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<title>Equations Solve Karna Gauss-Jordan Method Se</title>
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<h1>Linear Equations Ko Solve Karna</h1>
<h2>(a) Sawaal (Problem)</h2>
<p>Yeh equations ko Gauss-Jordan method ka istemal karke solve karo:</p>
<div class="equations">
x + 4y - z = -5
x + y - 6z = -12
3x - y - z = 4
</div>
<h2>Gauss-Jordan Elimination Ke Steps</h2>
<p>Sabse pehle, hum in equations ka augmented matrix (sanjojit avyuh) banayenge:</p>
<div class="matrix-display"><code>[ 1 4 -1 | -5 ]
[ 1 1 -6 | -12 ]
[ 3 -1 -1 | 4 ]</code></div>
<h3>Step 1: Pehle pivot ke neeche zeros banana</h3>
<p>Pehla pivot R1,C1 mein 1 hai. Iske neeche ke elements (R2,C1 aur R3,C1) ko zero karenge.</p>
<p class="operation">R2 → R2 - R1 (Row 2 mein se Row 1 ko minus karo)</p>
<p class="operation">R3 → R3 - 3*R1 (Row 3 mein se Row 1 ka 3 guna minus karo)</p>
<div class="matrix-display"><code>[ <span class="highlight">1</span> 4 -1 | -5 ]
[ 0 -3 -5 | -7 ] <-- (1-1=0, 1-4=-3, -6-(-1)=-5, -12-(-5)=-7)
[ 0 -13 2 | 19 ] <-- (3-3*1=0, -1-3*4=-13, -1-3*(-1)=2, 4-3*(-5)=19)</code></div>
<h3>Step 2: Dusra pivot (R2,C2) ko 1 banana</h3>
<p>Ab R2,C2 wale element (-3) ko 1 banana hai.</p>
<p class="operation">R2 → R2 / (-3) (Row 2 ko -3 se divide karo)</p>
<div class="matrix-display"><code>[ 1 4 -1 | -5 ]
[ 0 <span class="highlight">1</span> 5/3 | 7/3 ]
[ 0 -13 2 | 19 ]</code></div>
<h3>Step 3: 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 - 4*R2 (Row 1 mein se Row 2 ka 4 guna minus karo)</p>
<p class="operation">R3 → R3 + 13*R2 (Row 3 mein Row 2 ka 13 guna add karo)</p>
<div class="matrix-display"><code>[ 1 0 -23/3 | -43/3 ] <-- R1: [1, 4-4*1, -1-4*(5/3) | -5-4*(7/3)] = [1,0,-23/3|-43/3]
[ 0 1 5/3 | 7/3 ]
[ 0 0 71/3 | 148/3 ] <-- R3: [0, -13+13*1, 2+13*(5/3) | 19+13*(7/3)] = [0,0,71/3|148/3]</code></div>
<h3>Step 4: Teesra pivot (R3,C3) ko 1 banana</h3>
<p>Ab R3,C3 wale element (71/3) ko 1 banana hai.</p>
<p class="operation">R3 → R3 * (3/71) (Row 3 ko 3/71 se multiply karo)</p>
<div class="matrix-display"><code>[ 1 0 -23/3 | -43/3 ]
[ 0 1 5/3 | 7/3 ]
[ 0 0 <span class="highlight">1</span> | 148/71 ]</code></div>
<h3>Step 5: 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 + (23/3)*R3 (Row 1 mein Row 3 ka 23/3 guna add karo)</p>
<p class="operation">R2 → R2 - (5/3)*R3 (Row 2 mein se Row 3 ka 5/3 guna minus karo)</p>
<div class="matrix-display"><code>[ 1 0 0 | 117/71 ] <-- R1: [-23/3 + (23/3)*1 = 0], [-43/3 + (23/3)*(148/71) = 117/71]
[ 0 1 0 | -81/71 ] <-- R2: [5/3 - (5/3)*1 = 0], [7/3 - (5/3)*(148/71) = -81/71]
[ 0 0 1 | 148/71 ]</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 = 117/71 <br>
y = -81/71 <br>
z = 148/71
</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 + 4y - z = -5</h3>
<p>(117/71) + 4(-81/71) - (148/71) = (117 - 324 - 148) / 71 = (117 - 472) / 71 = -355 / 71 = <strong>-5</strong> (Sahi hai!)</p>
<h3>Equation 2: x + y - 6z = -12</h3>
<p>(117/71) + (-81/71) - 6(148/71) = (117 - 81 - 888) / 71 = (36 - 888) / 71 = -852 / 71 = <strong>-12</strong> (Sahi hai!)</p>
<h3>Equation 3: 3x - y - z = 4</h3>
<p>3(117/71) - (-81/71) - (148/71) = (351 + 81 - 148) / 71 = (432 - 148) / 71 = 284 / 71 = <strong>4</strong> (Sahi hai!)</p>
<p>Solution sahi hai.</p>
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