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

Order. Rows gino (neeche): $2$ horizontal lines hain. Columns gino (aaray): $3$ vertical lines hain. To order hai $==2\times 3==$.

$a_{11}$ — "R phir C": row $1$ tak neeche, phir column $1$ tak aaray. Yeh top-left number hai, $a_{11}=8$.

$a_{13}$ — row $1$, column $3$: top row $[8-25]$ mein raho, $3$ri entry lo: $a_{13}=5$.

$a_{22}$ — row $2$, column $2$: bottom row $[04-6]$ par jao, $2$ri entry lo: $a_{22}=4$.

Order likha jaata hai rows $\times$ columns, to $4\times 1$ ka matlab hai $4$ rows aur $1$ column.

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

Ek column, kai rows $\Rightarrow$ yeh ek column matrix hai (column vector). Dekho Types of Matrices.

Order. Rows (neeche): $3$. Columns (aaray): $4$. Order $=3\times 4$.

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

$d_{34}$ — row $3$, column $4$: bottom row hai $[609-3]$, $4$thi entry $=-3$. To $d_{34}=-3$.

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

Square? Square matrix ke liye rows $=$ columns chahiye. Yahan $3\ne 4$, to square nahi — yeh rectangular hai.

Rule $m_{ij}=3i-j$ ek position $(i,j)$ leta hai aur wahan rehne waala number return karta hai.

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

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

Main diagonal wahan hoti hai jahan $i=j$ ho (figure mein green cells dekho — woh top-left se bottom-right tak chalti hain).

$a_{ij}=i^2-j$ aur $i=j$ ke saath:

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

Sum $=0+2+6=8$. (Is diagonal sum ka ek naam hai, trace — tum isse phir Determinant ke saath miloge.)

"Symmetric" ka matlab hai row aur column indices swap karne par number nahi badalta — yaani $S$ apne hi transpose ke barabar hai. Har off-diagonal pair $(i,j)$ vs $(j,i)$ check karo:

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

Har mirror pair match karta hai, to $S$ symmetric hai.

Piecewise rule: har cell ke liye, pehle $i$ aur $j$ compare karo ki kaun sa formula lagega, phir compute karo.

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 $i+j=2+3=5$

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

Dono $2\times 2$ hain, to shapes pehle se match karti hain — ab element-by-element agreement maango:

• Position $(1,1)$: $a=2$
• Position $(1,2)$: $3=3$ ✓ (kuch solve nahi karna)
• Position $(2,1)$: $5=5$ ✓
• Position $(2,2)$: $b=-4$

To $a=2$ aur $b=-4$.

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

Diye gaye information se equation banao: $n^2-n=56.$ Rearrange karo: $n^2-n-56=0$. Factor karo: $(n-8)(n+7)=0$, jisse $n=8$ ya $n=-7$ milta hai.

Order ek positive whole number hona chahiye, to $n=-7$ reject karo aur $n=8$ rakho.

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

Har column ek landing point hai jo vertically likha gaya hai. Column $1$ woh hai jahan "right" jaata hai: $(0,1)$ ban jaata hai $\left[\begin{array}{c}0\\ 1\end{array}\right]$. Column $2$ woh hai jahan "up" jaata hai: $(-1,0)$ ban jaata hai $\left[\begin{array}{c}-1\\ 0\end{array}\right]$. Inhe side by side rakho: $T=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ $t_{21}$ — row $2$, column $1$: bottom-left entry $=1$. To $t_{21}=1$.

(Yeh rotation matrix hai jise tum Matrix Multiplication aur System of Linear Equations mein baad mein miloge — yahan hum ise sirf padhte aur banate hain.)

$a_{ij}$ ka pehla subscript hamesha row select karta hai. Order $m\times n$ mein rows pehle aate hain, phir columns. $m=n$ wali matrix ko square matrix kehte hain. Do matrices equal tab hi hoti hain jab unka same order ho aur equal elements hon. Matrix ki main diagonal un entries ka set hai jahan ==$i=j$== ho.