Chapter 3

Determinants

3.1 Basic Techniques and Properties

Let A be an n × n matrix. The determinant of A, denoted as det (A) is a number. If thematrix is a 2×2 matrix, this number is very easy to find.

Definition 3.1.1 Let A =

(a b

c d

). Then

det (A) ≡ ad− cb.

The determinant is also often denoted by enclosing the matrix with two vertical lines. Thus

det

(a b

c d

)=

∣∣∣∣∣ a b

c d

∣∣∣∣∣ .Example 3.1.2 Find det

(2 4

−1 6

).

From the definition this is just (2) (6)− (−1) (4) = 16.Assuming the determinant has been defined for k × k matrices for k ≤ n − 1, it is now

time to define it for n× n matrices.

Definition 3.1.3 Let A = (aij) be an n×n matrix. Then a new matrix called the cofactormatrix, cof (A) is defined by cof (A) = (cij) where to obtain cij delete the ith row and thejth column of A, take the determinant of the (n− 1)× (n− 1) matrix which results, (This

is called the ijth minor of A. ) and then multiply this number by (−1)i+j

. To make theformulas easier to remember, cof (A)ij will denote the ijth entry of the cofactor matrix.

Now here is the definition of the determinant given recursively.

Theorem 3.1.4 Let A be an n× n matrix where n ≥ 2. Then

det (A) =

n∑j=1

aij cof (A)ij =

n∑i=1

aij cof (A)ij . (3.1)

The first formula consists of expanding the determinant along the ith row and the secondexpands the determinant along the jth column.

Note that for a n× n matrix, you will need n! terms to evaluate the determinant in thisway. If n = 10, this is 10! = 3, 628 , 800 terms. This is a lot of terms.

In addition to the difficulties just discussed, why is the determinant well defined? Whyshould you get the same thing when you expand along any row or column? I think youshould regard this claim that you always get the same answer by picking any row or columnwith considerable skepticism. It is incredible and not at all obvious. However, it requiresa little effort to establish it. This is done in the section on the theory of the determinantwhich follows.

Notwithstanding the difficulties involved in using the method of Laplace expansion,certain types of matrices are very easy to deal with.

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Chapter 3Determinants3.1 Basic Techniques and PropertiesLet A be an n x n matrix. The determinant of A, denoted as det (A) is a number. If thematrix is a 2X2 matrix, this number is very easy to find.Definition 3.1.1 Let A= ( “ ; . ThenCcdet (A) = ad — cb.The determinant is also often denoted by enclosing the matrix with two vertical lines. Thusb bdet ( “ ) -|°c dc d; 2 4Example 3.1.2 Find det ( .—1 6From the definition this is just (2) (6) — (—1) (4) = 16.Assuming the determinant has been defined for k x k matrices for k <n —1, it is nowtime to define it for n x n matrices.Definition 3.1.3 Let A = (a;;) be ann xn matrix. Then a new matrix called the cofactormatrix, cof (A) is defined by cof (A) = (ci;) where to obtain c;; delete the i” row and thei” column of A, take the determinant of the (n—1) x (n—1) matrix which results, (Thisis called the ijt” minor of A. ) and then multiply this number by (-1)'*9, To make theformulas easier to remember, cof (A); will denote the ij” entry of the cofactor matrix.Now here is the definition of the determinant given recursively.Theorem 3.1.4 Let A be ann X n matrix where n > 2. Thendet (A) = S° ai; cof (A); = © aij cof (A); - (3.1)j=l i=lThe first formula consists of expanding the determinant along the it’ row and the secondexpands the determinant along the j*” column.Note that for an x n matrix, you will need n! terms to evaluate the determinant in thisway. If n = 10, this is 10! = 3,628,800 terms. This is a lot of terms.In addition to the difficulties just discussed, why is the determinant well defined? Whyshould you get the same thing when you expand along any row or column? I think youshould regard this claim that you always get the same answer by picking any row or columnwith considerable skepticism. It is incredible and not at all obvious. However, it requiresa little effort to establish it. This is done in the section on the theory of the determinantwhich follows.Notwithstanding the difficulties involved in using the method of Laplace expansion,certain types of matrices are very easy to deal with.83