix} \]

Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Question 12:

You are given matrices \( M \) and \( N \): \[ M = \begin{pmatrix} 6 & 0 \\ -1 & 2 \end{pmatrix}, \quad N = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} \] Step-by-step solutions:
(a) Calculate MN
Multiply \( M \times N \): \[ \begin{pmatrix} 6 & 0 \\ -1 & 2 \end{pmatrix} \times \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} \] 1. Multiply the first row by the first column: \[ (6 \times 1) + (0 \times 3) = 6 + 0 = 6 \] 2. Multiply the first row by the second column: \[ (6 \times -2) + (0 \times 5) = -12 + 0 = -12 \] 3. Multiply the second row by the first column: \[ (-1 \times 1) + (2 \times 3) = -1 + 6 = 5 \] 4. Multiply the second row by the second column: \[ (-1 \times -2) + (2 \times 5) = 2 + 10 = 12 \] So, the result of \( MN \) is: \[ \begin{pmatrix} 6 & -12 \\ 5 & 12 \end{pmatrix} \]
(b) Calculate NM
Multiply \( N \times M \): \[ \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} \times \begin{pmatrix} 6 & 0 \\ -1 & 2 \end{pmatrix} \] 1. Multiply the first row by the first column: \[ (1 \times 6) + (-2 \times -1) = 6 + 2 = 8 \] 2. Multiply the first row by the second column: \[ (1 \times 0) + (-2 \times 2) = 0 – 4 = -4 \] 3. Multiply the second row by the first column: \[ (3 \times 6) + (5 \times -1) = 18 – 5 = 13 \] 4. Multiply the second row by the second column: \[ (3 \times 0) + (5 \times 2) = 0 + 10 = 10 \] So, the result of \( NM \) is: \[ \begin{pmatrix} 8 & -4 \\ 13 & 10 \end{pmatrix} \]
(c) Calculate \( M^2 \)
Multiply \( M \times M \): \[ \begin{pmatrix} 6 & 0 \\ -1 & 2 \end{pmatrix} \times \begin{pmatrix} 6 & 0 \\ -1 & 2 \end{pmatrix} \] 1. Multiply the first row by the first column: \[ (6 \times 6) + (0 \times -1) = 36 + 0 = 36 \] 2. Multiply the first row by the second column: \[ (6 \times 0) + (0 \times 2) = 0 + 0 = 0 \] 3. Multiply the second row by the first column: \[ (-1 \times 6) + (2 \times -1) = -6 – 2 = -8 \] 4. Multiply the second row by the second column: \[ (-1 \times 0) + (2 \times 2) = 0 + 4 = 4 \] So, the result of \( M^2 \) is: \[ \begin{pmatrix} 36 & 0 \\ -8 & 4 \end{pmatrix} \]
Question 12(d)
We need to calculate \( N^2 \), where: \[ N = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} \] To find \( N^2 \), multiply \( N \times N \): \[ \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} \times \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix} \] Step-by-step solution:
1. Multiply the first row by the first column: \[ (1 \times 1) + (-2 \times 3) = 1 – 6 = -5 \]
2. Multiply the first row by the second column: \[ (1 \times -2) + (-2 \times 5) = -2 – 10 = -12 \] 3. Multiply the second row by the first column: \[ (3 \times 1) + (5 \times 3) = 3 + 15 = 18 \] 4. Multiply the second row by the second column: \[ (3 \times -2) + (5 \times 5) = -6 + 25 = 19 \] So, the result of \( N^2 \) is: \[ \begin{pmatrix} -5 & -12 \\ 18 & 19 \end{pmatrix} \]

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Solutions to Exercise 8e Matrix NEW General Mathematics SS3

Conclusion:

This complete solution guide for Exercise 8e in NEW General Mathematics for Senior Secondary School 3 (SS3) is your key to understanding matrix multiplication step-by-step. With these detailed explanations, you can easily follow along and excel in your studies. Be sure to bookmark this page and share it with fellow students who might find these solutions helpful.

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