5.3 basic properties of determinants

Instructor Dr. Rehana Naz 1 Mathematical Economics I

Lecture 2

Sections 5.1, 5.3 from Fundamental methods of Mathematical Economics, McGraw Hill

2005, 4th Edition. By A. C. Chiang & Kevin Wainwright are covered.

5.3 Basic Properties of determinants


Linear system: A finite set of linear equations in the variables x1,x2,...xn is called a system of

linear equations or linear system.

Introduction to systems of linear equations:

An arbitrary system of m linear equations in n unknowns of form

+ + + = + + + =

+ + + =

Where x1,x2,...xn are unknowns and a’s and d’s denote constants, is known as non-

homogeneous system provided ’ 𝑙𝑙 (at least one of them is non-zero).

This system can be expressed as 𝐴 =


where

𝐴 = [ ……… ] , = [ ] , = [ ]

A is called matrix of coefficients.

The double subscripted parameter symbol represents the coefficients appearing in the ith

equation and attached to the jth variable. For example, is the coefficient appearing in the

second equation, attached to the variable . The parameter which is unattached to any

variable, on the other hand, represents the constant term in the ith equation. For instance,

is the constant term in the first equation. All subscripts are therefore keyed to the specific

locations of the variables and parameters.

Examples of Linear Models from Economics

Example 1 (National-income model)

= 𝐶 + 𝐼 + 𝐺 , 𝐶 = + > , < <

Where Y and C stand for the endogenous (dependent) variables national income and

(planned) consumption expenditure, respectively, and 𝐼 and 𝐺 represent the exogenously

(independent) determined investment and government expenditures.

Example 2 (National-income model) = 𝐶 + 𝐼 + 𝐺 , 𝐶 = + − 𝑇 > , < < [T: taxes] 𝑇 = + , > , < < [t: income tax rate]

where Y , C and T are endogenous variables, and 𝐼 and 𝐺 represent the exogenous

variables.


Example 3 (National-income model)

= 𝐶 + 𝐼 + 𝐺, 𝐶 = + − 𝑇 > , < <

𝐺 = 𝑔 , < 𝑔 < [G: Government expenditure]

where Y , C and G are endogenous variables, and 𝐼 , and 𝑇 represent the exogenous

variables.

Linear dependence of two vectors:

Two vectors 𝑣 and 𝑣 are linearly dependent if one vector is multiple of other; otherwise they are

linearly independent. For example

1. The two row vectors 𝑣′ = [ ] and 𝑣′ = [ ] are linearly dependent because 𝑣′ = [ ] = [ ] = 𝑣′ . 2. The two row vectors 𝑣′ = [ ] and 𝑣′ = [− ] are linearly independent because one

vector is not multiple of other vector.

3. The two column vectors 𝑣 = [− ] and 𝑣 = [− ] are linearly dependent because − 𝑣 =− [− ] = [− ] = 𝑣 . 4. The two column vectors 𝑣 = [ ] and 𝑣 = [ ] are linearly independent because one

column vector is not multiple of other column vector.

Example 1: Are the rows linearly independent in following coefficient matrix

𝐴 = [ ]? Hence determine it singular or nonsingular?

Solution: We can write 𝐴 = [ ] = [𝑣′𝑣′ ] 𝑣′ = [ ] and 𝑣′ = [ ]. The two row vectors are linearly dependent because

𝑣′ = [ ] = [ ] = 𝑣′ . Hence matrix A is singular.

Example 2: Are the rows linearly independent in following coefficient matrix


𝐴 = [ ]? Hence determine it singular or nonsingular?

Solution: We can write 𝐴 = [ ] = [𝑣′𝑣′ ] 𝑣′ = [ ] and 𝑣′ = [ ]. The two row vectors are linearly independent hence matrix A is nonsingular.

Determinantal Criterion for linear independence:

An × matrix coefficient A (where AX=d, a linear system) can be considered as an

ordered set of row vectors, i.e., as a column vector whose elements are themselves row vetors:

𝐴 = [ ] = [𝑣′𝑣′𝑣′

]

Where

𝑣′ = [ ] 𝑣′ = [ ]

𝑣′ = [ ].

1. |𝐴| ≠ if and only if row vectors 𝑣′ , 𝑣′ , 𝑣′ are linearly independent.

2. |𝐴| = if and only if row vectors 𝑣′ , 𝑣′ , 𝑣′ are linearly dependent

3. For a square matrix row independence if and only if column independence.

Example 3: Check row vectors are linearly independent or not for following 3×3 matrix

𝐴 = [ ] = [𝑣′𝑣′𝑣′ ]

Solution: The determinant of A is

|𝐴| = ≠

Hence row vectors are linearly independent.


Determinantal Criterion for nonsingularity:

A square matrix A is said to be non-singular if and only if its determinant value is non-zero. A

square matrix A is said to be singular if and only if |A| = 0.

Rank of a square matrix: For a n×n if |𝐴| ≠ then rank of A is equal to n otherwise rank is

less than n.

Example 4: Check following 4×4 matrix is singular or non-singular

𝐴 = [ ]

Solution: The determinant of A is

|𝐴| = ≠

Hence given matrix is non-singular.

Example 5: Given the national income model = 𝐶 + 𝐼 + 𝐺 , 𝐶 = + > , < < where Y and C stand for endogenous variables are national income and consumption expenditure,

respectively, and 𝐼 and 𝐺 represent the exogenously determined investment and government

expenditure.

(a) Rewrite national-income model in the AX=d (with Y as the first variable in vector X).

(b) Are the rows linearly independent in the coefficient matrix A. Hence determine it singular or

nonsingular?

(c) Test whether the coefficient matrix A is non-singular using determinant criterion.

Solution: (a) System can be rewritten as − 𝐶 = 𝐼 + 𝐺 − + 𝐶 = [ −− ] [𝐶] = [𝐼 + 𝐺 ] where 𝐴 = [ −− ] , = [𝐶] , = [𝐼 + 𝐺 ]


(b) 𝐴 = [ −− ] = [𝑣′𝑣′ ]

where 𝑣′ = [ − ] and 𝑣′ = [− ]. The two row vectors are linearly independent

because one vector is not multiple of other vector. Hence A is non-singular matrix.

(c) |𝐴| = | −− | = − ≠ . Hence A is non-singular matrix.

Example 6: Given the following model = 𝐶 + 𝐼 + 𝐺 , 𝐶 = + − 𝑇 > , < < [T: taxes] 𝑇 = + , > , < < [t: income tax rate]

where Y , C and T are endogenous variables, and 𝐼 and 𝐺 represent the exogenous variables.

(a) Rewrite national-income model in the AX=d (with Y as the first variable in vector X).

(b) Test whether the coefficient matrix A is non-singular.

Solution: (a) System can be rewritten as − 𝐶 = 𝐼 + 𝐺 , − + 𝐶 + 𝑇 = − + 𝑇 =

[ −−− ] [𝐶𝑇] = [𝐼 + 𝐺 ]

where 𝐴 = [ −−− ] , = [𝐶𝑇] , = [𝐼 + 𝐺 ]

(b) |𝐴| = | −−− | Expanding along first row yields

|𝐴| = | | − − |−− | + = − + |𝐴| = − + ≠ . Hence A is non-singular matrix.

Question: Test whether the following matrix is nonsingular. What can you say about its rank?


(i) B = [ − ]

(ii) C = [ −− ]


Sections 5.4 from Fundamental methods of Mathematical Economics, McGraw Hill

2005, 4th Edition.

By A. C. Chiang & Kevin Wainwright are covered.

Inverse of a matrix:

The inverse of a square matrix A, denoted by A-1, is the matrix so that the product of A and A-1 is the Identity

matrix.

− = − = 𝐼

The requirements for a matrix to have an inverse are

1. The matrix must be square.

2. The determinant is non-zero that is det( A)≠0.

A square matrix which has an inverse is called invertible, and a square matrix without an inverse is called

noninvertible.

The inverse of a square matrix is unique.

Inverse of a 2×2 matrix

If A=[𝑎 𝑎𝑎 𝑎 ] then

the matrix inverse is

− = | | [ 𝑎 −𝑎−𝑎 𝑎 ] = 𝑎 𝑎 − 𝑎 𝑎 [ 𝑎 −𝑎−𝑎 𝑎 ]

Adjoint of a matrix: If A is any n×n matrix and Cij is the cofactor of aij, then the matrix

[ 𝑛𝑛𝑛 𝑛 𝑛𝑛]

is called the matrix of cofactors from A. The transpose of this matrix is called the adjoint of A and is denoted by

adj(A).

Example 1: Let

http://mathworld.wolfram.com/Determinant.html

http://www.mathwords.com/i/invertible_matrix.htm

http://www.mathwords.com/s/singular_matrix.htm


= [ ] then find matrix of cofactors of A and adj(A).

Solution: The matrix of cofactors is

[ | | − | | | |− | | | | − | | | | − | | | |]

[ − −− − ] Taking the transpose of matrix of cofactors, we obtain adj(A) that is

𝑎𝑑𝑗 = [ −− − − ]

Inverse of a matrix using its adjoint

If A is an invertible matrix, then

− = det 𝑎𝑑𝑗

Example 2: Find − for matrix A in Example 1.

Solution: Expanding using first row of A

det(A)=(-13)(1)+(3)(-2)+(18)(2)=17

𝑎𝑑𝑗 = [ −− − − ]


− = det 𝑎𝑑𝑗

− = [ −− − − ] = [ − / / /− / − / // / − / ]

You can verify

− = − = 𝐼

Example 3: Let

= [ − −− − ] then find − .


[ | −− − | − | −− | | − |− |−− − | | − | − | −− | |− − | − | − | | − |]

[ − −− − − ] Taking the transpose of matrix of cofactors, we obtain adj(A) that is

𝑎𝑑𝑗 = [ − −−− − ] Expanding using first row of A


det(A)=(2)(-2)+(-3)(6)+(1)(-4)=-26

− = − [ − −−− − ] Example 4: Let

= [ −− ] then find − .


[ −− ] And 𝑎𝑑𝑗 = [ −− ] det(A)=(3)(12)+(2)(6)+(-1)(-16)=64

− = [ −− ] Notes:

1. Non-square matrices do not have inverses.

2. Not all square matrices have inverses.

Inverse of a product: If A and B are invertible matrices of same size then AB is

invertible and

− = − −


Inverse of a triangular matrix : A triangular matrix is invertible if and only if its

diagonal entries are all nonzero.

The Inverse of an invertible lower triangular matrix is lower triangular and the inverse

of an invertible upper triangular matrix is upper triangular.

Example 5: Let

= [ ] then find − .

Solution: The cofactor matrix for A is

[−− − ] so the adjoint is

[ − −− ] det A =(1)(4)(6)= 24, we get

− = [ − −− ] = [ − / − // − // ]

Notice that the successive diagonal entries of − are

𝑎 , 𝑎 , 𝑎


Inverse of a diagonal matrix: The inverse of an n×n diagonal matrix is a diagonal matrix

and the successive diagonal entries of − are

𝑎 , 𝑎 , 𝑎 , … , 𝑎𝑛𝑛

Example 6: Find − where

= [ ]

Solution: det(A)=(1)(3)(2)(1)=6

The cofactors of A are

[ ]

and

𝑎𝑑𝑗 = [ ]

− = [ ]

= [ / / ]

the successive diagonal entries of − are

𝑎 , 𝑎 , 𝑎 , 𝑎


Thus we can write the inverse of a diagonal matrix by just taking reciprocals of its diagonal

entries.

Example 7: Check if following matrices are inverses of each other or not

(i) = [ ] ; = [ − ] (ii) = [ ] ; = [ ]

Solution:

(i) = [ ] [ − ] = [ ] = 𝐼

Thus A and B are inverses of each other.

(ii) = [ ] [ ] = [ ] ≠ 𝐼

Thus A and C are not inverses of each other.

5.3 basic properties of determinants

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