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<title>Dusra Sawaal: Equations Solve Karna Gauss-Jordan Method Se</title>
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<h1>Linear Equations Ko Solve Karna (Part 2)</h1>
<h2>(a) Sawaal (Problem Statement)</h2>
<p>Gauss-Jordan method ka istemal karke yeh equations solve karo:</p>
<div class="equations">
2x - 6y + 8z = 24
5x + 4y - 3z = 2
3x + y + 2z = 16
</div>
<h2>Gauss-Jordan Elimination Ke Steps</h2>
<p>Sabse pehle, in equations ka augmented matrix banayenge:</p>
<div class="matrix-display"><code>[ 2 -6 8 | 24 ]
[ 5 4 -3 | 2 ]
[ 3 1 2 | 16 ]</code></div>
<h3>Step 1: Pehla pivot (R1,C1) ko 1 banana</h3>
<p>Pehla element (R1,C1) abhi 2 hai, isko 1 banana hai.</p>
<p class="operation">R1 → R1 / 2 (Row 1 ko 2 se divide karo)</p>
<div class="matrix-display"><code>[ <span class="highlight">1</span> -3 4 | 12 ]
[ 5 4 -3 | 2 ]
[ 3 1 2 | 16 ]</code></div>
<h3>Step 2: Pehle pivot ke neeche zeros banana</h3>
<p>Ab R1,C1 wale pivot (1) ke neeche ke elements (R2,C1 aur R3,C1) ko zero karenge.</p>
<p class="operation">R2 → R2 - 5*R1</p>
<p class="operation">R3 → R3 - 3*R1</p>
<div class="matrix-display"><code>[ 1 -3 4 | 12 ]
[ 0 19 -23 | -58 ] <span class="comment"><-- R2: [5-5*1, 4-5*(-3), -3-5*4 | 2-5*12] = [0, 19, -23 | -58]</span>
[ 0 10 -10 | -20 ] <span class="comment"><-- R3: [3-3*1, 1-3*(-3), 2-3*4 | 16-3*12] = [0, 10, -10 | -20]</span></code></div>
<h3>Step 3: Dusra pivot (R2,C2) ko 1 banana (Thoda Smart Work)</h3>
<p>Dekho, Row 3 (R3) ko 10 se divide karke simplify kar sakte hain:</p>
<p class="operation">R3 → R3 / 10</p>
<div class="matrix-display"><code>[ 1 -3 4 | 12 ]
[ 0 19 -23 | -58 ]
[ 0 1 -1 | -2 ] <span class="comment"><-- Simplified R3</span></code></div>
<p>Ab R2 aur R3 ko swap (badal) kar lete hain taaki R2,C2 mein 1 aa jaaye.</p>
<p class="operation">R2 ↔ R3</p>
<div class="matrix-display"><code>[ 1 -3 4 | 12 ]
[ 0 <span class="highlight">1</span> -1 | -2 ]
[ 0 19 -23 | -58 ]</code></div>
<p>Ab R2,C2 wala pivot 1 ho gaya!</p>
<h3>Step 4: Dusre pivot ke upar aur neeche zeros banana</h3>
<p>Ab R2,C2 wale pivot (1) ke upar (R1,C2) aur neeche (R3,C2) zero banana hai.</p>
<p class="operation">R1 → R1 + 3*R2</p>
<p class="operation">R3 → R3 - 19*R2</p>
<div class="matrix-display"><code>[ 1 0 1 | 6 ] <span class="comment"><-- R1: [1, -3+3*1, 4+3*(-1) | 12+3*(-2)] = [1, 0, 1 | 6]</span>
[ 0 1 -1 | -2 ]
[ 0 0 -4 | -20 ] <span class="comment"><-- R3: [0, 19-19*1, -23-19*(-1) | -58-19*(-2)] = [0, 0, -4 | -20]</span></code></div>
<h3>Step 5: Teesra pivot (R3,C3) ko 1 banana</h3>
<p>Ab R3,C3 wale element (-4) ko 1 banana hai.</p>
<p class="operation">R3 → R3 / (-4)</p>
<div class="matrix-display"><code>[ 1 0 1 | 6 ]
[ 0 1 -1 | -2 ]
[ 0 0 <span class="highlight">1</span> | 5 ]</code></div>
<h3>Step 6: Teesre pivot ke upar zeros banana</h3>
<p>Ab R3,C3 wale pivot (1) ke upar (R1,C3 aur R2,C3) zero banana hai.</p>
<p class="operation">R1 → R1 - R3</p>
<p class="operation">R2 → R2 + R3</p>
<div class="matrix-display"><code>[ 1 0 0 | 1 ] <span class="comment"><-- R1: [1-0, 0-0, 1-1 | 6-5] = [1, 0, 0 | 1]</span>
[ 0 1 0 | 3 ] <span class="comment"><-- R2: [0+0, 1+0, -1+1 | -2+5] = [0, 1, 0 | 3]</span>
[ 0 0 1 | 5 ]</code></div>
<p>Yeh matrix ab Reduced Row Echelon Form (RREF) mein hai.</p>
<h2>Hal (Solution)</h2>
<p>RREF matrix se humein solution milta hai:</p>
<div class="solution">
x = 1 <br>
y = 3 <br>
z = 5
</div>
<h2>Jaanch (Verification)</h2>
<p>Ab x, y, aur z ki values ko original equations mein daal kar check karte hain:</p>
<h3>Equation 1: 2x - 6y + 8z = 24</h3>
<p>2(1) - 6(3) + 8(5) = 2 - 18 + 40 = -16 + 40 = <strong>24</strong> (Sahi hai!)</p>
<h3>Equation 2: 5x + 4y - 3z = 2</h3>
<p>5(1) + 4(3) - 3(5) = 5 + 12 - 15 = 17 - 15 = <strong>2</strong> (Sahi hai!)</p>
<h3>Equation 3: 3x + y + 2z = 16</h3>
<p>3(1) + (3) + 2(5) = 3 + 3 + 10 = 6 + 10 = <strong>16</strong> (Sahi hai!)</p>
<p>Solution bilkul sahi hai!</p>
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