USE OF FACTOR THEOREM TO FIND THE VALUE OF DETERMINANT

If by putting x = a the value of a determinant vanishes then (x−a) is a factor of the determinant.

Example : Prove that

Hence, (a – b) (b – c) (c – a) is factor of determinant. But the given determinant is of fifth order so

Since this is an identity so in order to find the values of λ and µ. Let a = 0, b = 1, c = – 1

Self Practice Problems

1. Find the value of ∆ =

3. Prove that

b

4. Show that

ab

DET. &MATRICES/Page: 5of 54

We have multiplied here rows by rows but we can also multiply rows by columns, columns by rows and columns by columns.

If ∆ = |a_ij| is a detereminant of order n, then the value of the determinant |A_ij| = ∆^{n – 1}. This is also known as power cofactor formula.

Example : Find the value of

Solution. Given determinant can be splitted into product of two determinants

i.e.

DET. &MATRICES/Page: 6of 54

Self Practice Problems 1. Find the value of ∆

Here function of r can be the elements of only one row or column. None of the elements other then that row or column should be dependent on r. If more than one column or row have elements dependent on r then first expand the determinant and then find the summation.

Example : Evaluate

∑

DET. &MATRICES/Page: 7of 54

Solution. On expansion of determinent, we get

D_r = (r –1) (3 – r) + 7 + r² + 4r = 8r + 4 ⇒

∑

Self Practice Problem

1. Evaluate

∑

Note : If more than one row or one column are function of x then first expand the determinant and then integrate it.

DET. &MATRICES/Page: 8of 54

Example : Let α be a repeated root of quadratic equation f(x) = 0 and A(x), B(x) and C(x) be polynomial of degree 3, 4 and 5 respectively, then show that

)

α divisible by f(x).

Solution. Let g(x) = expression of degree 3. Also f(x) = 0 have repeated root α. So g(x) is divisible by f(x).

Example : Prove that F depends only on x₁, x₂ and x₃

and simplify F.

Solution :

DET. &MATRICES/Page: 9of 54

⇒ 1 0

0

1 = A A = 0.

Differentiating the given determinant w.r.t x, we get

) Self Practice Problem

1. If (i) Two Variables

(a) Consistent Equations: Definite & unique solution. [ intersecting lines ] (b) Inconsistent Equation: No solution. [ Parallel line ]

(c) Dependent equation: Infinite solutions. [ Identical lines ] Let a₁x + b₁y + c₁ = 0 & a₂x + b₂y + c₂ = 0 then:

a = ≠ ⇒ Given equations are inconsistent &

2

a = = ⇒ Given equations are dependent (ii) Three Variables

Let, a₁x + b₁y + c₁z = d₁... (I)

(iii) Consistency of a system of Equations

(a) If D ≠ 0 and alteast one of D_1, D_2, D₃ ≠ 0, then the given system of equations are consistent and have unique non trivial solution.

(b) If D ≠ 0 & D₁ = D₂ = D₃ = 0^, then the given system of equations are consistent and have trivial solution only.

(c) If D = D₁ = D₂ = D₃ = 0, then the given system of equations have either infinite solutions or no solution.

(Refer Example & Self Practice Problem with*)

(d) If D = 0 but atleast one of D_1, D_2, D₃ is not zero then the equations are inconsistent and have no solution.

(e) If a given system of linear equations have Only Zero Solution for all its variables then the given equations are said to have TRIVIAL SOLUTION.

(iv) Three equation in two variables :

If x and y are not zero, then condition for a₁x + b₁y + c₁ = 0 ; a₂x + b₂y + c₂ = 0 &

DET. &MATRICES/Page: 10of 54

Example: Find the nature of solution for the given system of equations.

x + 2y + 3z = 1 2x + 3y + 4z = 3 3x + 4y + 5z = 0

Solution. Let D =

5 4 3

4 3 2

3 2 1

apply C₁ → C₁ – C₂ , C₂ → C₂ – C₃ D =

5 1 1

4 1 1

3 1 1

−

−

−

−

−

−

= 0 D = 0

Now, D₁ =

5 4 0

4 3 3

3 2 1

C₃ → C₃ – C₂ D₁ =

1 4 0

1 3 3

1 2 1

R₁ → R₁ – R₂ , R₂ → R₂ – R₃

D₁ =

1 4 0

0 1 3

0 1 2

−

−

−

= 5

D = 0 But D₁ ≠ 0 Hence no solution

*Example : Solve the following system of equations x + y + z = 1

2x + 2y + 2z = 3 3x + 3y + 3z = 4

Solution. ∴ D =

3 3 3

2 2 2

1 1 1

= 0 D₁ = 0, D₂ = 0, D₃ = 0

∵ Let z = t

x + y = 1 – t 2x + 2y = 3 – 2t

Since both the lines are parallel hence no value of x and y Hence there is no solution of the given equation.

*Example : Solve the following system of equations x + y + z = 2

2x + 2y + 2z = 4 3x + 3y + 3z = 6

Solution. ∴ D =

3 3 3

2 2 2

1 1 1

= 0 D₁ = 0, D₂ = 0, D₃ = 0

All the cofactors of D, D₁, D₂ and D₃ are all zeros, hence the system will have infinite solutions.

Let z = t₁, y = t₂ ⇒ x = 2 – t₁ – t₂ where t₁, t₂ ∈ R.

Example : Consider the following system of equations x + y + z = 6

x + 2y + 3z = 10 x + 2y + λz = µ

Find values of λ and µ if such that sets of equation have

(i) unique solution (ii) infinite solution

(iii) no solution Solution. x + y + z = 6

x + 2y + 3z = 10 x + 2y + λz = µ

D = 1 2 λ 3 2 1

1 1 1

Here for λ = 3 second and third rows are identical hence D = 0 for λ = 3.

10

DET. &MATRICES/Page: 11of 54

D₁ =

λ µ 2

3 2 10

1 1 6

D₂ =

λ 1 µ

3 10 1

1 6 1

D₃ = 2 µ 1

10 2 1

6 1 1

If λ = 3 then D₁ = D₂ = D₃ = 0 for µ = 10

(i) For unique solution D ≠ 0 i.e. λ ≠ 3

(ii) For infinite solutions

D = 0 ⇒ λ = 3

D₁ = D₂ = D₃ = 0 ⇒ µ = 10.

(iii) For no solution

D = 0 ⇒ λ = 3

Atleast one of D₁, D₂ or D₃ is non zero ⇒ µ ≠ 10.

Self Practice Problems

*1. Solve the following system of equations x + 2y + 3z = 1

2x + 3y + 4z = 2 3x + 4y + 5z = 3

Ans. x = 1 + t y = –2t z = t where t ∈ R

2. Solve the following system of equations x + 2y + 3z = 0

2x + 3y + 4z = 0

x – y – z = 0 Ans. x = 0, y = 0, z = 0

3. Solve: (b+c)(y+z)− ax = b − c, (c+a)(z+x)− by = c − a, (a+b)(x+y)− cz = a − b where a+b+c≠0.

Ans. x = c b a b c

−

+ + , y = a c a b c

−

+ + , z = b a a b c

− + +

4. Let 2x + 3y + 4 = 0 ; 3x + 5y + 6 = 0, 2x² + 6xy + 5y² + 8x + 12y + 1 + t = 0, if the system of equations in x and y are consistent then find the value of t. Ans. t = 7

1 3 . 1 3 . 1 3 .

1 3 . A p p lic a t io n of D e t e rm i n an t s:A p p lic a t io n of D e t e rm i n an t s:A p p lic a t io n of D e t e rm i n an t s:A p p lic a t io n of D e t e rm i n an t s:

Following examples of short hand writing large expressions are:

(i) Area of a triangle whose vertices are (x_r, y_r); r = 1, 2, 3 is:

D =1

2 x y 1

1 y x

1 y x

3 3

2 2

1 1

If D = 0 then the three points are collinear.

(ii) Equation of a straight line passing through (x₁, y₁) & (x_2, y₂) is

1 y x

1 y x

1 y x

2 2

1

1 = 0

(iii) The lines: a₁x + b₁y + c₁ = 0... (1) a₂x + b₂y + c₂ = 0... (2) a₃x + b₃y + c₃ = 0... (3) are concurrent if,

3 3 3

2 2 2

1 1 1

c b a

c b a

c b a

= 0.

Condition for the consistency of three simultaneous linear equations in 2 variables.

(iv) ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of straight lines if:

abc + 2fgh − af² − bg² − ch² = 0 =

c f g

f b h

g h a

DET. &MATRICES/Page: 12of 54

Matrices

Any rectangular arrangement of numbers (real or complex) (or of real valued or complex valued expressions) is called a matrix. If a matrix has m rows and n columns then the order of matrix is said to be m by n (denoted as m × n).

(ii) Capital letters of English alphabets are used to denote a matrix.

1. Basic Definitions

(i) Row matrix : A matrix having only one row is called as row matrix (or row vector). General form of row matrix is A = [a₁₁, a₁₂, a₁₃, ...., a_1n]

(ii) Column matrix : A matrix having only one column is called as column matrix. (or column vector)

Column matrix is in the form A = form of a square matrix is

A =

(v) Upper triangular matrix :

A = [a_ij]_{m × n} is said to be upper triangular, if a_ij = 0 for i > j (i.e., all the elements below the diagonal elements are zero).

(vi) Lower triangular matrix : A = [a_ij]_{m × n} is said to be a lower triangular matrix, if a_ij = 0 for i < j. (i.e., all the elements above the diagonal elements are zero.)

(vii) Diagonal matrix : A square matrix [a_ij]_n is said to be a diagonal matrix if a_ij = 0 for i ≠ j.

(i.e., all the elements of the square matrix other than diagonal elements are zero) Note : Diagonal matrix of order n is denoted as Diag (a₁₁, a₂₂, ...a_nn).

(viii) Scalar matrix :Scalar matrix is a diagonal matrix in which all the diagonal elements are same A = [a_ij]_n is a scalar matrix, if (i) a_ij = 0 for i ≠ j and (ii) a_ij = k for i = j.

(ix) Unit matrix (Identity matrix) :

Unit matrix is a diagonal matrix in which all the diagonal elements are unity. Unit matrix of order 'n' is denoted by Ι_n (or Ι). the same order (i.e., number of rows of A & B are same and also the number of columns).

(xi) Equality of matrices : Two matrices A and B are said to be equal if they are comparable and all the corresponding elements are equal.

Let A = [a_ij] _{m × n} & B = [b_ij]_{p × q} A = B iff (i) m = p, n = q

(ii) a_ij = b_ij ∀ i & j.

(xii) Multiplication of matrix by scalar :

Let λ be a scalar (real or complex number) & A = [a_ij]_{m × n} be a matrix. Thus the product λA is defined as λA = [b_ij]_{m × n} where b_ij = λa_ij ∀ i & j.

Note : If A is a scalar matrix, then A = λΙ, where λ is the diagonal element.

(xiii) Addition of matrices : Let A and B be two matrices of same order (i.e. comparable matrices).

12

DET. &MATRICES/Page: 13of 54

Then A + B is defined to be.

A + B = [a_ij]_{m × n} + [b_ij]_{m × n}.

= [c_ij]_{m × n} where c_ij = a_ij + b_ij ∀ i & j.

(xiv) Substraction of matrices : Let A & B be two matrices of same order. Then A – B is defined as A + – B where – B is (– 1) B.

(xv) Properties of addition & scalar multiplication :

Consider all matrices of order m × n, whose elements are from a set F (F denote Q, R or C).

(xvi) Multiplication of matrices : Let A and B be two matrices such that the number of columns of A is same as number of rows of B. i.e., A = [a_ij]_{m × p} & B = [b_ij]_{p × n}. column vector of B.

Note - 1: The product AB is defined iff number of columns of A equals number of rows of B. A is called as premultiplier & B is called as post multiplier. AB is defined ⇒/ BA is defined.

Note - 2 : In general AB ≠≠≠≠ BA, even when both the products are defined.

Note - 3 : A (BC) = (AB) C, whenever it is defined.

(xvii) Properties of matrix multiplication :

Consider all square matrices of order 'n'. Let M_n (F) denote the set of all square matrices of order n. (where F is Q, R or C). Then

(a) A, B ∈ M_n (F) ⇒ AB ∈ M_n (F) (b) In general AB ≠ BA

(c) (AB) C = A(BC)

(d) Ι_n, the identity matrix of order n, is the multiplicative identity.

AΙ_n = A = Ι_n A ∀ A ∈ M_n (F)

(e) For every non singular matrix A (i.e., |A| ≠ 0) of M_n (F) there exist a unique (particular) matrix B ∈ M_n (F) so that AB = Ι_n = BA. In this case we say that A & B are multiplicative n & m respectively.

(ii) For a square matrix A, A² denotes AA, A³ denotes AAA etc.

Solved Example # 1

Let A =

By definition A & B are equal if they have the same order and all the corresponding elements are equal.

Thus we have sin θ = Solved Example # 2

f(x) is a quadratic expression such that



for three unequal numbers a, b, c. Find f(x).

Solution. The given matrix equation implies



DET. &MATRICES/Page: 14of 54

Self Practice Problems :

1. If A(θ) = 

Then which of the products ABC, ACB, BAC, BCA, CAB, CBA are defined. Calculate the product whichever is defined. Ans. only CAB is defined. CAB = [25 100]

2. Transpose of a Matrix

Let A =[a_ij]_{m × n}. Then the transpose of A is denoted by A′( or A^T) and is defined as A′ = [b_ij]_{n × m} where b_ij = a_ji ∀ i & j.

i.e. A′ is obtained by rewriting all the rows of A as columns (or by rewriting all the columns of A as rows).

(vii) Symmetric & skew symmetric matrix : A square matrix A is said to be symmetric if A′ = A i.e. Let A = [a_ij]_n. A is symmetric iff a_ij = a_ji ∀ i & j.

is a symmetric matrix.

B =

is a skew symmetric matrix.

Note-1 In a skew symmetric matrix all the diagonal elements are zero. (∵ aii = – a_ii ⇒ a_ii = 0) Note-2 For any square matrix A, A + A′ is symmetric & A – A′ is skew symmetric.

Note- 3 Every square matrix can be uniqualy expressed as sum of two square matrices of which one is symmetric and other is skew symmetric.

A = B + C, where B = skew symmetric (where B is any square matrix whose order is same as that of A).

Solution. Case - ΙΙΙΙ A is symmetric ⇒ A′ = A

Self Practice Problems :

1. For any square matrix A, show that A′A & AA′ are symmetric matrices.

2. If A & B are symmetric matrices of same order, than show that AB + BA is symmetric and AB – BA is skew symmetric.

3. Submatrix, Minors, Cofactors & Determinant of a Matrix

(i) Submatrix : Let A be a given matrix. The matrix obtained by deleting some rows or columns of A is called as submatrix of A.

eg. A =

are all submatrices of A.

14

DET. &MATRICES/Page: 15of 54

(ii) Determinant of a square matrix :

Let A = [a]_1×1 be a 1×1 matrix. Determinant A is defined as |A| = a.

e.g. A = [– 3]_1×1 |A| = – 3 Let A = 



 



 d c

b

a , then |A| is defined as ad – bc.

e.g. A = 









−1 4 3 5

, |A| = 23 (iii) Minors & Cofactors :

Let A = [a_ij]_n be a square matrix. Then minor of element a_ij, denoted by M_ij is defined as the determinant of the submatrix obtained by deleting i^th row & j^th column of A. Cofactor of element a_ij, denoted by C_ij (or A_ij) is defined as C_ij = (– 1)^{i + j} M_ij.

e.g. 1 A = 



 



 d c

b a M₁₁ = d = C₁₁ M₁₂ = c, C₁₂ = – c M₂₁ = b, C₂₁ = – b M₂₂ = a = C₂₂

e.g. 2 A =

















z y x

r q p

c b a

M₁₁ = z y

r

q = qz – yr = C₁₁₁.

M₂₃ = y x

b

a = ay – bx, C₂₃ = – (ay – bx) = bx – ay etc.

(iv) Determinant of any order :

Let A = [a_ij]_n be a square matrix (n > 1). Determinant of A is defined as the sum of products of elements of any one row (or any one column) with corresponding cofactors.

e.g.1 A =

















33 32 31

23 22 21

13 12 11

a a a

a a a

a a a

|A| = a₁₁C₁₁ + a₁₂ C₁₂ + a₁₃C₁₃ (using first row).

= a₁₁

33 32

23 22

a a

a

a – a₁₂

33 31

23 21

a a

a

a + a₁₃

32 31

22 21

a a

a a

|A| = a₁₂ C₁₂ + a₂₂ C₂₂ + a₃₂C₃₂ (using second column).

= – a₁₂

33 31

23 21

a a

a

a + a₂₂

33 31

13 11

a a

a

a – a₃₂

23 21

13 11

a a

a

a .

(v) Some properties of determinant (a) |A| = |A′| for any square matrix A.

(b) If two rows are identical (or two columns are identical) then |A| = 0.

(c) Let λ be a scalar. Than λ |A| is obtained by multiplying any one row (or any one column) of |A| by λ

Note : |λA| = λⁿ |A|, when A = [a_ij]_n.

(d) Let A = [a_ij]_n. The sum of the products of elements of any row with corresponding cofactors of any other row is zero. (Similarly the sum of the products of elements of any column with corresponding cofactors of any other column is zero).

(e) If A and B are two square matrices of same order, then |AB| = |A| |B|.

Note : As |A| = |A′|, we have |A| |B| = |AB′| (row - row method)

|A| |B| = |A′B| (column - column method)

|A| |B| = |A′B′| (column - row method)

(vi) Singular & non singular matrix : A square matrix A is said to be singular or non singular according as |A| is zero or non zero respectively.

(vii) Cofactor matrix & adjoint matrix :Let A = [a_ij]_n be a square matrix. The matrix obtained by replacing each element of A by corresponding cofactor is called as cofactor matrix of A, denoted as cofactor A. The transpose of cofactor matrix of A is called as adjoint of A, denoted as adj A.

DET. &MATRICES/Page: 16of 54

i.e. if A = [a_ij]_n

then cofactor A = [c_ij]_n when c_ij is the cofactor of a_ij ∀ i & j.

Adj A = [d_ij]_n where d_ij = c_ji ∀ i & j.

(viii) Properties of cofactor A and adj A:

(a) A . adj A = |A| Ι_n = (adj A) A where A = [a_ij]_n. (b) |adj A| = |A|^{n – 1}, where n is order of A.

In particular, for 3 × 3 matrix, |adj A| = |A|²

(c) If A is a symmetric matrix, then adj A are also symmetric matrices.

(d) If A is singular, then adj A is also singular.

(ix) Inverse of a matrix (reciprocal matrix) : Let A be a non singular matrix. Then the matrix

| A

|

1 adj A is the multiplicative inverse of A (we call it inverse of A) and is denoted by A^–1. We have A (adj A) = |A| Ι_n = (adj A) A

⇒ A 







 adjA

| A

|

1 = Ι_n = 







 adjA

| A

|

1 A, for A is non singular

⇒ A^–1 =

| A

|

1 adj A.

Remarks :

1. The necessary and sufficient condition for existence of inverse of A is that A is non singular.

2. A^–1 is always non singular.

3. If A = dia (a₁₁, a₁₂, ..., a_nn) where a_ii≠ 0 ∀ i, then A^–1 = diag (a₁₁^{– 1}, a₂₂^–1, ...., a_nn^–1).

4. (A^–1)′ = (A′)^–1 for any non singular matrix A. Also adj (A′) = (adj A)′. 5. (A^–1)^–1 = A if A is non singular.

6. Let k be a non zero scalar & A be a non singular matrix. Then (kA)^–1 = k 1

A^–1.

7. |A^–1| = |A| 1

for |A| ≠ 0.

8. Let A be a nonsingular matrix. Then AB = AC ⇒ B = C & BA = CA ⇒ _{B= C.}

9. A is non-singular and symmetric ⇒ _A^–1 is symmetric.

10. In general AB = 0 does not imply A = 0 or B = 0. But if A is non singular and AB = 0, then B = 0.

Similarly B is non singular and AB = 0 ⇒ A = 0. Therefore, AB = 0 ⇒ either both are singular or one of them is 0.

Solved Example # 4

For a 3×3 skew symmetric matrix A, show that adj A is a symmetric matrix.

Solution.

A =

















−

−

−

0 c b

c 0 a

b a 0

cof A =

















−

−

−

−

2 2 2

a ab ca

ab b

bc

ca bc c

adj A = (cof A)′ =

















−

−

−

−

2 2 2

a ab ca

ab b

bc

ca bc c

which is symmetric.

Solved Example # 5

For two nonsingular matrices A & B, show that adj (AB) = (adj B) (adj A) Solution.

We have (AB) (adj (AB)) = |AB| Ι_n

= |A| |B| Ι_n A^–1 (AB)(adj (AB)) = |A| |B| A^–1

⇒ B adj (AB) = |B| adj A (∵ A^–1 = |A| 1

adj A)

⇒ B^–1 B adj (AB) = |B| B^–1 adj A

⇒ adj (AB) = (adjB) (adj A)

16

DET. &MATRICES/Page: 17of 54

Self Practice Problems :

1. If A is nonsingular, show that adj (adj A) = |A|^{n – 2} A.

2. Prove that adj (A^–1) = (adj A)^–1.

3. For any square matrix A, show that |adj (adj A) | = ^|A|(n−1)2. 4. If A and B are nonsingular matrices, show that (AB)^–1 = B^–1 A^–1. 4. System of Linear Equations & Matrices

Consider the system

a₁₁ x₁ + a₁₂x₂ + ... + a_1nx_n = b₁

Then the above system can be expressed in the matrix form as AX = B.

The system is said to be consistent if it has atleast one solution.

(i) System of linear equations and matrix inverse:

If the above system consist of n equations in n unknowns, then we have AX = B where A is a square matrix. If A is nonsingular, solution is given by X = A^–1B.

If A is singular, (adj A) B = 0 and all the columns of A are not proportional, then the system has infinite many solution.

If A is singular and (adj A) B ≠ 0, then the system has no solution (we say it is inconsistent).

(ii) Homogeneous system and matrix inverse:

If the above system is homogeneous, n equations in n unknowns, then in the matrix form it is AX = O. (∵ in this case b1 = b₂ = ... b_n = 0), where A is a square matrix.

If A is nonsingular, the system has only the trivial solution (zero solution) X = 0

If A is singular, then the system has infinitely many solutions (including the trivial solution) and hence it has non trivial solutions.

(iii) Rank of a matrix :

Let A = [a_ij]_m×n. A natural number ρ is said to be the rank of A if A has a nonsingular submatrix of order ρ and it has no nonsingular submatrix of order more than ρ. Rank of zero matrix is regarded to be zero.

eg. A =

3 as a non singular submatrix.

The square matrices of order 3 are



and all these are singular. Hence rank of A is 2.

(iv) Elementary row transformation of matrix :

The following operations on a matrix are called as elementary row transformations.

(a) Interchanging two rows.

(b) Multiplications of all the elements of row by a nonzero scalar.

(c) Addition of constant multiple of a row to another row.

Note : Similar to above we have elementary column transformations also.

Remark :

1. Elementary transformation on a matrix does not affect its rank.

2. Two matrices A & B are said to be equivalent if one is obtained from other using elementary transformations. We write A ≈ B.

(v) Echelon form of a matrix : A matric is said to be in Echelon form if it satisfy the followings:

(a) The first non-zero element in each row is 1 & all the other elements in the corresponding column (i.e. the column where 1 appears) are zeroes.

(b) The number of zeroes before the first non zero element in any non zero row is less than the number of such zeroes in succeeding non zero rows.

Result : Rank of a matrix in Echelon form is the number of non zero rows (i.e. number of rows with atleast one non zero element.)

Remark :

1. To find the rank of a given matrix we may reduce it to Echelon form using elementary row transformations and then count the number of non zero rows.

DET. &MATRICES/Page: 18of 54

(vi) System of linear equations & rank of matrix:

Let the system be AX = B where A is an m × n matrix, X is the n-column vector & B is the m-column vector. Let [AB] denote the augmented matrix (i.e. matrix obtained by accepting elements of B as n + 1^th column & first n columns are that of A).

ρ(A) denote rank of A and ρ([AB]) denote rank of the augmented matrix.

Clearly ρ(A) ≤ ρ([AB]).

(a) If ρ(A) < ρ([AB]) then the system has no solution (i.e. system is inconsistent).

(b) If ρ(A) = ρ([AB]) = number of unknowns, then the system has unique solution.

(and hence is consistent)

(c) If ρ(A) = ρ([AB]) < number of unknowns, then the systems has infinitely many solutions (and so is consistent).

(vii) Homogeneous system & rank of matrix :

Let the homogenous system be AX = 0, m equations in 'n' unknowns. In this case B = 0 and so ρ(A) = ρ([AB]).

Hence if ρ(A) = n, then the system has only the trivial solution. If ρ(A) < n, then the system has infinitely many solutions.

Solved Example # 6 Solve the system

1

using matrix inverse.

Solution.

Solved Example # 7

Test the consistancy of the system

8.

. Also find the solution, if any..

Solution.

DET. &MATRICES/Page: 19of 54

This is in Echelon form.

ρ(AB) = 2 = ρ(A) < number of unknowns Hence there are infinitely many solutions n – ρ = 1.

Hence we can take one of the variables any value and the rest in terms of it.

Let z = r, where r is any number.

Self Practice Problems:

1. A =

. Find the inverse of A using |A| and adj A. Also find A^–1 by solving a system of equations.

2. Find real values of λ and µ so that the following systems has

(i) unique solution (ii) infinite solution (iii) No solution.

x + y + z = 6 x + 2y + 3z = 1 x + 2y + λz = µ

Ans. (i) λ ≠ 3, µ ∈ R (ii) λ = 3, µ = 1 (iii) λ = 3, µ ≠ 1 3. Find λ so that the following homogeneous system have a non zero solution

x + 2y + 3z = λx 3x + y + 2z = λy 2x + 3y + z = λz Ans. λ = 6

5. More on Matrices

(i) Characteristic polynomial & Characteristic equation :

Let A be a square matrix. Then the polynomial | A – xΙ | is called as characteristic polynomial of A & the equation | A – xΙ | = 0 is called as characteristic equation of A.

Remark :

Every square matrix A satisfy its characteristic equation (Cayley - Hamilton Theorem).

i.e. a₀ xⁿ + a, x^{n – 1} + ... + a_{n – 1}x + a_n = 0 is the characteristic equation of A, then a₀Aⁿ + a₁A^{n – 1} + ... + a_{n – 1} A + a_n Ι = 0

(ii) More Definitions on Matrices : (a) Nilpotent matrix:

A square matrix A is said to be nilpotent ( of order 2) if, A² = O.

A square matrix is said to be nilpotent of order p, if p is the least positive integer such that A^p = O.

(b) Idempotent matrix:

A square matrix A is said to be idempotent if, A² = A.

DET. &MATRICES/Page: 20of 54

e.g. 







 1 0

0

1 is an idempotent matrix.

(c) Involutory matrix:

A square matrix A is said to be involutory if A² = Ι, Ι being the identity matrix.

e.g. A = 







 1 0

0

1 is an involutory matrix.

(d) Orthogonal matrix:

A square matrix A is said to be an orthogonal matrix if, A′ A = Ι = A′A.

(e) Unitary matrix:

A square matrix A is said to be unitary if A( A )′ = Ι, where A is the complex conjugate of A.

Solved Example # 8

If A =

















−

− 1 0 0

0 1 2

0 2 1

, show that 5A^–1 = A² + A – 5Ι.

Solution.

We have the characteristic equation of A.

| A – xΙ | = 0

i.e.

















−

−

−

−

−

x 1 0 0

0 x 1 2

0 2 x 1

= 0 i.e. x³ + x² – 5x – 5 = 0.

Using cayley - Hamilton theorem.

A³ + A² – 5A – 5Ι = 0

⇒ 5Ι = A³ + A² – 5A Multiplying by A^–1, we get

5A^–1 = A² + A – 5Ι Solved Example # 9

Show that a square matrix A is involutory, iff (Ι – A) (Ι + A) = 0 Solution.

Let A be involutory Then A²= Ι

(Ι – A) (Ι + A) = ΙΙ + ΙA – AΙ – A²

= Ι + A – A – A²

= Ι – A²

= 0

Conversly, let (Ι – A) (Ι + A) = 0

⇒ ΙΙ + ΙA – AΙ – A² = 0

⇒ Ι + A – A – A² = 0

⇒ Ι – A² = 0

⇒ A is involutory Self Practice Problems

1. If A is idempotent, show that B = Ι – A is idempotent and that AB = BA = 0.

2. If A is a nilpotent matrix of index 2, show that A (Ι + A)ⁿ = A for all n ∈ N.

3. A is a skew symmetric matrix, such that A² + Ι = 0. Show that A is orthogonal and is of even order.

4. Let A =

















−

−

−

0 a b

a 0 c

b c 0

. If A³ + λA = 0, find λ.

Ans. a² + b² + c².

20

In document XII Mathematics IIT JEE Advanced Study Package 2014 15 (Page 44-61)