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Determinants

Module by: Thanos Antoulas. E-mail the author

Summary: (Blank Abstract)

Say that we have the following matrix and that we want to find its determinant.

A=( a 1 1 a 1 2 … a 1 n a 2 1 a 2 2 … a 2 n ⋮⋮⋱⋮ a n 1 a n 2 … a n n ) A a 1 1 a 1 2 … a 1 n a 2 1 a 2 2 … a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 … a n n
(1)

Calculating the determinant of a matrix is a recursive process. Basically, we start by choosing any one row or column. The determinant will then be found with respect to this row or column. What this means is that we will find a sum of the products of this row or column's values and sub-determinants formed by blocking out the row and column of the particular value.

Why is this choice of row or column left to us instead of always being defined as, say, the first row? The reason is that by choosing this row or column wisely, we can sometimes reduce the amount of work we do. For example, if a certain row or column contains a few zeros, choosing it as the row/column that we take the determinant with respect to would be a smart move. As the values of this chosen row or column will be multiplied by sub-determinants of the matrix in question, a value of 0 in one of these products would mean that we have one less matrix whose determinant we need to calculate.

In the case of the matrix above, we'll compute the determinant with respect to the first column. The final equation for the determinant is:

detA= a 1 1 ⁢-11+1⁢det A 1 1 + a 2 1 ⁢-12+1⁢det A 1 2 +…+ a n 1 ⁢-1n+1⁢det A 1 n A a 1 1 -1 11 A 1 1 a 2 1 -1 21 A 1 2 … a n 1 -1 n1 A 1 n
(2)

Here, A i j A i j means the matrix formed by eliminating the ii-the column and the jj-th row of AA.

Let's just look at the first term in Equation 2. It is basically the first element of AA's first column times the determinant of the matrix formed by the elimination of the first row and first column of AA. There is also a (-1)r+c (-1) r c term included. This serves to make the signs of all of the terms in the determinant equation fluctuate back and forth. The next term is the same, except that we have moved on to the second element in the first column of AA. As this element holds a position in the second row and first column of AA, the sub-determinant in this term is obtained by hiding the second row and first column of AA.

In a generic 3×3 3 3 example, we would find the following solution for the determinant:

det( a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 )= a 1 1 ⁢det( a 2 2 a 2 3 a 3 2 a 3 3 )− a 2 1 ⁢det( a 1 2 a 1 3 a 3 2 a 3 3 )+ a 3 1 ⁢det( a 1 2 a 1 3 a 2 2 a 2 3 ) a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 a 1 1 a 2 2 a 2 3 a 3 2 a 3 3 a 2 1 a 1 2 a 1 3 a 3 2 a 3 3 a 3 1 a 1 2 a 1 3 a 2 2 a 2 3
(3)

To find the determinants of the 2×2 2 2 sub-determinants, we could again apply the rule of Equation 2, keeping in mind that the determinant of a scalar value is simply that scalar value. However, it is easier to remember the following solution

det( ab cd )=a⁢d−b⁢c a b c d ad bc
(4)

Example 1

To clarify, take the following example of finding the determinant of a numeric 2×2 2 2 matrix.

A=( 1-12 311 -2-20 ) A 1 -1 2 3 1 1 -2 -2 0
(5)

First we need to choose a row or column to take the determinant with respect to. We notice that the element in the third row and third column is a zero. Knowing that choosing a row or column that contains a zero will reduce our workload, we will choose the third column. Then, by applying Equation 2, we get

detA=2×-14⁢det( 31 -2-2 )+1×-15⁢det( 1-1 -2-2 )+0 A 2 -1 4 3 1 -2 -2 1 -1 5 1 -1 -2 -2 0
(6)

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