Determinants

Suppose A is an n x n matrix.  Associated with A is a number called the determinant of A and is denoted by det A or |A|.  Symbolically, we distinguish a matrix A from the determinant of A by replacing the parentheses by vertical bars:

 

    \[ A =\begin{bmatrix} a_{11} & a_{12}& \dots & a_{1n}\\ a_{21} & a_{22}& \dots & a_{2n}\\ \vdots & & & \vdots\\ a_{n1} & a_{n2}& \dots & a_{nn}\\ \end{bmatrix}\]

and

    \[det A =\begin{vmatrix} a_{11} & a_{12}& \dots & a_{1n}\\ a_{21} & a_{22}& \dots & a_{2n}\\ \vdots & & & \vdots\\ a_{n1} & a_{n2}& \dots & a_{nn}\\ \end{vmatrix}\]

 

Thus, every square matrix A, is assigned a particular scalar quantity called the determinant of A, denoted by det A or |A|.

 

A determinant of an n x n matrix is said to be a determinant of order n.

 

Let's begin by defining the determinants of  1 x 1, 2 x 2 , and 3 x 3 matrices.

 

The Determinant of Order One

For a 1 x 1 matrix A = (a) we have det A = |a| = a.

For example,  if A = (-5), then det A = |-5| = -5.

Note: In this case, the vertical bars || around a number do not mean the absolute value of the number.

 

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The Determinant of Order Two

The determinant of a 2 x 2 matrix is said to be a determinant of order 2.

Consider matrix A:

    \[ A =\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\]

 

The determinant of

    \[ A =\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\]

is the number

    \[det A =\begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}\]

 

Determinant of Order Two
Determinant of Order Two

 

As a mnemonic, a determinant of order 2 is thought to be the product of the main diagonal entries of A minus the product of the other diagonal entries.

 

What is a mnemonic?
PEMDAS Mnemonic

 

For example, if matrix A is the 2 x 2 matrix

 

    \[ A =\begin{bmatrix} 5 & -7\\ 2 & 9 \end{bmatrix}\]

,

 

then det A is

 

    \[\begin{vmatrix} A = \end{vmatrix} \begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}\]

The Determinant of Order Three

The determinant of a 3 x 3 matrix is said to be a determinant of order 3.

Consider matrix A:

    \[ A =\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}\]

The determinant of

    \[ A =\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}\]

is the number

    \[det A =\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}\]

    \[=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}\]

Note that there are six products, each consisting of three elements in the matrix.

Three of the products appear with a positive sign (they preserve their sign) and three with a negative sign (they change their sign).

The expression in (3) can be written in a more tractable form. By factoring, we have

    \[det A = a_{11}(a_{22}a_{33} - a_{23}a_{32}) + a_{12}(-a_{21}a_{33} + a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})\]

But in view of (1), each term in parentheses is recognized as the determinant of a 2 x 2 matrix:

    \[det A = a_{11}\begin{vmatrix} a_{22} & a_{23}\\ a_{32} & a_{33} \end{vmatrix} + a_{12} \begin{bmatrix} - \begin{vmatrix} a_{21} & a_{23}\\ a_{31} & a_{33} \end{vmatrix} \end{bmatrix}\]

    \[+ a_{13} \begin{vmatrix} a_{21} & a_{22}\\ a_{31} & a_{32} \end{vmatrix}\]

.

Observe that each determinant in (4) is a determinant of  submatrix of the matrix A and corresponds to its coefficient in the following manner:

    \[a_{11}\]

. is the coefficient of the determinant of the submatrix obtained by deleting the first row and first column of A;

    \[a_{12}\]

. is the coefficient of the negative of the determinant of the submatrix obtained by deleting the first row and second column of A; and finally,

    \[ a_{13}\]

. is the coefficient of the determinant of the submatrix obtained by deleting the first row and third colomn of A. In other words, the coefficients in (4) are simply the entries of the first row of A. We say that det A has been expanded by cofactors along the first row, with the cofactors of

    \[a_{11}, a_{12} and a_{13}\]

being the determinants

    \[C_{11} = \begin{vmatrix}a_{22} & a_{23}\\a_{32} & a_{33}\end{vmatrix}\]

    \[C_{12} = \begin{vmatrix}a_{21} & a_{31}\\a_{23} & a_{33}\end{vmatrix}\]

    \[C_{13} = \begin{vmatrix}a_{21} & a_{31}\\a_{22} & a_{32}\end{vmatrix}\]

Thus (4) is

    \[det A = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}\]

In general, the cofactor of

    \[a_{ij}\]

is the determinant

    \[C_{ij} = (-1) \]

where

    \[M_{ij}\]

is the determinant of the submatrix obtained by deleting the ith row and the jth row and the jth column of A. The determinant

    \[M_{ij}\]

is called a minor determinant. A cofactor is a signed minor determinant; that is,

    \[C_{ij} = M_{ij}\]

when

    \[i + j\]

is even and

    \[C_{ij} = -M_{ij}\]

when

    \[i + j\]

is odd.

A

    \[3 x 3\]

matrix has nine cofactors:

    \[C_{11} = M_{11}\]

    \[C_{12} = -M_{12}\]

    \[C_{13} = M_{13}\]

    \[C_{21} = -M_{21}\]

    \[C_{22} = M_{22}\]

    \[C_{23} = -M_{23}\]

    \[C_{31} = -M_{31}\]

    \[C_{32} = -M_{32}\]

    \[C_{33} = M_{33}\]

Inspection of the above array shows that the sign factor +1 or -1 associated with a cofactor can be obtained from the checkerboard pattern:
+ - +
- + -
+ - +
3 x 3 matrix
Now observe that (3) can be rearranged and factored again as

    \[det A = - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{22}(a_{11}a_{33} - a_{13}a_{31}) - a_{32}(a_{11}a_{23} - a_{13}a_{21})\]

    \[= a_{12} \begin{bmatrix} - \begin{vmatrix}a_{21} & a_{23}\\a_{31} & a_{33}\end{vmatrix} \end{bmatrix}\]

    \[ + a_{22} \begin{vmatrix}a_{11} & a_{13}\\a_{31} & a_{33}\end{vmatrix}\]

    \[+ a_{32} \begin{bmatrix} - \begin{vmatrix}a_{11} & a_{13}\\a_{21} & a_{23}\end{vmatrix} \end{bmatrix}\]

    \[= a_{12} C_{12} + a_{22} C_{22} + a_{32} C_{32},\]

which is the cofactor expansion of det A along the second column. It is left as an exercise to show from (3) that det A can also be expanded by cofactors along the third row:

    \[det A = a_{31}C_{31} + a_{32}C_{32} + a_{33}C_{33}.\]

We are, of course, suggesting in (5), (8) and (9) the following general result:The determinant of a 3 x 3 matrix can be evaluated  by expanding det A by expanding det A by cofactors along any row or along any column.

Cofactor Expansion Along the First Row

Evaluate the determinant of

    \[ A = \begin{bmatrix} 2&4&7\\6&0&3\\ 1&5&3\end{bmatrix}\]

Using cofactor expansion along the first row gives

    \[det A = \begin{vmatrix} 2&4&7\\6&0&3\\ 1&5&3\end{vmatrix} = 2C_{11} + 4C_{12} + 7C_{13}.\]

Now, the cofactors of the entries in the first row of A are

    \[C_{11} = ( - 1)^{1+1} \begin{vmatrix} 2&4&7\\6&0&3\\ 1&5&3\end{vmatrix} = ( - 1) ^{1+1} \begin{vmatrix} 0&3\\5&3\end{vmatrix}\]

    \[C_{12} = ( - 1)^{1+2} \begin{vmatrix} 2&4&7\\6&0&3\\ 1&5&3\end{vmatrix} = ( - 1) ^{1+2} \begin{vmatrix} 6&3\\1&3\end{vmatrix}\]

    \[C_{13} = ( - 1) ^{1+3}\begin{vmatrix} 2&4&7\\6&0&3\\ 1&5&3\end{vmatrix} = ( - 1) ^{1+3} \begin{vmatrix} 6&0\\1&5\end{vmatrix}\]

where the dashed lines indicate the row and column that are deleted. Thus,

    \[det A = 2(-1) ^{1+1} \begin{vmatrix} 0&3\\5&3\end{vmatrix} + 4(-1) ^{1+2} \begin{vmatrix} 6&3\\1&3\end{vmatrix} + 7(-1) ^{1+3} \begin{vmatrix} 6&0\\1&5\end{vmatrix}\]

    \[= 2[0(3) - 3(5)] - 4[6(3) - 3(1)] + 7[6(5) - 0(1)] = 120\]

If a matrix has a row (or a column) containing many zero entries, then wisdom dictates that we evaluate the determinant of the matrix using co-factor expansion along that row (or column). Thus, in Example 1, had we expanded the determinant of A using cofactors along, say, the second row, then

    \[det A = 6C_{21}+0C_{22} + 3C_{23} = 6C_{21} + 3C_{23}\]

    \[=6(-1)^{1+2} \begin{vmatrix} 4&7\\5&3\end{vmatrix}+3(-1)^{2+3}\begin{vmatrix} 2&4\\1&5\end{vmatrix}\]

    \[-6(-23) -3(6) = 120\]

Cofactor Expansion Along the Third Column

Evaluate the determinant of

    \[ A = \begin{bmatrix} 6&5&0\\-1&8&-7\\ -2&4&0\end{bmatrix}\]

Solution: Since there are two zeros in the third column, we expand by cofactors of that column:

    \[det A = \begin{vmatrix} 6&5&0\\-1&8&-7\\-2&4&0\end{vmatrix} = 0C_{13} + (-7)C_{23} + 0C_{33}\]

    \[=(-7)(-1)^{2+3} \begin{vmatrix} 6&5&0\\1&8&7\\-2&4&0\end{vmatrix}=(-7)(-1)^{2+3}\begin{vmatrix} 6&5\\-2&4\end{vmatrix}\]

    \[=7[6(4)-5(-2)]=238\]

Carrying the above ideas one step further, we can evaluate the determinant of a 4 X 4 matrix by multiplying the entries in a row (or column) by their corresponding cofactors and adding the products. In this case, each cofactor is a signed minor determinant of an appropriate 3 X 3 submatrix. The following theorem, which we shall give without proof, states that the determinant of any n X n matrix A can be evaluated by means of cofactors.

Cofactor Expansion of a Determinant

    \[Let A =(a_{i}_{j})_{n\times n}\]

be an

    \[n\times n\]

matrix.

For each 1 ≤ i ≤ n, the cofactor of det A along the ith row is:

    \[det A = a_{i}_{1}C_{i}_{1} + a_{i}_{2}C_{i}_{2} + ... + a_{i}_{n}C_{i}_{n}\]

For each 1 ≤ j ≤ n, the cofactor expansion of det A along the jth column is:

    \[det A = a_{1}_{j}C_{1}_{j} + a_{2}_{j}C_{2}_{j} + ... + a_{n}_{j}C_{n}_{j}\]

The sign factor pattern for the cofactors illustrated in (7) extends to matrices of order greater than 3.

.                                 +  -  +  -  +  ...

+  -  +  -                     -  +  -  +  -  ...

-  +  -  +                     +  - +  -  +  ...

+  -  +  -                     -  +  -  +  -  ...

-  +  -  +                     +  -  +  -  +  ...

.                                  :   :   :   :   :

4 X 4 matrix         n X n matrix

Cofactor Expansion Along the Fourth Row

Evaluate the determinant of

    \[ A = \begin{bmatrix} 5&1&2&4\\-1&0&2&3\\1&1&6&1\\1&0&0&-4\end{bmatrix}\]

Solution: Since the matrix has two zero entries in its fourth row, we choose to expand det A by cofactors along that row:

    \[det A = \begin{vmatrix} 5&1&2&4\\-1&0&2&3\\1&1&6&1\\1&0&0&-4\end{vmatrix} = (1)C_{41} + (0)C_{42} + (0)C_{43}+ (-4)C_{44}\]

(10)

where

    \[C_{41} = ( - 1)^{4+1} \begin{vmatrix} 1&2&4\\0&2&3\\ 1&6&1\end{vmatrix}\]

and

    \[C_{44}=( - 1) ^{4+4} \begin{vmatrix} 5&1&2\\-1&0&2\\1&1&6\end{vmatrix}\]

We then expand both these determinants by cofactors along the second row:

    \[C_{41}=( - 1)\begin{vmatrix} 1&2&4\\0&2&3\\1&6&1\end{vmatrix}\]

=

    \[-(0(-1) ^{2+1}\begin{vmatrix} 2&4\\6&1\end{vmatrix}+2(-1)^{2+2}\begin{vmatrix} 1&4\\1&1\end{vmatrix}+3(-1)^{2+3}\begin{vmatrix} 1&2\\1&6\end{vmatrix}\]

= 18

    \[C_{44}=\begin{vmatrix} 5&1&2\\-1&0&2\\1&1&6\end{vmatrix}\]

=

    \[(-1)(-1) ^{2+1}\begin{vmatrix} 1&2\\1&6\end{vmatrix}+0(-1)^{2+2}\begin{vmatrix} 5&2\\1&6\end{vmatrix}+2(-1)^{2+3}\begin{vmatrix} 5&1\\1&1\end{vmatrix}\]

= -4

Therefore (10) becomes

    \[det A = \begin{vmatrix} 5&1&2&4\\-1&0&2&3\\1&1&6&1\\1&0&0&-4\end{vmatrix} = (1)(18) + (-4)(-4) = 34\]

You should verify this result by expanding det A by cofactors along the second column

Remarks

In previous mathematics courses you may have seen the following memory device, analogous to (2), for evaluating a determinant of order 3:

.                                                           multiply                   multiply                                     (11)

    \[\begin{vmatrix} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{vmatrix}\begin{vmatrix} a_{11} & a_{12}\\a_{21} & a_{22}\\a_{31} & a_{32}\end{vmatrix}\]

(i) Add the products of the entries on the arrows that go from left to right.

(ii) Subtract from the number in (i) the sum of the products of the entries on the arrows that go from right to left.

A word of caution is in order here. The memory device given in (11), though easily adapted to matrices larger than 3 X 3, does not give the correct results. There are no mnemonic devices for evaluating the determinants of order 4 or greater.

Complex Numbers

Definition

A complex number is any number of the form z = a + ib where a and b are real numbers and i is the imaginary unit. The imaginary unit i is defined by the property = -1. using the imaginary unit, a general complex number is build out of two real numbers.

Terminology

The number i in z = a + ib is called the imaginary unit. The real number x in z = x + iy is called the real part of z; the real number y is called the imaginary part of z. The real and imaginary parts of a complex number z are abbreviated Re(z) and Im(z), respectively. For example, if z = 4 -9i, then Re(z) = 4 and Im(z) = -9. A real constant multiple of the imaginary unit is called a pure imaginary number. For example, z = 6i is a pure imaginary number. Two complex numbers are equal if their real and imaginary parts are equal. Since this simple concept is sometimes useful, we formalize the last statement in the next definition.

 

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.