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Sec 1 Math Weekly Assessment Week 4 Answers 2026 - Math Egypt

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Mathematics Assessments.

Week 4: Weekly Assessments.

Sec 1.

This page provides detailed answers and comprehensive explanations for the Sec 1 Mathematics Weekly Assessment – Week 4. Students will master matrix multiplication rules, find the order of resulting matrices, solve trigonometric equations in the first quadrant, and determine conditions for parallel and perpendicular vectors.

Sec 1 Math Weekly Assessment Week 4 Answers 2026 | First Secondary Math Egypt

Updated: March 2026

First Group

(1) If matrix A of order $ 2 \times 2 $ and matrix B of order $ 2 \times 1 $, find the order of matrix AB.

Answer: $ 2 \times 1 $

(2) If $ A $ $ = $ $ \begin{pmatrix} 3 \\ 5 \end{pmatrix} $, $ B $ $ = $ $ \begin{pmatrix} 1 & 3 \end{pmatrix} $, find AB.

Answer: $ \begin{pmatrix} 3 & 9 \\ 5 & 15 \end{pmatrix} $

(3) Solve the equation $ \sin \theta $ $ = $ $ \frac{\sqrt{2}}{2} $, $ \theta \in [0, \frac{\pi}{2}[ $.

Answer: $ \theta $ $ = $ $ \frac{\pi}{4} $

(4) If $ \vec{A} $ $ = $ $ (1, 3) $, $ \vec{B} $ $ = $ $ (m, 12) $, find m when $ \vec{A} \parallel \vec{B} $.

Answer: $ m $ $ = $ $ 4 $

(5) If $ \vec{A} $ $ = $ $ (3, 1) $, $ \vec{B} $ $ = $ $ (6, m) $, find m when $ \vec{A} \perp \vec{B} $.

Answer: $ m $ $ = $ $ -18 $

Second Group

(1) If matrix A of order $ 3 \times 3 $ and matrix B of order $ 3 \times 1 $, find the order of matrix AB.

Answer: $ 3 \times 1 $

(2) If $ A $ $ = $ $ \begin{pmatrix} 1 \\ 2 \end{pmatrix} $, $ B $ $ = $ $ \begin{pmatrix} 5 & 4 \end{pmatrix} $, find AB.

Answer: $ \begin{pmatrix} 5 & 4 \\ 10 & 8 \end{pmatrix} $

(3) Solve the equation $ \sin \theta $ $ = $ $ \frac{\sqrt{3}}{2} $, $ \theta \in [0, \frac{\pi}{2}[ $.

Answer: $ \theta $ $ = $ $ \frac{\pi}{3} $

(4) If $ \vec{A} $ $ = $ $ (m, 1) $, $ \vec{B} $ $ = $ $ (16, 4) $, find m when $ \vec{A} \parallel \vec{B} $.

Answer: $ m $ $ = $ $ 4 $

(5) If $ \vec{A} $ $ = $ $ (m, 6) $, $ \vec{B} $ $ = $ $ (6, 3) $, find m when $ \vec{A} \perp \vec{B} $.

Answer: $ m $ $ = $ $ -3 $

Third Group

(1) If matrix A of order $ 2 \times 4 $ and matrix B of order $ 4 \times 3 $, find the order of matrix AB.

Answer: $ 2 \times 3 $

(2) If $ A $ $ = $ $ \begin{pmatrix} 1 \\ 4 \end{pmatrix} $, $ B $ $ = $ $ \begin{pmatrix} 2 & 1 \end{pmatrix} $, find AB.

Answer: $ \begin{pmatrix} 2 & 1 \\ 8 & 4 \end{pmatrix} $

(3) Solve the equation $ \sin \theta $ $ = $ $ \frac{1}{2} $, $ \theta \in [0, \frac{\pi}{2}[ $.

Answer: $ \theta $ $ = $ $ \frac{\pi}{6} $

(4) If $ \vec{A} $ $ = $ $ (5, 2) $, $ \vec{B} $ $ = $ $ (m, 8) $, find m when $ \vec{A} \parallel \vec{B} $.

Answer: $ m $ $ = $ $ 20 $

(5) If $ \vec{A} $ $ = $ $ (m, -1) $, $ \vec{B} $ $ = $ $ (3, 9) $, find m when $ \vec{A} \perp \vec{B} $.

Answer: $ m $ $ = $ $ 3 $

Mastery & Detailed Explanations

01 First Group: In-Depth Analysis

Q1: Matrix Multiplication Order Rule

To multiply A (m × n) by B (n × p), the number of columns in A must equal the number of rows in B.
Given: A is $ 2 \times \mathbf{2} $ and B is $ \mathbf{2} \times 1 $.
Since the inner numbers (2) match, multiplication is defined.
The order of AB is the outer numbers: $ \mathbf{2 \times 1} $.

Q2: Outer Product Calculation

When multiplying a column vector by a row vector:
$ \begin{pmatrix} 3 \\ 5 \end{pmatrix} \begin{pmatrix} 1 & 3 \end{pmatrix} $ $ = $ $ \begin{pmatrix} 3 \times 1 & 3 \times 3 \\ 5 \times 1 & 5 \times 3 \end{pmatrix} $ $ = $ $ \mathbf{\begin{pmatrix} 3 & 9 \\ 5 & 15 \end{pmatrix}} $.

Q3: Special Angle Sine

$ \sin \theta $ $ = $ $ \frac{\sqrt{2}}{2} $. The angle in the first quadrant whose sine is $ \frac{\sqrt{2}}{2} $ is 45°.
In radian measure: $ \mathbf{\theta} $ $ = $ $ \mathbf{\frac{\pi}{4}} $.

Q4: Condition for Parallel Vectors

Two vectors $ (x_1, y_1) $ and $ (x_2, y_2) $ are parallel if $ x_1 y_2 - x_2 y_1 $ $ = $ $ 0 $.
For $ (1, 3) $ and $ (m, 12) $: $ (1 \times 12) - (3 \times m) $ $ = $ $ 0 $.
$ 12 - 3m $ $ = $ $ 0 $ $ \Rightarrow $ $ 3m $ $ = $ $ 12 $ $ \Rightarrow $ $ \mathbf{m} $ $ = $ $ \mathbf{4} $.

Q5: Condition for Perpendicular Vectors

Two vectors are perpendicular if their dot product is zero: $ x_1 x_2 + y_1 y_2 $ $ = $ $ 0 $.
For $ (3, 1) $ and $ (6, m) $: $ (3 \times 6) + (1 \times m) $ $ = $ $ 0 $.
$ 18 + m $ $ = $ $ 0 $ $ \Rightarrow $ $ \mathbf{m} $ $ = $ $ \mathbf{-18} $.

Pro Practice Example

"Remember: For parallelism, slopes $ y/x $ are equal. For perpendicularity, product of slopes is -1."

02 Second Group: Key Concepts

Q1: Matrix Order 3x3 by 3x1

Matrix A: $ 3 \times \mathbf{3} $. Matrix B: $ \mathbf{3} \times 1 $.
The inner dimensions match. The resulting order is $ \mathbf{3 \times 1} $.

Q2: Calculating Matrix AB

$ \begin{pmatrix} 1 \\ 2 \end{pmatrix} \begin{pmatrix} 5 & 4 \end{pmatrix} $ $ = $ $ \begin{pmatrix} 1 \times 5 & 1 \times 4 \\ 2 \times 5 & 2 \times 4 \end{pmatrix} $ $ = $ $ \mathbf{\begin{pmatrix} 5 & 4 \\ 10 & 8 \end{pmatrix}} $.

Q3: Sine Value for 60°

$ \sin \theta $ $ = $ $ \frac{\sqrt{3}}{2} $. The angle in the first quadrant is 60°.
In radians: $ \mathbf{\theta} $ $ = $ $ \mathbf{\frac{\pi}{3}} $.

Q4: Solving for m (Parallel)

$ \frac{1}{m} $ $ = $ $ \frac{4}{16} $ $ \Rightarrow $ $ 4m $ $ = $ $ 16 $ $ \Rightarrow $ $ \mathbf{m} $ $ = $ $ \mathbf{4} $.

Q5: Solving for m (Perpendicular)

$ m(6) + 6(3) $ $ = $ $ 0 $ $ \Rightarrow $ $ 6m + 18 $ $ = $ $ 0 $ $ \Rightarrow $ $ 6m $ $ = $ $ -18 $ $ \Rightarrow $ $ \mathbf{m} $ $ = $ $ \mathbf{-3} $.

03 Third Group: Advanced Steps

Q1: Order of 2x4 and 4x3

Matrix A: $ 2 \times \mathbf{4} $. Matrix B: $ \mathbf{4} \times 3 $.
Resulting order is $ \mathbf{2 \times 3} $.

Q2: Calculating Matrix AB

$ \begin{pmatrix} 1 \\ 4 \end{pmatrix} \begin{pmatrix} 2 & 1 \end{pmatrix} $ $ = $ $ \begin{pmatrix} 1 \times 2 & 1 \times 1 \\ 4 \times 2 & 4 \times 1 \end{pmatrix} $ $ = $ $ \mathbf{\begin{pmatrix} 2 & 1 \\ 8 & 4 \end{pmatrix}} $.

Q3: Sine Value for 30°

$ \sin \theta $ $ = $ $ \frac{1}{2} $. The angle is 30°.
In radians: $ \mathbf{\theta} $ $ = $ $ \mathbf{\frac{\pi}{6}} $.

Q4: Solving for m (Parallel)

$ \frac{2}{5} $ $ = $ $ \frac{8}{m} $ $ \Rightarrow $ $ 2m $ $ = $ $ 40 $ $ \Rightarrow $ $ \mathbf{m} $ $ = $ $ \mathbf{20} $.

Q5: Solving for m (Perpendicular)

$ m(3) + (-1)(9) $ $ = $ $ 0 $ $ \Rightarrow $ $ 3m - 9 $ $ = $ $ 0 $ $ \Rightarrow $ $ 3m $ $ = $ $ 9 $ $ \Rightarrow $ $ \mathbf{m} $ $ = $ $ \mathbf{3} $.

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