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PlanetPhysics/Determinant

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In attempting to solve a linear system of equations for x1, x2 and x3

a1x1+b1x2+c1x3=d1a2x1+b2x2+c2x3=d2a3x1+b3x2+c3x3=d3

one is led in a very natural way to consider the square array

|a11a21a31a12a22a32a13a23a33|

We have written aij=aji,i,j=1,2,3. The solution of (1) requires that we attach a numerical value to the matrix of elements (2). We do this in the following way: We attach 33=27 numerical values to a set of ϵijk,i,j,k=1,2,3. If at least two of the superscripts in ϵijk are the same, the value of ϵijk is zero. Thus ϵ223=ϵ131=ϵ333=0, etc. If the i, j, k are all different, the value of ϵijk is to be +1 or −1 according to whether it takes an even or odd number of permutations to rearrange the ijk into the natural order 123. Let us lood at ϵ321 and hence at the arrangement 321. Permuting the integers 2 and 3 permutes 321 into 231, then permuting 3 and 1 permutes 231 into 213, and finally 213 permutes into 123 if we interchange the integers 2 and 1. Three (an odd number) permutations were required to permute 321 into 123. Thus ϵ123=−1. We have

ϵ123=ϵ312=ϵ231=+1 ϵ213=ϵ321=ϵ132=−1

We now define

|a11a21a31a12a22a32a13a23a33|≡ϵijkai1aj2ak3

The letters i, j, k are indices of summation. Equation (3) defines the determinant of the square matrix of elements (2) Its numerical value is given by the right-hand side of (3). It consists, in general, of 3!=3⋅2⋅1=6 terms, each term a product of three elements, one element from each row and column of (3). Expand (3) to get

ϵijkai1aj2ak3=(a11a22a33+a31a12a23+a21a32a13)−(a21a12a33+a31a22a13+a11a32a23)

Only 3!=6 terms occur in the exapansion of (3) since there are 3! permutations of 123. All other values of ϵijk are zero.

We can define ϵijk in exactly the same manner in which the ϵijk were defined. We leave it to the reader to show that

ϵijkai1aj2ak3=ϵijka1ia2ja3k

The generalization of second and third order determinants (the order of a determinant is the number of rows or columns of the determinant) to the nth order determinants is simple. We define the ϵi1i2⋯in to have the following numerical values: ϵi1i2⋯in=0 if at least two of the superscripts are the same. The values of the superscripts range from 1 to n. If the i1,i2,…,in are distinct, the value of ϵi1i2⋯in is to be +1 or −1 depending on whether an even or odd number of permutations is required to rearrange i1,i2,…,in into the natural order 123⋯n. The numerical value (determinant) of the square array of elements ‖aji‖, i,j=1,2,…,n, is defined as

|aji|=|a11a21…an1a12a22…an2…………a1na2n…ann|

|aji|=ϵi1i2⋯inai11ai22⋯ainn

|aji|=ϵi1i2⋯ina1i1a2i2⋯anin

where the ϵi1i2⋯in are defined in precisely the same manner in which the ϵi1i2⋯in are defined. In general, (4) consists of n! terms, each term a product of elements, one element fom each row and column of |aji|.

To facilitate writing, we shall deal with third order determinants, but it will be obvious to the reader that any theorem derived for third order determinants will apply to determinants of any finite order. Le us consider

Δ=|a11a21a31a12a22a32a13a23a33|=ϵijkai1aj2ak3=ϵijka1ia2ja3k

We can obtain a new third order determinant by interchanging the first and third row of Δ. This yields

Δ′=|a13a23a33a12a22a32a11a21a31|=ϵijkai3aj2ak1

But ϵijkai3aj2ak1=ϵijkak1aj2ai1=ϵkjiai1aj2ak3, since i amd k are dummy indices. We see that every term of (6) is the same as every term in (5) with the exception that ϵijk is replaced by ϵkji. Since ϵijk=−ϵkji, we conclude that Δ=−Δ′. we thus obtain the following theorems:

THEOREM 1.1 . Interchanging two rows (or columns) of a determinant changes the sign of the determinant.

THEOREM 1.1 . If two rows (or columns) of a determinant are the same, the value of the determinant is zero.

We note that

Δ″=|la11la21la31a12a22a32a13a23a33|=ϵijk(lai1)aj2ak3=lΔ

THEOREM 1.3 . If a row (or column) of a determinant is multiplied by a factor l, the value of the determinant is thereby multiplied by l.

Let us now investigate the determinant

Δ‴=|a11+la13a21+la23a3l+la33a12a22a32a13a23a33|

Δ‴=ϵijk(ai1+lai3)aj2ak3

Δ‴=ϵijkai1aj2ak3+lϵijkai3aj2ak3

Δ‴=Δ

since ϵijkai3aj2ak3=0 from Theorem (1.2). Hence we obtain the following theorem:

THEOREM 1.4 . The value of a determinant remains unchanged if to the elements of any row (or column) is added a scalar multiple of the corresponding elements of another row (or column).

The theorems derived above are very useful in evaluating a determinant.

References

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[1] Lass, Harry. "Elements of pure and applied mathematics" New York: McGraw-Hill Companies, 1957.

This entry is a derivative of the Public domain work [1].