https://file.aiquickdraw.com/mathbot/file/1725526928623k8lbkidd.pdf

Here are the step-by-step solutions for the first few questions from the exam you uploaded:

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**1. Construct a 2 x 3 matrix whose elements are given by \(a_{ij} = i - 2j\)**

**Solution:**

- The matrix is of order \(2 \times 3\), meaning it has 2 rows and 3 columns.
- Each element of the matrix is determined by the formula \(a_{ij} = i - 2j\), where \(i\) is the row number and \(j\) is the column number.

For \(i = 1\) (first row) and \(j = 1, 2, 3\):
- \(a_{11} = 1 - 2(1) = 1 - 2 = -1\)
- \(a_{12} = 1 - 2(2) = 1 - 4 = -3\)
- \(a_{13} = 1 - 2(3) = 1 - 6 = -5\)

For \(i = 2\) (second row) and \(j = 1, 2, 3\):
- \(a_{21} = 2 - 2(1) = 2 - 2 = 0\)
- \(a_{22} = 2 - 2(2) = 2 - 4 = -2\)
- \(a_{23} = 2 - 2(3) = 2 - 6 = -4\)

Thus, the matrix is:

\[
\begin{bmatrix}
-1 & -3 & -5 \\
0 & -2 & -4
\end{bmatrix}
\]

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**2. Examine if the relation \(R\) in the set \(A = \{1, 2, 3, \dots, 12\}\) defined by \(R = \{(x, y): 2x - y = 0\}\) is reflexive, symmetric, and transitive.**

**Solution:**

1. **Reflexive:**
   A relation is reflexive if \( (x, x) \in R \) for all \(x \in A\).
   
   - For \(R\) to be reflexive, \(2x - x = 0 \), which simplifies to \(x = 0\). Since no element in the set \(A\) satisfies this condition, \(R\) is **not reflexive**.

2. **Symmetric:**
   A relation is symmetric if \( (x, y) \in R \implies (y, x) \in R \).
   
   - From the definition, \( (x, y) \in R \) means \(2x - y = 0 \implies y = 2x\). For symmetry, we must have \(2y - x = 0\), but that simplifies to \(x = 2y\), which is not necessarily true for all pairs. Therefore, \(R\) is **not symmetric**.

3. **Transitive:**
   A relation is transitive if \( (x, y) \in R \) and \( (y, z) \in R \) implies \( (x, z) \in R \).
   
   - If \(2x - y = 0\) and \(2y - z = 0\), then substituting \(y = 2x\) into the second equation, we get \(2(2x) - z = 0 \implies z = 4x\). Thus, \(2x - z = 0\) is not always true, so \(R\) is **not transitive**.

Therefore, the relation \(R\) is **neither reflexive, symmetric, nor transitive**.

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**3. If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), show that \(|2A| = 4|A|\).**

**Solution:**

1. **Determinant of \(A\):**
   For a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), the determinant is given by:
   \[
   |A| = ad - bc
   \]

2. **Matrix \(2A\):**
   The matrix \(2A\) is obtained by multiplying each element of \(A\) by 2:
   \[
   2A = \begin{bmatrix} 2a & 2b \\ 2c & 2d \end{bmatrix}
   \]
   
   The determinant of \(2A\) is:
   \[
   |2A| = (2a)(2d) - (2b)(2c) = 4(ad - bc) = 4|A|
   \]

Hence, \(|2A| = 4|A|\).

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Would you like further details on any of these solutions or have any questions? Here are 5 questions that build on the material:

1. What is the significance of a matrix being symmetric or skew-symmetric?
2. How do you determine if a matrix is invertible using its determinant?
3. Can a non-reflexive relation ever be symmetric or transitive?
4. What are the properties of a bijective function in mathematics?
5. How is the determinant of a matrix affected when multiplied by a scalar?

**Tip:** In matrix operations, multiplying a matrix by a scalar affects the determinant exponentially, not linearly.

Detailed step-by-step solutions for matrix construction, examining relations, and solving determinants. Learn how to construct a matrix using a formula, analyze the reflexivity, symmetry, and transitivity of relations, and calculate the determinant of matrices. Suitable for Grades 11-12 students.

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