for what values of k does the system 2x-y+4z = 8 3x-ky+z=1 5x-y+kz=2 have a unique solution?
R rrbandali New member Joined Mar 9, 2012 Messages 1 Mar 9, 2012 #1 for what values of k does the system 2x-y+4z = 8 3x-ky+z=1 5x-y+kz=2 have a unique solution?
pka Elite Member Joined Jan 29, 2005 Messages 11,990 Mar 9, 2012 #2 rrbandali said: for what values of k does the system 2x-y+4z = 8 3x-ky+z=1 5x-y+kz=2 have a unique solution? Click to expand... Here is the determinate: \(\displaystyle \left| {\begin{array}{rrr} 2&{ - 1}&4 \\ 3&{ - k}&1 \\ 5&{ - 1}&k \end{array}} \right| = - 2{k^2} + 23k - 15\)
rrbandali said: for what values of k does the system 2x-y+4z = 8 3x-ky+z=1 5x-y+kz=2 have a unique solution? Click to expand... Here is the determinate: \(\displaystyle \left| {\begin{array}{rrr} 2&{ - 1}&4 \\ 3&{ - k}&1 \\ 5&{ - 1}&k \end{array}} \right| = - 2{k^2} + 23k - 15\)
