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<title>Gauss Elimination Method (Naya Sawaal)</title>
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<h1>Gauss Elimination Method Se Hal</h1>
<h2>(a) Sawaal (Problem Statement)</h2>
<p>Gauss Elimination method ka istemal karke yeh system of equations solve karo:</p>
<div class="equations">
10x + y + 2z = 13
3x + 10y + z = 14
2x + 3y + 10z = 15
</div>
<h2>Gauss Elimination Ke Steps</h2>
<p>Pehle, augmented matrix banate hain:</p>
<div class="matrix-display"><code>[ 10 1 2 | 13 ]
[ 3 10 1 | 14 ]
[ 2 3 10 | 15 ]</code></div>
<h3>Step 1: Pehla pivot (R1,C1) ko 1 banana</h3>
<p>Pehla element (R1,C1) 10 hai, isko 1 banana hai.</p>
<p class="operation">R1 → R1 / 10</p>
<div class="matrix-display"><code>[ <span class="highlight">1</span> 0.1 0.2 | 1.3 ] <span class="comment"><-- (10/10=1, 1/10=0.1, 2/10=0.2, 13/10=1.3)</span>
[ 3 10 1 | 14 ]
[ 2 3 10 | 15 ]</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 - 3*R1</p>
<p class="operation">R3 → R3 - 2*R1</p>
<div class="matrix-display"><code>[ 1 0.1 0.2 | 1.3 ]
[ 0 9.7 0.4 | 10.1 ] <span class="comment"><-- R2: [3-3*1, 10-3*0.1, 1-3*0.2 | 14-3*1.3] = [0, 9.7, 0.4 | 10.1]</span>
[ 0 2.8 9.6 | 12.4 ] <span class="comment"><-- R3: [2-2*1, 3-2*0.1, 10-2*0.2 | 15-2*1.3] = [0, 2.8, 9.6 | 12.4]</span></code></div>
<h3>Step 3: Dusra pivot (R2,C2) ko 1 banana</h3>
<p>Ab R2,C2 wala element 9.7 hai, isko 1 banana hai.</p>
<p class="operation">R2 → R2 / 9.7</p>
<div class="matrix-display"><code>[ 1 0.1 0.2 | 1.3 ]
[ 0 <span class="highlight">1</span> 0.4/9.7 | 10.1/9.7 ] <span class="comment"><-- (0.4/9.7 ≈ 0.041237, 10.1/9.7 ≈ 1.041237)</span>
[ 0 2.8 9.6 | 12.4 ]</code></div>
<p class="comment">Decimal values ko thoda round off karke likh sakte hain, ya fractions mein bhi rakh sakte hain accuracy ke liye. Hum yahan calculation mein exact values use karenge, display mein approximate.</p>
<div class="matrix-display"><code>[ 1 0.1 0.2 | 1.3 ]
[ 0 1 0.041237 | 1.041237 ] <span class="comment">(Approx values for display)</span>
[ 0 2.8 9.6 | 12.4 ]</code></div>
<h3>Step 4: Dusre pivot ke neeche zero banana</h3>
<p>Ab R2,C2 wale pivot (1) ke neeche (R3,C2) zero banana hai.</p>
<p class="operation">R3 → R3 - 2.8*R2</p>
<div class="matrix-display"><code>[ 1 0.1 0.2 | 1.3 ]
[ 0 1 0.041237 | 1.041237 ]
[ 0 0 9.484536 | 9.484536 ] <span class="comment"><-- R3_C3: 9.6 - 2.8*(0.4/9.7) = (9.6*9.7 - 2.8*0.4)/9.7 = (93.12 - 1.12)/9.7 = 92/9.7 ≈ 9.484536</span>
<span class="comment"> R3_Const: 12.4 - 2.8*(10.1/9.7) = (12.4*9.7 - 2.8*10.1)/9.7 = (120.28 - 28.28)/9.7 = 92/9.7 ≈ 9.484536</span></code></div>
<p>Matrix ab Row Echelon Form (REF) mein hai. Ab teesre pivot ko 1 banana hai.</p>
<h3>Step 5: Teesra pivot (R3,C3) ko 1 banana</h3>
<p>R3,C3 mein ab 92/9.7 hai.</p>
<p class="operation">R3 → R3 / (92/9.7) (Yaani R3 * (9.7/92))</p>
<div class="matrix-display"><code>[ 1 0.1 0.2 | 1.3 ]
[ 0 1 0.041237 | 1.041237 ] <span class="comment">(0.4/9.7 | 10.1/9.7)</span>
[ 0 0 <span class="highlight">1</span> | 1 ]</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 (accurate fractions use karte hue):</p>
<div class="back-substitution">
Equation 1: x + (1/10)y + (2/10)z = 13/10 <br>
Equation 2: y + (4/97)z = 101/97 <br>
Equation 3: z = 1
</div>
<p><strong>Equation 3 se:</strong></p>
<p>z = <strong>1</strong></p>
<p><strong>z ki value Equation 2 mein daalo:</strong></p>
<p>y + (4/97)(1) = 101/97</p>
<p>y = 101/97 - 4/97</p>
<p>y = 97/97</p>
<p>y = <strong>1</strong></p>
<p><strong>y aur z ki values Equation 1 mein daalo:</strong></p>
<p>x + (1/10)(1) + (2/10)(1) = 13/10</p>
<p>x + 1/10 + 2/10 = 13/10</p>
<p>x + 3/10 = 13/10</p>
<p>x = 13/10 - 3/10</p>
<p>x = 10/10</p>
<p>x = <strong>1</strong></p>
<h2>Hal (Solution)</h2>
<div class="solution">
x = 1 <br>
y = 1 <br>
z = 1
</div>
<h2>Jaanch (Verification)</h2>
<p>Values ko original equations mein daal kar check karte hain:</p>
<h3>Original Equation 1: 10x + y + 2z = 13</h3>
<p>10(1) + (1) + 2(1) = 10 + 1 + 2 = <strong>13</strong> (Sahi hai!)</p>
<h3>Original Equation 2: 3x + 10y + z = 14</h3>
<p>3(1) + 10(1) + (1) = 3 + 10 + 1 = <strong>14</strong> (Sahi hai!)</p>
<h3>Original Equation 3: 2x + 3y + 10z = 15</h3>
<p>2(1) + 3(1) + 10(1) = 2 + 3 + 10 = <strong>15</strong> (Sahi hai!)</p>
<p>Solution bilkul sahi hai! Ekdum perfect!</p>
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