**Introduction**

Let us consider the
2x2 matrix given on the left. Its determinant as per the rule given above is AD-BC.

However, look at the
diagram given right. It can be explained as follows.

**Rotation Method**

Rotate left the
second row to one position and then multiply the entries in the respective
columns. Subtract the second column product from that of the first.

This process can be
extended into higher orders also. This
is done by carefully completing all the rotations depending on the number of
rows. Let us see the case of a 3x3 matrix.

At first, rotate left
the second row to one position and third row to two positions. The matrix thus
obtained is given on the right.

Thus we get A= (A1.B2.C3)+(A2.B3.C1)+(A3.B1.C2).

Now we rotate left
the second row to one more position and the third row to two more positions.
The matrix thus obtained is as follows.

Multiply the values
in each column separately and then add the three products. Thus we get B=(A1.B3.C2)+(A2.B1.C3)+(A3.B2.C1).

The determinant of
the 3x3 matrix is then

A-B = (A1.B2.C3+A2.B3.C1+A3.B1.C2)-( A1.B3.C2+A2.B1.C3+A3.B2.C1).

**Cylindrical Rotation Method**

This process is
better understood if we can represent the matrix cylindrically. Consider a
cylinder that has three horizontal sections which can be rotated freely with
respect to a central vertical axis. Entries in the matrix are given on the
exterior of the cylinder.

After the first set
of two types of left rotations, the cylinder looks like what is on the left.

Multiplying the
entries in the columns separately we get

A=(A1.B2.C3)+(A2.B3.C1)+(A3.B1.C2).

After the second set
of two types of left rotations, the cylinder looks like what is on the right.

B = (A1.B3.C2)+(A2.B1.C3)+(A3.B2.C1).

The determinant of
the 3x3 matrix is then

A-B = (A1.B2.C3+A2.B3.C1+A3.B1.C2)-( A1.B3.C2+A2.B1.C3+A3.B2.C1).

**Palm Method**

This can be
visualized in yet another way also.

Let us use the three
central fingers on the left palm to represent a 3x3 matrix.

Instead of the first
set of rotations, multiply the entries from left as indicated by the dark
lines, starting from the diagonal to get the value A.

**Conclusion**

An advantage of the
above mentioned process is the elimination of the repeated use of plus(+) and minus(–)
which is sometimes disturbing for beginners and non-Mathematics students. Can this method be extended to higher order matrices?