Find the number of different matrices with determinant value 0, such that each of its entries is either or , but there are no two rows that are identical.
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This problem is really an interesting example to show that we need not do algebra problems by algebra alone, or geometry problems by geometry alone, and that mathematics as a whole is a subject and there can really be no sharp boundaries between the disciplines. I love how this problem can be posed in an algebraic way, interpreted with the help of vectors, visualized with the help of 3-D geometry, and answered with the help of combinatorics.
We note that any such matrix, is the scalar triple product of vectors, a , b , c where the x,y,z-components of the three vectors are +1 or -1.
But the scalar triple product of three vectors is the volume of the parallelopiped with three adjacent sides along the three vectors, and if the volume is 0, that means that the 3 vectors are coplanar.
Any point whose position vector has x,y,z-components +1 or -1 is a vertex of the cube of side length 2 units,centred at the origin and sides parallel to the axes.
So we are looking for 3 coplanar vectors among the 8 such vectors possible. Taking the origin as O, we observe that any triple of 3 coplanar vectors from the origin to the vertices will lie on the plane (containing 4 vectors) determined by two opposite face diagonals and the corresponding sides as shown below.
Now such a plane can be chosen in 6 distinct ways, 2 for each pair of opposite faces, and from the 4 vectors on each plane, 3 vectors can be chosen in 4 C 3 = 4 ways, so the total number of ways of choosing three such vectors is 4 ∗ 6 = 2 4 ways.
But in the matrix the 3 vectors can be arranged from top to bottom in 3 ! = 6 different ways, so the answer is 2 4 ∗ 6 = 1 4 4 .