Exercises — Matrix definition — rows, columns, order, elements

Order. Count rows (down): there are $2$ horizontal lines. Count columns (across): $3$ vertical lines. So the order is $==2\times 3==$.

$a_{11}$ — "R then C": down to row $1$, then across to column $1$. That is the top-left number, $a_{11}=8$.

$a_{13}$ — row $1$, column $3$: stay in the top row $[8-25]$, take the $3$rd entry: $a_{13}=5$.

$a_{22}$ — row $2$, column $2$: go to bottom row $[04-6]$, take the $2$nd entry: $a_{22}=4$.

Order is written rows $\times$ columns, so $4\times 1$ means $4$ rows and $1$ column.

Total elements $=m\times n=4\times 1=4$.

One column, several rows $\Rightarrow$ this is a column matrix (a column vector). See Types of Matrices.

Order. Rows (down): $3$. Columns (across): $4$. Order $=3\times 4$.

Total elements $=3\times 4=12$.

$d_{34}$ — row $3$, column $4$: bottom row is $[609-3]$, $4$th entry $=-3$. So $d_{34}=-3$.

$d_{23}$ — row $2$, column $3$: middle row $[25-41]$, $3$rd entry $=-4$. So $d_{23}=-4$.

Square? A square matrix needs rows $=$ columns. Here $3\ne 4$, so not square — it is rectangular.

The rule $m_{ij}=3i-j$ takes a position $(i,j)$ and returns the number that lives there.

$m_{13}$: here $i=1$, $j=3$, so $m_{13}=3(1)-3=3-3=0$.

$m_{21}$: here $i=2$, $j=1$, so $m_{21}=3(2)-1=6-1=5$.

The main diagonal is where $i=j$ (look at the green cells in the figure — they run top-left to bottom-right).

With $a_{ij}=i^2-j$ and $i=j$:

• $a_{11}=1^2-1=0$
• $a_{22}=2^2-2=4-2=2$
• $a_{33}=3^2-3=9-3=6$

Sum $=0+2+6=8$. (This diagonal sum has a name, the trace — you meet it again with the Determinant.)

"Symmetric" means swapping row and column indices leaves the number unchanged — i.e. $S$ equals its own transpose. Check each off-diagonal pair $(i,j)$ vs $(j,i)$:

• $s_{12}=7$ and $s_{21}=7$ ✓
• $s_{13}=-1$ and $s_{31}=-1$ ✓
• $s_{23}=5$ and $s_{32}=5$ ✓

Every mirror pair matches, so $S$ is symmetric.

A piecewise rule: for each cell, first compare $i$ and $j$ to pick which formula applies, then compute.

Row 1 ($i=1$):

• $(1,1)$: $i=j\Rightarrow$ use $i+j=1+1=2$
• $(1,2)$: $i<j\Rightarrow$ use $i+j=1+2=3$
• $(1,3)$: $i<j\Rightarrow$ use $i+j=1+3=4$

Row 2 ($i=2$):

• $(2,1)$: $i>j\Rightarrow$ use $i-j=2-1=1$
• $(2,2)$: $i=j\Rightarrow$ use $i+j=2+2=4$
• $(2,3)$: $i<j\Rightarrow$ use $i+j=2+3=5$

$P=\left[\begin{array}{ccc}2& 3& 4\\ 1& 4& 5\end{array}\right]$

Both are $2\times 2$, so the shapes already match — now demand element-by-element agreement:

• Position $(1,1)$: $a=2$
• Position $(1,2)$: $3=3$ ✓ (nothing to solve)
• Position $(2,1)$: $5=5$ ✓
• Position $(2,2)$: $b=-4$

So $a=2$ and $b=-4$.

Total elements $=n^2$. Diagonal elements $=n$. So off-diagonal $=n^2-n$.

Set up the equation from the given information: $n^2-n=56.$ Rearrange: $n^2-n-56=0$. Factor: $(n-8)(n+7)=0$, giving $n=8$ or $n=-7$.

An order must be a positive whole number, so we reject $n=-7$ and keep $n=8$.

Order $=8\times 8$, total elements $=8^2=64$ (check: $64-8=56$ off-diagonal ✓).

Each column is a landing point written vertically. Column $1$ is where "right" goes: $(0,1)$ becomes $\left[\begin{array}{c}0\\ 1\end{array}\right]$. Column $2$ is where "up" goes: $(-1,0)$ becomes $\left[\begin{array}{c}-1\\ 0\end{array}\right]$. Place them side by side: $T=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ $t_{21}$ — row $2$, column $1$: bottom-left entry $=1$. So $t_{21}=1$.

(This is the rotation matrix you meet in Matrix Multiplication and System of Linear Equations later — here we only read and build it.)

The first subscript of $a_{ij}$ always selects the row. The order $m\times n$ lists rows before columns. A matrix with $m=n$ is called a square matrix. Two matrices are equal only if they have the same order and equal elements. The main diagonal of a matrix is the set of entries where ==$i=j$==.