Q.Construct a $3 \times 4$ matrix, whose elements are given by:

Concept: Matrix Construction — generate each entry by substituting the row index $i$ and column index $j$ into the given formula.

(i) $a_{ij} = \frac{1}{2} |-3i+j|$

For $i=1,2,3$ and $j=1,2,3,4$:

• Row 1: $j=1 \to \frac12|-3+1|=1$; $j=2 \to \frac12|-3+2|=\frac12$; $j=3 \to \frac12|-3+3|=0$; $j=4 \to \frac12|-3+4|=\frac12$
• Row 2: $j=1 \to \frac12|-6+1|=\frac52$; $j=2 \to \frac12|-6+2|=2$; $j=3 \to \frac12|-6+3|=\frac32$; $j=4 \to \frac12|-6+4|=1$
• Row 3: $j=1 \to \frac12|-9+1|=4$; $j=2 \to \frac12|-9+2|=\frac72$; $j=3 \to \frac12|-9+3|=3$; $j=4 \to \frac12|-9+4|=\frac52$

(ii) $a_{ij} = 2i-j$

• Row 1: $2(1)-1=1$, $2(1)-2=0$, $2(1)-3=-1$, $2(1)-4=-2$ …

Concept and Intuition

A matrix is just a rectangular array of numbers arranged in rows and columns. When a formula for $a_{ij}$ is given, it tells you exactly how to compute the entry at the intersection of row $i$ and column $j$. The indices $i$ and $j$ are your coordinates — like a grid reference on a map.

For a $3 \times 4$ matrix, $i$ runs from 1 to 3 (rows), and $j$ runs from 1 to 4 (columns). So we have 12 entries to compute for each part. The absolute value in part (i) ensures all entries are non-negative, and the $\frac12$ factor may produce fractions. Part (ii) is a simple linear expression, so all entries will be integers.

Step-by-step solution

Part (i): $a_{ij} = \frac{1}{2} |-3i + j|$

1. Set up the grid.

We'll compute for each row $i = 1, 2, 3$ and each column $j = 1, 2, 3, 4$.

2. Row 1 ($i=1$).

The expression becomes $\frac12 | -3(1) + j | = \frac12 | j - 3 |$.

• $j=1$: $\frac12 |1-3| = \frac12 \times 2 = 1$
• $j=2$: $\frac12 |2-3| = \frac12 \times 1 = \frac12$
• $j=3$: $\frac12 |3-3| = \frac12 \times 0 = 0$
• $j=4$: $\frac12 |4-3| = \frac12 \times 1 = \frac12$

So row 1: $\begin{bmatrix} 1 & \frac12 & 0 & \frac12 \end{bmatrix}$

3. Row 2 ($i=2$).

Now $\frac12 | -6 + j | = \frac12 | j - 6 |$.

• $j=1$: $\frac12 |1-6| = \frac12 \times 5 = \frac52$
• $j=2$: $\frac12 |2-6| = \frac12 \times 4 = 2$
• $j=3$: $\frac12 |3-6| = \frac12 \times 3 = \frac32$
• $j=4$: $\frac12 |4-6| = \frac12 \times 2 = 1$

Row 2: $\begin{bmatrix} \frac52 & 2 & \frac32 & 1 \end{bmatrix}$

4. Row 3 ($i=3$).

$\frac12 | -9 + j | = \frac12 | j - 9 |$.

• $j=1$: $\frac12 |1-9| = \frac12 \times 8 = 4$
• $j=2$: $\frac12 |2-9| = \frac12 \times 7 = \frac72$
• $j=3$: $\frac12 |3-9| = \frac12 \times 6 = 3$
• $j=4$: $\frac12 |4-9| = \frac12 \times 5 = \frac52$

Row 3: $\begin{bmatrix} 4 & \frac72 & 3 & \frac52 \end{bmatrix}$

5. Assemble the matrix for (i).

Part (ii): $a_{ij} = 2i - j$

1. Again, go row by row. …

Method: Building a Rectangular Matrix From an Element Rule

Use this to construct any $m\times n$ matrix given a formula for $a_{ij}$, especially when the rule contains an absolute value or a linear expression that can turn negative.

Steps

Step 1: Set the index ranges from the order.

For a $3\times4$ matrix, $i=1,2,3$ (rows) and $j=1,2,3,4$ (columns) — twelve entries. Work row by row to stay organised.

Step 2: Substitute carefully, respecting absolute values and signs. …

Common Mistakes

Mistake 1: Dropping the absolute value and leaving negative entries.

Why it's wrong: $\tfrac12\lvert -3i+j\rvert$ is always $\ge 0$; a negative entry means the modulus was ignored. Correct approach: evaluate the bracket, then take the magnitude before scaling.

Mistake 2: Confusing which index is the row vs the column.

Why it's wrong: computing $a_{ij}$ with $i,j$ swapped fills the wrong cells. Correct approach: hold $i$ as the row while sweeping $j$ across the columns. …