= Solution
Replacing row one by $R_1-R_2-R_3-R_4$ does not change the <determinant>: adding a multiple of another row preserves a <determinant>. It produces first row $(-2,0,0,0)$, whose <cofactor> minor is the three-dimensional <identity matrix>. Thus \b[the <determinant> is $-2$].
The columns are homogeneous coordinate <vectors> $(1,v_i)^\top$ 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 $v_1$. That absolute <determinant> is the <parallelepiped> <volume>. A <tetrahedron> occupies one sixth of that <parallelepiped>, so \b[$V=|-2|/3!=1/3$] in cubic coordinate units.
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