Exercise 3.1 · Q4: Construct a $2\times2$ matrix from a rule for $a_{ij}$

Basics you need

Part (i): $a_{ij} = \frac{(i+j)^2}{2}$

1. Compute the four entries

   $$\begin{aligned} a_{11} &= \frac{(1+1)^2}{2} = \frac{4}{2} = 2 \\ a_{12} &= \frac{(1+2)^2}{2} = \frac{9}{2} \\ a_{21} &= \frac{(2+1)^2}{2} = \frac{9}{2} \\ a_{22} &= \frac{(2+2)^2}{2} = \frac{16}{2} = 8 \end{aligned}$$
   Each entry uses its own address: $a_{12}$ is row 1, column 2, so $i = 1$⁠, $j = 2$⁠. Because $i + j = j + i$⁠, swapping $i$ and $j$ does not change the value, so $a_{12} = a_{21}$ here. We still compute both, since in general they differ (see Part (ii)).
   Brackets first: add $i + j$⁠, then square, then halve. Basics: substitution

2. Assemble the matrix

   Place each value at its address, row 1 on top:

   $$A = \begin{bmatrix} 2 & \frac{9}{2} \\[1mm] \frac{9}{2} & 8 \end{bmatrix}.$$
   Row 1 holds $a_{11}, a_{12}$ and row 2 holds $a_{21}, a_{22}$⁠. The matrix is unchanged by reflection in its diagonal, a pattern called symmetric that the chapter studies in Section 3.6.
   $A=[a_{ij}]_{2\times2}$ has $a_{ij}$ in row $i$⁠, column $j$⁠. Basics: the address

Part (ii): $a_{ij} = \frac{i}{j}$

1. Compute the four entries

   $$\begin{aligned} a_{11} &= \frac{1}{1} = 1 \\ a_{12} &= \frac{1}{2} \\ a_{21} &= \frac{2}{1} = 2 \\ a_{22} &= \frac{2}{2} = 1 \end{aligned}$$
   Here the order of $i$ and $j$ matters: $\frac{i}{j}$ has the row number on top and the column number underneath. So $a_{12} = \frac12$ but $a_{21} = 2$⁠. Mixing up the subscripts would swap these two entries and give the wrong matrix. Every $j$ is 1 or 2, never 0, so no division by zero can occur.
   Row number $i$ on top, column number $j$ below. Basics: rows and columns

2. Assemble the matrix

   $$A = \begin{bmatrix} 1 & \frac{1}{2} \\[1mm] 2 & 1 \end{bmatrix}.$$
   Unlike Part (i), the entries above and below the diagonal are different ($\frac12$ and $2$⁠), because $\frac{i}{j}$ changes when $i$ and $j$ are swapped: it becomes its reciprocal.

Part (iii): $a_{ij} = \frac{(i+2j)^2}{2}$

1. Compute the four entries

   $$\begin{aligned} a_{11} &= \frac{(1+2\cdot1)^2}{2} = \frac{9}{2} \\ a_{12} &= \frac{(1+2\cdot2)^2}{2} = \frac{25}{2} \\ a_{21} &= \frac{(2+2\cdot1)^2}{2} = \frac{16}{2} = 8 \\ a_{22} &= \frac{(2+2\cdot2)^2}{2} = \frac{36}{2} = 18 \end{aligned}$$
   Inside the bracket the column number is doubled before adding: $i + 2j$⁠, not $2(i+j)$⁠. For $a_{12}$⁠: $i = 1$⁠, $j = 2$⁠, so $i + 2j = 1 + 4 = 5$ and $5^2 = 25$⁠. For $a_{21}$⁠: $i = 2$⁠, $j = 1$⁠, so $i + 2j = 2 + 2 = 4$ and $4^2 = 16$⁠. The two off-diagonal entries differ because $i$ and $j$ play different roles.
   Order of operations: multiply $2j$ first, then add $i$⁠, then square. Basics: substitution

2. Assemble the matrix

   $$A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\[1mm] 8 & 18 \end{bmatrix}.$$
   Moving one step right (column $j$ up by 1) adds 2 inside the bracket; moving one step down (row $i$ up by 1) adds only 1. That is why the entries grow faster along a row than down a column.

Every fact used, and where it comes from