Solving systems using matrix inversion

2.6.12 · Maths › Matrices & Determinants — Introduction

Ek linear system jaise $2x+3y=5,x-y=1$ ko EK matrix equation $AX=B$ mein pack kiya ja sakta hai, aur ek hi baar mein $X=A^{-1}B$ se solve kiya ja sakta hai.

The Big Picture

KYU seekhein yeh? Kyunki yeh "3 equations elimination se solve karo" ko "ek inverse compute karo aur multiply karo" mein convert kar deta hai — ek mechanical, mistake-proof recipe jo yeh bhi bataati hai ki unique solution kab exist karta hai.

Step 1 — Ek system ko $AX=B$ ke roop mein likhna

YEH system ke barabar KAISE hai? $A$ ki row $i$ ko column $X$ se multiply karo: aapko $a_{i1}{x}_1+\cdots +a_{in}{x}_n$ milta hai, aur ise $B$ ki entry $b_i$ ke barabar set karne se equation $i$ bilkul waisi hi milti hai. Toh $AX=B$ woh system hai, bas naye packaging mein.

Step 2 — Scratch se solution derive karna

Hume $X$ chahiye. Shuru karte hain $AX=B.$

Hum "divide" KYU kar sakte hain? Matrices ki koi division nahi hoti, lekin inverse ZAROOR hota hai. Suppose karein $A^{-1}$ exist karta hai (yani $A^{-1}A=I$). Dono sides ko LEFT se $A^{-1}$ se multiply karo — left se, kyunki matrix multiplication commutative nahi hoti aur $A$, $X$ ke left mein hai: $A^{-1}(AX)=A^{-1}B.$ Associativity use karo: $(A^{-1}A)X=A^{-1}B.$ Kyunki $A^{-1}A=I$ (identity) aur $IX=X$:

$\det A\ne 0$ condition KYU? Kyunki $A^{-1}=\frac{1}{\det A}\mathrm{adj}(A)$, aur $\det A=0$ se divide karna illegal hai. Geometrically $\det A=0$ matlab $A$ space ko collapse kar deta hai, toh information lost ho jaati hai aur aap ise undo nahi kar sakte.

Consistency: jab $\det A=0$ ho toh kya hota hai?

KYU? $AX=B$ ko $\mathrm{adj}(A)$ se multiply karo: $(\mathrm{adj}A)AX=(\mathrm{adj}A)B$. Lekin $(\mathrm{adj}A)A=(\det A)I$. Agar $\det A=0$ toh left side $O$ hai, jo force karta hai ki $(\mathrm{adj}A)B=O$; agar yeh violate hota hai, toh koi $X$ exist nahi kar sakta.

Worked Example 1 — ek clean $2\times 2$

Solve karo $2x+3y=5$, $x-y=1$.

$A=\left[\begin{array}{cc}2& 3\\ 1& -1\end{array}\right],B=\left[\begin{array}{c}5\\ 1\end{array}\right].$

Step A: $\det A=2(-1)-3(1)=-5$. Yeh step KYU? Non-zero ⇒ unique solution exist karta hai, invert karna safe hai.

Step B: $A^{-1}=\frac{1}{-5}\left[\begin{array}{cc}-1& -3\\ -1& 2\end{array}\right]$. KYU? $2\times 2$ rule: diagonal swap karo, off-diagonal negate karo, $\det$ se divide karo.

Step C: $X=A^{-1}B=\frac{1}{-5}\left[\begin{array}{cc}-1& -3\\ -1& 2\end{array}\right]\left[\begin{array}{c}5\\ 1\end{array}\right]=\frac{1}{-5}\left[\begin{array}{c}-5-3\\ -5+2\end{array}\right]=\frac{1}{-5}\left[\begin{array}{c}-8\\ -3\end{array}\right]=\left[\begin{array}{c}8/5\\ 3/5\end{array}\right].$ Order $A^{-1}B$ KYU? Kyunki $X=A^{-1}B$ — $B$ right side par hai.

Check: $2(8/5)+3(3/5)=16/5+9/5=25/5=5.$ ✓

Worked Example 2 — ek $3\times 3$

Solve karo $x+y+z=6$, $y+3z=11$, $x+z=2z$ (yani $x-z=0$).

$A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 3\\ 1& 0& -1\end{array}\right],B=\left[\begin{array}{c}6\\ 11\\ 0\end{array}\right].$

Step A: $\det A$. Row 1 ke saath expand karo: $1(1\cdot (-1)-3\cdot 0)-1(0\cdot (-1)-3\cdot 1)+1(0\cdot 0-1\cdot 1)=1(-1)-1(-3)+1(-1)=-1+3-1=1.$ KYU? $\det =1\ne 0$ ⇒ unique solution, aur 1 se divide karne se numbers clean rehte hain.

Step B: cofactors → adj. Compute karo (cofactor matrix ka transpose): $\mathrm{adj}A=\left[\begin{array}{ccc}-1& 1& 2\\ 3& -2& -3\\ -1& 1& 1\end{array}\right].$ Transpose KYU? $\mathrm{adj}(A{)}_{ij}=C_{ji}$; transpose bhoolna sabse bada #1 slip hai.

Step C: $A^{-1}=\frac{1}{1}\mathrm{adj}A$, toh $X=A^{-1}B=\left[\begin{array}{ccc}-1& 1& 2\\ 3& -2& -3\\ -1& 1& 1\end{array}\right]\left[\begin{array}{c}6\\ 11\\ 0\end{array}\right]=\left[\begin{array}{c}-6+11+0\\ 18-22-0\\ -6+11+0\end{array}\right]=\left[\begin{array}{c}5\\ -4\\ 5\end{array}\right].$

Wait — row 2 check karo: $y+3z=-4+15=11$ ✓, $x+y+z=5-4+5=6$ ✓, $x-z=5-5=0$ ✓. Solution $(5,-4,5)$.

Common Mistakes (Steel-manned)

Active-Recall Flashcards

Connections

• Determinants — woh $\det A\ne 0$ gate jo solvability decide karta hai.
• Adjoint and Inverse of a Matrix — yahan use hone wala $A^{-1}$ build karta hai.
• Cramer's Rule — alternative solver, same $\det A\ne 0$ condition.
• Cofactors and Minors — $\mathrm{adj}(A)$ ke ingredients.
• Consistency of Linear Systems — $\det =0$ branch.
• Gaussian Elimination — large systems ke liye practical rival.

Concept Map