Talk:Reduced row echelon form: Difference between revisions
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The leading coefficient in the (3, 5)-th position is not the only nonzero element in its column. |
The leading coefficient in the (3, 5)-th position is not the only nonzero element in its column. |
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== Critique of... something? == |
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Hello RosettaCode ...<br> |
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Your algorithm/function does NOT work ...<br> |
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The "SWAP" is not being done. ...<br> |
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Try this matrix to see the problem. ...<br> |
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//===================================<br> |
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Linear Algebra with Applications by W. Keith Nicholson, University of Calgary<br> |
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page 12 - Systems of Linear Equations<br> |
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Solve the following system of equations.<br> |
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3x + y − 4z = −1<br> |
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x + 10z = 5<br> |
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4x + y + 6z = 1<br> |
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Solution. The corresponding augmented matrix is<br> |
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3 1 −4 −1<br> |
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1 0 10 5<br> |
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4 1 6 1<br> |
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Create the first leading one by interchanging rows 1 and 2<br> |
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1 0 01 5<br> |
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3 1 −4 −1<br> |
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4 1 6 1<br> |
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Now subtract 3 times row 1 from row 2, and subtract 4 times row 1 from row 3. The result is<br> |
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1 0 10 5<br> |
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0 1 −34 −16<br> |
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0 1 −34 −19<br> |
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Now subtract row 2 from row 3 to obtain<br> |
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1 0 10 5<br> |
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0 1 −34 −16<br> |
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0 0 0 −3<br> |
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This means that the following reduced system of equations<br> |
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x + 10z = 5<br> |
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y − 34z = −16<br> |
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0 +0 +0 = −3<br> |
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is equivalent to the original system. In other words, the two have the same solutions. <br> |
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But this last system clearly has no solution (the last equation requires that x, y and z satisfy 0x+0y+0z = −3,<br> |
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and no such numbers exist). <br> |
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Hence the original system has no solution.<big>Big text</big><br> |
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//=========================== |
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[[User:Umariani]] (Moved from main page by --[[User:Thundergnat|Thundergnat]] ([[User talk:Thundergnat|talk]]) 11:06, 20 July 2023 (UTC)) |