Find all the values of k so that the set{ ( 1,−3,2),(−3,9,−6),(5,−7,k)} form the basis for R³
For a set to form a basis in R3 it is sufficient that the determinant "det \\begin{vmatrix}\n 1 & -3 & 2 \\\\\n -3 & 9 & -6 \\\\\n 5 & -7 & k\n\\end{vmatrix}" was not equal to zero.
"det \\begin{vmatrix}\n 1 & -3 & 2 \\\\\n -3 & 9 & -6 \\\\\n 5 & -7 & k\n\\end{vmatrix}\n= 9\\cdot k+ 2\\cdot (-3)\\cdot (-7) + (-3)\\cdot (-6)\\cdot 5 - 2\\cdot 9 \\cdot 5 - (-3) \\cdot (-3) \\cdot k - 1\\cdot (-6)\\cdot (-7) = 0"
So there is no such k that these three vectors form a basis.
It is clear from the fact, that the first two vectors are linear dependent as 3*(1, -3, 2) + 1*(-3,9, -6) = 0
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