Replacing row one by does not change the determinant: adding a multiple of another row preserves a determinant. It produces first row , whose cofactor minor is the three-dimensional identity matrix. Thus the determinant is .
The columns are homogeneous coordinate vectors of the four vertices. Subtracting the first column from each other column preserves the determinant and reduces its magnitude to the absolute determinant of the three edge vectors based at . That absolute determinant is the parallelepiped volume. A tetrahedron occupies one sixth of that parallelepiped, so in cubic coordinate units.
A unit vector along the line is . Reflection preserves the orthogonal projection onto that line and negates its perpendicular component. Therefore andAs checks, , , and . Its determinant is minus one, as expected for a planar reflection matrix.
Gaussian elimination gives the row operations , , , producing the lower three rows , , . Subtract the new row two from row three and add three times row two to row four; the remaining nonzero row is . Move it above the zero row and finish elimination to obtainThe pivot columns are one, two and three. Row operations preserve the linear dependence relations among columns, so the corresponding columns of the original matrix form a basis for the column space, which is the image of the linear map:In particular, the matrix rank is three. Columns of the reduced matrix itself would generally give a different subspace, so the original columns must be used.
The reduced row echelon form gives , , and . Taking gives every vector in the null space as . Thus a basis for the kernel of the linear map isDirect multiplication gives . The rank-nullity theorem also checks the count: matrix rank three plus nullity one equals the domain dimension four.
No. The nonzero vector is in the kernel of the linear map, so although . A linear map is injective exactly when its kernel consists only of zero.
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