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# Determinants

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#### Determinants - Lesson Summary

Consider the system of linear equations:

l1x + m1y = p l2x + m2y = q

If l1/l2 ≠ m1/m2  the system of equations can have a unique solution.

If l1/l2 = m1/m2 ≠ p/q the system of equations can have a no solution.

If l1/l2 = m1/m2 = p/q the system of equations can have infinitely many solution.

A system of linear equations can be expressed in the form of a matrix.

The given system can be represented in matrix form as: l 1 m 1 l 2 m 2   x y   = p q

Consider the 2x2 matrix, l 1 m 1 l 2 m 2

The real number, l1m2 - l2m1, which is associated with the matrix l 1 m 1 l 2 m 2 , can be obtained by

multiplying the diagonal elements and then subtracting the products.

Determinant of matrix l 1 m 1 l 2 m 2   = l1m2 – l2m1

Let A =Set of square matrices B =Set of numbers (real and complex)

Let a function, f: A → B, be defined by f(X) = b such that X ∈ A and b ∈ B. Then f(X) is called the determinant of X.

It is denoted by |A|, det(A) or Δ, and is read as determinant of A.

If A =  a 1 b 1 a 2 b 2   , then |A| = det A = a 1 b 1 a 2 b 2 .

Note:

Modulus exists only for real numbers not for matrices. Hence, |A| is read as determinant of A, but not as modulus of A.

Determinants exist only for square matrices.

Determinant of a matrix of order one

Let A = [a11]

Determinant of matrix A = |a11| = a11

Determinant of a matrix of order two

Let  A = a 11 a 12 a 21 a 22

Determinant of matrix A = a 11 a 12 a 21 a 22 = a11a22 - a12a21

Determinant of a matrix of order three

Consider the general form of a 3x3 matrix, a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33

The determinant of a third order matrix can be determined by expanding it along a row or a column in terms of a second order determinant.This is known as the expansion of a determinant.

The determinant of matrix A can be calculated in 6 ways. They are:

(i) Expanding the determinant along R1 (1st row)

(ii) Expanding the determinant along R2 (2nd row)

(iii) Expanding the determinant along R3 (3rd row)

(iv) Expanding the determinant along C1 (1st column)

(v) Expanding the determinant along C2 (2nd column)

(vi) Expanding the determinant along C3 (3rd column)

Expanding the determinant along R1 (1st row):

Step1: Multiply the first element, i.e. a11, by (-1)1+1, and then by the second order determinant obtained by deleting the elements of the first row and the first column of matrix A.

The product corresponding is  (-1) 1+1 a 11 a 22 a 23 a 32 a 33 .

Step2: Multiply the second element, a12, by (-1)1+2 and then by the second order determinant obtained by deleting the entries of the first row and the second column.

The product is   (-1) 1+2 a 12 a 21 a 23 a 31 a 33 .

Step3: Multiply the third element, a13, by (-1)1+3 and then by the second order determinant obtained by deleting the entries of the first row and the third column.

The product is   (-1) 1+3 a 13 a 21 a 22 a 31 a 32

Step 4: The sum of the results of these three steps is the required determinant of A.

|A| =  (-1) 1+1 a 11 a 22 a 23 a 32 a 33 +   (-1) 1+2 a 12 a 21 a 23 a 31 a 33 +   (-1) 1+3 a 13 a 21 a 22 a 31 a 32

= a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31)

= a11a22a33 - a11a23a32 - a12a21a33 + a12a23a31 + a13a21a32 - a13a22a31

Expanding the determinant along R2 ( 2nd row):

Expanding along R2, we get

|A| =   (-1) 2+1 a 21 a 12 a 13 a 32 a 33 +   (-1) 2+2 a 22 a 11 a 13 a 31 a 33 +   (-1) 2+3 a 23 a 11 a 12 a 31 a 32

|A| = - a 21 a 12 a 13 a 32 a 33 + a 22 a 11 a 13 a 31 a 33 - a 23 a 11 a 12 a 31 a 32

= -a21(a12a33 - a13a32) + a22(a11a33 - a13a31) - a23(a11a32 - a12a31)

= -a21a12a33 + a21a13a32 + a22a11a33 - a22a13a31 - a23a11a32 + a23a12a31)

Expanding the determinant along R3 ( 3rd row):

Expanding along R3, we get

|A| =   (-1) 3+1 a 31 a 12 a 13 a 22 a 23 +   (-1) 3+2 a 32 a 11 a 13 a 21 a 23 +   (-1) 3+3 a 33 a 11 a 12 a 21 a 22

|A| =   a 31 a 12 a 13 a 22 a 23 – a 32 a 11 a 13 a 21 a 23 + a 23 a 11 a 12 a 21 a 22

= a31(a12a23 - a13a22) - a32(a11a23 - a13a21) - a33(a11a22 - a12a21)

|A| = a11a22a33 - a11a23a32 - a12a21a33 + a12a23a31 + a13a21a32 - a13a22a31

Expanding the determinant along C1 ( 1st column):

Expanding along C1, we get

|A| =   (-1) 1+1 a 11 a 22 a 23 a 32 a 33 +   (-1) 2+1 a 21 a 12 a 13 a 32 a 32 +   (-1) 3+1 a 31 a 12 a 13 a 22 a 23

|A| =   a 11 a 22 a 23 a 32 a 32 – a 21 a 12 a 13 a 32 a 33 + a 31 a 12 a 13 a 22 a 23

= a11(a22a33 - a23a32) - a21(a12a33 - a13a32) + a31(a12a23 - a13a22)

|A| = a11a22a33 - a11a23a32 - a21a12a33 + a21a13a32 + a31a12a23 - a31a13a22

Similarly the determinant of A will be the same along columns C 2 and C3.

Some important points:

(i) If any row (or column) contains maximum number of zero(s), then finding the determinant along that row (or column) makes the calculation easier.

(ii) If all the elements in a row (or a column) are zeros, then the determinant of that matrix is zero.

(iii) If M and N are two square matrices such that M = kN, where k is a real number, then |M| = kn|N|, where n is the order of matrices M and N.

(iv) At the time of expansion, aij can be multiplied by +1 or -1, depending upon whether (i + j) is even or odd, instead of multiplying it by (-1)i + j.