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<title>Gauss-Jordan Method Se Equations Solve Karna (x,y,z)</title>
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<h1>Gauss-Jordan Method (x,y,z Variables)</h1>
<h2>(a) Sawaal (Problem Statement)</h2>
<p>Gauss-Jordan method ka istemal karke yeh equations solve karo:</p>
<div class="equations">
x + 2y + z = 3
2x + 3y + 3z = 10
3x - y + 2z = 13
</div>
<h2>Gauss-Jordan Elimination Ke Steps</h2>
<p>Sabse pehle, augmented matrix banayenge:</p>
<div class="matrix-display"><code>[ 1 2 1 | 3 ]
[ 2 3 3 | 10 ]
[ 3 -1 2 | 13 ]</code></div>
<p>Pehla pivot (R1,C1) already 1 hai, bahut accha!</p>
<h3>Step 1: Pehle pivot ke neeche zeros banana</h3>
<p class="operation">R2 β†’ R2 - 2*R1</p>
<p class="operation">R3 β†’ R3 - 3*R1</p>
<div class="matrix-display"><code>[ <span class="highlight">1</span> 2 1 | 3 ]
[ 0 -1 1 | 4 ] <span class="comment"><-- R2: [2-2*1, 3-2*2, 3-2*1 | 10-2*3] = [0, -1, 1 | 4]</span>
[ 0 -7 -1 | 4 ] <span class="comment"><-- R3: [3-3*1, -1-3*2, 2-3*1 | 13-3*3] = [0, -7, -1 | 4]</span></code></div>
<h3>Step 2: Dusra pivot (R2,C2) ko 1 banana</h3>
<p>Ab R2,C2 wale element (-1) ko 1 banana hai.</p>
<p class="operation">R2 β†’ R2 * (-1)</p>
<div class="matrix-display"><code>[ 1 2 1 | 3 ]
[ 0 <span class="highlight">1</span> -1 | -4 ]
[ 0 -7 -1 | 4 ]</code></div>
<h3>Step 3: Dusre pivot ke upar aur neeche zeros banana</h3>
<p class="operation">R1 β†’ R1 - 2*R2</p>
<p class="operation">R3 β†’ R3 + 7*R2</p>
<div class="matrix-display"><code>[ 1 0 3 | 11 ] <span class="comment"><-- R1: [1-2*0, 2-2*1, 1-2*(-1) | 3-2*(-4)] = [1, 0, 3 | 11]</span>
[ 0 1 -1 | -4 ]
[ 0 0 -8 | -24 ] <span class="comment"><-- R3: [0+7*0, -7+7*1, -1+7*(-1) | 4+7*(-4)] = [0, 0, -8 | -24]</span></code></div>
<h3>Step 4: Teesra pivot (R3,C3) ko 1 banana</h3>
<p>Ab R3,C3 wale element (-8) ko 1 banana hai.</p>
<p class="operation">R3 β†’ R3 / (-8)</p>
<div class="matrix-display"><code>[ 1 0 3 | 11 ]
[ 0 1 -1 | -4 ]
[ 0 0 <span class="highlight">1</span> | 3 ]</code></div>
<h3>Step 5: Teesre pivot ke upar zeros banana</h3>
<p class="operation">R1 β†’ R1 - 3*R3</p>
<p class="operation">R2 β†’ R2 + R3</p>
<div class="matrix-display"><code>[ 1 0 0 | 2 ] <span class="comment"><-- R1: [1-3*0, 0-3*0, 3-3*1 | 11-3*3] = [1, 0, 0 | 2]</span>
[ 0 1 0 | -1 ] <span class="comment"><-- R2: [0+0, 1+0, -1+1 | -4+3] = [0, 1, 0 | -1]</span>
[ 0 0 1 | 3 ]</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 = 2 <br>
y = -1 <br>
z = 3
</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: x + 2y + z = 3</h3>
<p>(2) + 2(-1) + (3) = 2 - 2 + 3 = 0 + 3 = <strong>3</strong> (Sahi hai!)</p>
<h3>Equation 2: 2x + 3y + 3z = 10</h3>
<p>2(2) + 3(-1) + 3(3) = 4 - 3 + 9 = 1 + 9 = <strong>10</strong> (Sahi hai!)</p>
<h3>Equation 3: 3x - y + 2z = 13</h3>
<p>3(2) - (-1) + 2(3) = 6 + 1 + 6 = 7 + 6 = <strong>13</strong> (Sahi hai!)</p>
<p>Solution bilkul sahi hai! Ekdum mast!</p>
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