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<title>Gauss Elimination Method Se Equations Solve Karna</title>
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<h1>Gauss Elimination Method</h1>
<h2>(a) Sawaal (Problem Statement)</h2>
<p>Gauss Elimination method ka istemal karke yeh system of equations solve karo:</p>
<div class="equations">
6x₁ + 3x₂ + 2x₃ = 6
6x₁ + 4x₂ + 3x₃ = 0
20x₁ + 15x₂ + 12x₃ = 0
</div>
<h2>Gauss Elimination Ke Steps</h2>
<p>Pehle, augmented matrix banate hain:</p>
<div class="matrix-display"><code>[ 6 3 2 | 6 ]
[ 6 4 3 | 0 ]
[ 20 15 12 | 0 ]</code></div>
<h3>Step 1: Pehla pivot (R1,C1) ko 1 banana</h3>
<p>Pehla element (R1,C1) 6 hai, isko 1 banana hai.</p>
<p class="operation">R1 → R1 / 6</p>
<div class="matrix-display"><code>[ <span class="highlight">1</span> 1/2 1/3 | 1 ] <span class="comment"><-- (6/6=1, 3/6=1/2, 2/6=1/3, 6/6=1)</span>
[ 6 4 3 | 0 ]
[ 20 15 12 | 0 ]</code></div>
<h3>Step 2: Pehle pivot ke neeche zeros banana</h3>
<p>Ab R1,C1 wale pivot (1) ke neeche ke elements ko zero karenge.</p>
<p class="operation">R2 → R2 - 6*R1</p>
<p class="operation">R3 → R3 - 20*R1</p>
<div class="matrix-display"><code>[ 1 1/2 1/3 | 1 ]
[ 0 1 1 | -6 ] <span class="comment"><-- R2: [6-6*1, 4-6*(1/2), 3-6*(1/3) | 0-6*1] = [0, 1, 1 | -6]</span>
[ 0 5 16/3 | -20] <span class="comment"><-- R3: [20-20*1, 15-20*(1/2), 12-20*(1/3) | 0-20*1] = [0, 5, (36-20)/3 | -20] = [0, 5, 16/3 | -20]</span></code></div>
<p>Yahaan R2,C2 mein already 1 aa gaya, toh accha hai!</p>
<h3>Step 3: Dusre pivot ke neeche zero banana</h3>
<p>Ab R2,C2 wala pivot 1 hai. Iske neeche (R3,C2) zero banana hai.</p>
<p class="operation">R3 → R3 - 5*R2</p>
<div class="matrix-display"><code>[ 1 1/2 1/3 | 1 ]
[ 0 <span class="highlight">1</span> 1 | -6 ]
[ 0 0 1/3 | 10] <span class="comment"><-- R3: [0-5*0, 5-5*1, 16/3-5*1 | -20-5*(-6)] = [0, 0, (16-15)/3 | -20+30] = [0, 0, 1/3 | 10]</span></code></div>
<p>Matrix ab Row Echelon Form (REF) mein hai. Ab pivot elements ko 1 banana hai (teesre ko).</p>
<h3>Step 4: Teesra pivot (R3,C3) ko 1 banana</h3>
<p class="operation">R3 → R3 * 3</p>
<div class="matrix-display"><code>[ 1 1/2 1/3 | 1 ]
[ 0 1 1 | -6 ]
[ 0 0 <span class="highlight">1</span> | 30 ]</code></div>
<p>Matrix ab Row Echelon Form (REF) mein hai aur pivots 1 hain. Gauss Elimination yahan tak hota hai. Ab hum back-substitution karenge.</p>
<h2>Back-Substitution (Peeche se values nikalna)</h2>
<p>Matrix se equations wapas likhte hain:</p>
<div class="back-substitution">
Equation 1: x₁ + (1/2)x₂ + (1/3)x₃ = 1 <br>
Equation 2: x₂ + x₃ = -6 <br>
Equation 3: x₃ = 30
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<p><strong>Equation 3 se:</strong></p>
<p>x₃ = <strong>30</strong></p>
<p><strong>x₃ ki value Equation 2 mein daalo:</strong></p>
<p>x₂ + (30) = -6</p>
<p>x₂ = -6 - 30</p>
<p>x₂ = <strong>-36</strong></p>
<p><strong>x₂ aur x₃ ki values Equation 1 mein daalo:</strong></p>
<p>x₁ + (1/2)(-36) + (1/3)(30) = 1</p>
<p>x₁ - 18 + 10 = 1</p>
<p>x₁ - 8 = 1</p>
<p>x₁ = 1 + 8</p>
<p>x₁ = <strong>9</strong></p>
<h2>Hal (Solution)</h2>
<div class="solution">
x₁ = 9 <br>
x₂ = -36 <br>
x₃ = 30
</div>
<h2>Jaanch (Verification)</h2>
<p>Values ko original equations mein daal kar check karte hain:</p>
<h3>Original Equation 1: 6x₁ + 3x₂ + 2x₃ = 6</h3>
<p>6(9) + 3(-36) + 2(30) = 54 - 108 + 60 = 114 - 108 = <strong>6</strong> (Sahi hai!)</p>
<h3>Original Equation 2: 6x₁ + 4x₂ + 3x₃ = 0</h3>
<p>6(9) + 4(-36) + 3(30) = 54 - 144 + 90 = 144 - 144 = <strong>0</strong> (Sahi hai!)</p>
<h3>Original Equation 3: 20x₁ + 15x₂ + 12x₃ = 0</h3>
<p>20(9) + 15(-36) + 12(30) = 180 - 540 + 360 = 540 - 540 = <strong>0</strong> (Sahi hai!)</p>
<p>Solution bilkul sahi hai!</p>
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