Question 378575
Add 5 times the second row to the first to get the new matrix



<pre>
1              0        7/13     | -12/13
0              1       -9/13     | -5/13
</pre>


Now the matrix is in reduced row echelon form (ie we can't do any more steps)



So the first line means that *[Tex \LARGE x+\frac{7}{13}z=-\frac{12}{13}]. Multiply EVERY term by the LCD 13 to get *[Tex \LARGE 13x+7z=-12] and then solve for 'x' to get  *[Tex \LARGE x=\frac{-7z-12}{13}]



The second line means that *[Tex \LARGE y-\frac{9}{13}z=-\frac{5}{13}]. Multiply EVERY term by the LCD 13 to get *[Tex \LARGE 13y-9z=-5] and then solve for 'y' to get  *[Tex \LARGE y=\frac{9z-5}{13}]



So the solution as an ordered triple is in the form



*[Tex \LARGE \left(\frac{-7z-12}{13} \ , \ \ \frac{9z-5}{13} \ , \ \ z\right)] where 'z' is any arbitrary real number.



Because 'z' can be any real number, this means that the system has an infinite number of solutions. 



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Jim