Exercises — Matrix operations — addition, subtraction (conditions)

Donon matrices ka same order hona chahiye — same number of rows aur same number of columns. Tabhi ek matrix ki har entry $(i,j)$ ka doosri matrix mein ek partner $(i,j)$ hoga. "Order" ka matlab samajhne ke liye Matrix Notation and Terminology dekho.

Addition tabhi defined hai jab donon ke orders identical hon.

• (a) Dono $2\times 3$ → ✓ defined hai, result $2\times 3$ hoga.
• (b) $3\times 2$ vs $2\times 3$ → ✗ undefined (rows aur columns swap ho gaye hain, equal nahi hain).
• (c) Dono $4\times 1$ → ✓ defined hai, result $4\times 1$ hoga.
• (d) $1\times 5$ vs $5\times 1$ → ✗ undefined. Ek single row aur ek single column ek hi shape nahi hote.

$B$ ki har entry, $A$ ki matching entry ki sign-flip hai, isliye $B=-A$. $A+B=\left[\begin{array}{cc}3+(-3)& -2+2\\ 5+(-5)& 1+(-1)\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=O$ Yeh zero matrix hai, aur $B$, $A$ ka additive inverse hai.

Same order $2\times 3$ → defined hai. Cell by cell add karo: $A+B=\left[\begin{array}{ccc}2+4& -3+1& 5+(-1)\\ 1+(-3)& 0+7& -2+6\end{array}\right]=\left[\begin{array}{ccc}6& -2& 4\\ -2& 7& 4\end{array}\right]$

Dono $3\times 2$ → defined hai. Subtraction matlab "sign-flip add karo", isliye $d_{ij}=p_{ij}-q_{ij}$: $P-Q=\left[\begin{array}{cc}5-2& -1-3\\ 3-(-1)& 2-1\\ 0-5& 4-(-2)\end{array}\right]=\left[\begin{array}{cc}3& -4\\ 4& 1\\ -5& 6\end{array}\right]$ Double-negatives ka dhyan rakho: $3-(-1)=4$ aur $4-(-2)=6$.

$A-B=\left[\begin{array}{ccc}2-4& -3-1& 5-(-1)\\ 1-(-3)& 0-7& -2-6\end{array}\right]=\left[\begin{array}{ccc}-2& -4& 6\\ 4& -7& -8\end{array}\right]$ $B-A=\left[\begin{array}{ccc}2& 4& -6\\ -4& 7& 8\end{array}\right]$ $B-A$ ki har entry, $A-B$ ki entry ka negative hai: yeh anti-commutativity hai, $B-A=-(A-B)$. Subtraction commute nahi karti.

Matrix equality ka matlab hai ki har cell match karni chahiye. Jo cells unknowns contain karti hain unhe padho:

• Cell $(1,1)$: $x+4=7\Rightarrow x=3$.
• Cell $(2,2)$: $y+5=9\Rightarrow y=4$.

(Baaki cells consistency ke liye check karo: $2+(-2)=0$ ✓, $3+(-3)=0$ ✓.) Toh $x=3,y=4$. Dhyan do ki yeh ek System of Linear Equations hai jo ek matrix equation ke andar chhipa hua hai.

• Defined hai? $N^T$, $N$ ko $3\times 2$ se $2\times 3$ kar deta hai. Ab $M$ aur $N^T$ dono $2\times 3$ hain, isliye $M+N^T$ defined hai.
• $M+N$ jaisa hi hai? Nahi. $M+N$ to pehle se hi undefined hai ($2\times 3$ vs $3\times 2$). Isse bhi important, $N$ ko transpose karna har entry ka meaning badal deta hai: $N$ ki entry $(1,2)$ ("row-item 1, col-item 2") $N^T$ mein position $(2,1)$ par aati hai. Isliye $M+N^T$ ek valid lekin alag computation hai — yeh "$M$ aur $N$ ka sum" nahi hai. Transpose of a Matrix dekho.

Hum additive inverse chahte hain: $X=-A$, jo har sign flip karke milta hai. $X=\left[\begin{array}{cc}-1& 2\\ 0& -3\end{array}\right]$ Check: $A+X=\left[\begin{array}{cc}1-1& -2+2\\ 0+0& 3-3\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=O$ ✓. Dhyan do $-0=0$, isliye zero entry waise hi rehti hai.

Pehle scale karo, phir cell by cell add karo: $2A=\left[\begin{array}{cc}2& 4\\ 0& -2\end{array}\right],3B=\left[\begin{array}{cc}9& -3\\ 6& 12\end{array}\right]$ $2A+3B=\left[\begin{array}{cc}2+9& 4-3\\ 0+6& -2+12\end{array}\right]=\left[\begin{array}{cc}11& 1\\ 6& 10\end{array}\right]$

Left side: $A+B=\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]$, isliye $2(A+B)=\left[\begin{array}{cc}8& 2\\ 4& 6\end{array}\right]$. Right side: $2A=\left[\begin{array}{cc}2& 4\\ 0& -2\end{array}\right]$, $2B=\left[\begin{array}{cc}6& -2\\ 4& 8\end{array}\right]$, isliye $2A+2B=\left[\begin{array}{cc}8& 2\\ 4& 6\end{array}\right]$. Dono sides $\left[\begin{array}{cc}8& 2\\ 4& 6\end{array}\right]$ ke barabar hain ✓. Ye isliye match karte hain kyunki har entry ke liye $2(a_{ij}+b_{ij})=2a_{ij}+2b_{ij}$ hota hai — ordinary numbers ka distributive law, jo cell by cell apply hota hai.

Ise ordinary algebra ki tarah treat karo, lekin har step ek matrix operation hai:

1. Dono sides mein $A$ add karo: $2X=B+A$.
2. $B+A=\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]$.
3. Scalar $\frac{1}{2}$ se multiply karo: $X=\frac{1}{2}\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]=\left[\begin{array}{cc}2& \frac{1}{2}\\ 1& \frac{3}{2}\end{array}\right]$.

Check: $2X-A=\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}3& -1\\ 2& 4\end{array}\right]=B$ ✓.

Teeno matrices $3\times 2$ hain (same "shops × products" layout), isliye poora expression defined hai aur $3\times 2$ hi rahega. Pehle $M+T$ add karo: $M+T=\left[\begin{array}{cc}16& 6\\ 16& 4\\ 9& 13\end{array}\right]$ $R$ subtract karo: $N=(M+T)-R=\left[\begin{array}{cc}16-1& 6-0\\ 16-2& 4-1\\ 9-0& 13-3\end{array}\right]=\left[\begin{array}{cc}15& 6\\ 14& 3\\ 9& 10\end{array}\right]$ Har cell ka meaning "woh shop, woh product" bana rehta hai, aur yahi wajah hai ki element-wise combining yahan valid hai.

Teeno ka order $m\times n$ hai, isliye neeche ka har operation defined hai aur ek $m\times n$ matrix produce karta hai; humein bas yeh dikhana hai ki $(i,j)$ entries agree karti hain.

• Left side ki entry: $((A+B)-C{)}_{ij}=(a_{ij}+b_{ij})-c_{ij}$.
• Right side ki entry: $(A+(B-C){)}_{ij}=a_{ij}+(b_{ij}-c_{ij})$.

Ordinary real numbers ke liye, addition/subtraction associative hoti hai: $(a+b)-c=a+(b-c)$. Kyunki yeh har cell $(i,j)$ mein hold karta hai, dono matrices equal hain. ∎ Yahi deep reason hai ki sabhi matrix-addition laws kaam karte hain: ye sirf number laws hain jo har entry par simultaneously apply hoti hain.

$X=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ likho.

• $X+X=2X=\left[\begin{array}{cc}2a& 2b\\ 2c& 2d\end{array}\right]=\left[\begin{array}{cc}4& 6\\ 6& -2\end{array}\right]$, jisse milta hai $a=2,b=3,c=3,d=-1$.
• Symmetry condition $X=X^T$ ke liye $b=c$ chahiye; yahan $b=c=3$ ✓, toh yeh automatically satisfy ho jaata hai.

Unique answer: $X=\left[\begin{array}{cc}2& 3\\ 3& -1\end{array}\right]$.