Matrices and Determinants Theory_E

MATHS

Matrices and Determinants 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 written as m × n and we call it as order m by n The general m × n matrix is

 a11 a12   a 21 a 22  ..... ..... A=   a i1 a i2  ..... .....  a m1 am2

a13 a 23 ..... a i3 ..... a m3

a1n   ...... a 2 j ..... a 2n  ..... ..... ..... .....   ...... a ij ...... a in  ..... ..... ..... .....   ..... a mj ..... a mn 

......

a1j

.....

where aij denote the element of ith row & j th column. The above matrix is usually denoted as [aij]m × n . Notes : (i)

The elements a11, a22, a33,........ are called as diagonal elements. Their sum is called as trace of A denoted as tr(A)

(ii)

Capital letters of English alphabets are used to denote matrices.

(iii)

Order of a matrix : If a matrix has m rows and n columns, then we say that its order is "m by n", written as "m × n".

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, ...., a1n] This is a matrix of order "1 × n" (or a row matrix of order n)

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

 a11    a 21   Column matrix is in the form A =  ...    a m1  This is a matrix of order "m × 1" (or a column matrix of order m)

Square matrix : A matrix in which number of rows & columns are equal is called a square matrix. The general form of a square matrix is

 a11 a12 ....... a1n    a 21 a 22 ........ a 2n  A =  ....... ....... ....... .......    an1 a n2 ....... a nn 

which we denote as A = [aij]n.

This is a matrix of order "n × n" (or a square matrix of order n)

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1

MATHS

Zero matrix : A = [aij]m × n is called a zero matrix, if aij = 0  i & j. 0 0 0 e.g. : (i)   0 0 0

0 0 0  (ii)  0 0 0  0 0 0

   

Upper triangular matrix : A = [aij]m × n is said to be upper triangular, if aij = 0 for i > j (i.e., all the elements below the diagonal elements are zero).

a b c d e.g. : (i)  0 x y z   0 0 u v 

a b c    (ii)  0 x y   0 0 z 

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

a 0 0   b c 0 e.g. : (i)   x y z 

a 0 0 0   b c 0 0 (ii)   x y z 0 

Diagonal matrix : A square matrix [aij]n is said to be a diagonal matrix if aij = 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, ......ann).

a 0 0   0 b 0 e.g. : (i)   0 0 c 

   (ii)   

a 0 0 0  0 b 0 0 0 0 0 0  0 0 0 c 

Scalar matrix : Scalar matrix is a diagonal matrix in which all the diagonal elements are same. A = [aij]n is a scalar matrix, if (i) aij = 0 for i  j and (ii) aij = k for i = j. a 0 e.g. : (i)   0 a

a 0 0   0 a 0 (ii)   0 0 a 


2

MATHS 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 n (or ). i.e. A = [aij]n is a unit matrix when aij = 0 for i  j & aii = 1

eg.

1 0 0   0 1 0  . 0 0 1

 1 0 2 =   , 3 = 0 1

Comparable matrices : Two matrices A & B are said to be comparable, if they have the same order (i.e., number of rows of A & B are same and also the number of columns). 2 3 4 e.g. : (i) A =    3 1 2 

&

2 3 4 e.g. : (ii) C =    3 1 2 

&

3 4 2 B=   are comparable 0 1 3

3 0   4 1 D=  are not comparable  2 3 

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 = [aij] m × n & B = [bij]p × q A = B iff (i) m = p, n = q (ii) aij = bij  i & j.

 sin  1/ 2  1/ 2 sin       Example # 1 : Let A =  1/ 2 cos   & B = cos  cos  . Find  so that A = B.  cos  tan   cos   1      Solution :

By definition A & B are equal if they have the same order and all the corresponding elements are equal. Thus we have sin  =



 = (2n + 1) –

1 2

, cos = –

1 2

& tan  = – 1

 . 4

Multiplication of matrix by scalar : Let  be a scalar (real or complex number) & A = [aij]m × n be a matrix. Thus the product A is defined as A = [bij]m × n where bij = aij  i & j.

5   2 1 3   e.g. : A =  0 2 1  3   0 0  1  2 

  6 3  9  15    9  & – 3A  (–3) A =  0  6  3  0 0 3 6 

Note : If A is a scalar matrix, then A = , where  is a diagonal entry of A


3

MATHS

Addition of matrices : Let A and B be two matrices of same order (i.e. comparable matrices). Then A + B is defined to be. A + B = [aij]m × n + [bij]m × n. = [c ij]m × n where c ij = aij + bij  i & j.

 1 1   1 2    2 3  2 3 e.g. : A =  , B=   1 0   5 7

 0 1    , A+ B =  0 0    6 7 

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.

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). Let Mm × n (F) denote the set of all such matrices. Then (a) A  Mm × n (F) & B  Mm × n (F)  A + B  Mm × n(F) (b) A+B=B+A (c) (A + B) + C = A + (B + C) (d) O = [o]m × n is the additive identity. (e) For every A  Mm × n(F), – A is the additive inverse. (f)  (A + B) = A + B (g) A = A (h) (1 + 2) A = 1A + 2A

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 = [aij]m × p & B = [bij]p × n. p

Then AB = [c ij]m × n where c ij =

a

ik b kj

, which is the dot product of ith row vector of A and j th

k 1

column vector of B.

0 1 1 1   1 2 3 3 4 9 1 0 0 1 0 e.g. : A =   , B=  , AB =   2 3 1 1 3 7 2  1 1 2 0  Notes : (1)

(2) (3)

The product AB is defined iff the number of columns of A is equal to the number of rows / BA is of B. A is called as premultiplier & B is called as post multiplier. AB is defined  defined. In general AB  BA, even when both the products are defined. A (BC) = (AB) C, whenever it is defined.


4

MATHS Properties of matrix multiplication : Consider all square matrices of order 'n'. Let Mn (F) denote the set of all square matrices of order n. (where F is Q, R or C). Then (a) A, B  Mn (F)  AB  Mn (F) (b) In general AB  BA (c) (AB) C = A(BC) (d) n, the identity matrix of order n, is the multiplicative identity. An = A = n A  A  Mn (F) (e) For every non singular matrix A (i.e., |A|  0) of Mn (F) there exist a unique (particular) matrix B  Mn (F) so that AB = n = BA. In this case we say that A & B are multiplicative inverse of one another. In notations, we write B = A–1 or A = B–1. (f) If  is a scalar (A) B = (AB) = A(B). (g) A(B + C) = AB + AC  A, B, C  Mn (F) (h) (A + B) C = AC + BC  A, B, C  Mn (F). Notes : (1) (2)

Let A = [aij]m × n. Then An = A & m A = A, where n & m are identity matrices of order n & m respectively. For a square matrix A, A2 denotes AA, A3 denotes AAA etc.

Example # 2 : f(x) is a quadratic expression such that

Solution :

a 2 a 1  f (0)  2a  1  2      b b 1  f (1)  = 2b  1 for three unequal numbers a, b, c. Find f(x). c 2 c 1 f ( 1) 2c  1     The given matrix equation implies a 2 f (0)  af (1)  f ( 1) 2a  1  2    b f (0)  bf (1)  f ( 1) = 2b  1 c 2 f (0)  cf (1)  f ( 1) 2c  1      

x 2 f(0) + xf(1) + f(–1) = 2x + 1 for three unequal numbers a, b, c (i) is an identity f(0) = 0, f(1) = 2 & f(– 1) = 1 f(x) = x (ax + b) 2 = a + b & – 1 = – a + b.



b=

1 3 &a= 2 2



f(x) =

.....(i)

3 2 1 x + x. 2 2

Self practice problems :

(1)

cos   sin  If A() =   , verify that A() A() = A( + ).  sin  cos  

Hence show that in this case A(). A() = A() . A().

(2)

4 6  1  2 4     2  , B =  0 1 and C = [3 1 2]. Let A = 3 0  1  2 5   1 2 Then which of the products ABC, ACB, BAC, BCA, CAB, CBA are defined. Calculate the product whichever is defined. Answer (2) Only CAB is defined. CAB = [25 100]


5

MATHS Transpose of a matrix : Let A =[aij]m × n. Then the transpose of A is denoted by A( or AT) and is defined as A = [bij]n × m where bij = aji  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).

1 1 2 3 4     2 a b c d  , A =  3 e.g. : A =    x y z w  4

Results :

x   b y  c z   d w  a

(i)

For any matrix A = [aij]m × n, (A) = A

(ii)

Let  be a scalar & A be a matrix. Then (A) = A

(iii)

(A + B) = A + B & (A – B) = A – B for two comparable matrices A and B.

(iv)

(A1 ± A2 ± ..... ± An) = A1 ± A2 ± ..... ± An, where Ai are comparable.

(v)

Let A = [aij]m × p & B = [bij]p × n , then (AB) = BA

(vi)

(A1 A2 .......An)= An. An – 1 ...........A2 . A1, provided the product is defined.

Symmetric & skew-symmetric matrix : i.e.

A square matrix A is said to be symmetric if A = A Let A = [aij]n. A is symmetric iff aij = aji  i & j.

A square matrix A is said to be skew-symmetric if A = – A i.e. Let A = [aij]n. A is skew-symmetric iff aij = – aji  i & j.

e.g.

a h g   A = h b f  is a symmetric matrix. g f c 

x y o   B =  x o z  is a skew-symmetric matrix.  y  z 0  Notes : (1) (2) (3)

In a skew-symmetric matrix all the diagonal elements are zero. ( aii = – aii  aii = 0) For any square matrix A, A + A is symmetric & A – A is skew-symmetric. Every square matrix can be uniqualy expressed as a sum of two square matrices of which one is symmetric and the other is skew-symmetric. A = B + C, where B =

1 1 (A + A) & C = (A – A). 2 2

Example # 3 : Show that BAB is symmetric or skew-symmetric according as A is symmetric or skewsymmetric (where B is any square matrix whose order is same as that of A).


6

MATHS Solution :

Case - A is symmetric  A = A (BAB) = (B)AB = BAB  BAB is symmetric. Case -  A is skew-symmetric  A = – A (BAB) = (B)AB = B ( – A) B = – (BAB)  BAB is skew-symmetric

Self practice problems : (3)

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

(4)

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

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 b c d    A = x y z w  p q r s 

Then

a c    a b d  x z , p q s  ,  p r  

a b c     x y z are all submatrices of A. p q r 

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 a b Let A =   , then |A| is defined as ad – bc.  c d

e.g.

 5 3 A=   , |A| = 23  1 4 

Minors & Cofactors : Let  be a determinant. Then minor of element aij, denoted by Mij, is defined as the determinant of the submatrix obtained by deleting ith row & j th column of . Cofactor of element aij, denoted by Cij, is defined as Cij = (– 1)i + j Mij. e.g. 1

=

a b c d

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

a b c e.g. 2

= p q r x y z

M11 =

q r y z

= qz – yr = C111.


7

MATHS M23 =

a b

= ay – bx, C23 = – (ay – bx) = bx – ay

x y

etc.

Determinant of any order : Let A = [aij]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

a11 a12  A = a 21 a 22 a 31 a 32

a13   a 23  a 33 

|A| = a11C11 + a12 C12 + a13C13 (using first row). = a11

a 22

a 23

a 32

a 33

a 21 a 23

– a12

a 31 a 33

+ a13

a 21 a 22 a 31 a 32

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

a 21 a 23 a 31 a 33

a11 a13

+ a22

a 31 a 33

– a32

a11

a13

a 21 a 23

.

Transpose of a determinant : The transpose of a determinant is the determinant of transpose of the corresponding matrix.

a1 D = a2 a3

b1

c1

b2

c2

b3

c3



D  T

a1 a 2

a3

b1 b 2

b3

c1 c 2

c3

Properties of determinant : (1)

|A| = |A| for any square matrix A.

i.e.

the value of a determinant remains unaltered, if the rows & columns are inter changed,

i.e.

a1 b1 a D = 2 b2 a3 b3

(2)

e.g.

(3)

c1 a1 a 2 c 2  b1 b 2 c3 c1 c 2

a3 b 3 = D c3

If any two rows (or columns) of a determinant be interchanged, the value of determinant is changed in sign only.

a1 b1 c 1 Let D1 = a 2 b 2 c 2 a 3 b3 c 3

a2 b2

c2

& D2 = a1 b1 c 1 a 3 b3 c 3

Then D2 = – D1

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


8

MATHS a1 b1 c 1 D = a 2 b2 c 2 a 3 b3 c 3

Ka1 Kb1 Kc1 b2 c2 and E = a 2 a3 b3 c3

Then E= KD

(4)

|A| = n |A|, when A = [aij]n.

(5)

A skew-symmetric matrix of odd order has deteminant value zero.

(6)

If a determinant has all the elements zero in any row or column, then its value is zero,

0 0 0 D = a 2 b2 c 2 a 3 b3 c 3

i.e. (7)

If a determinant has any two rows (or columns) identical (or proportional), then its value is zero,

a1

b1

i.e.

(10)

= 0.

If each element of any row (or column) can be expressed as a sum of two terms then the determinant can be expressed as the sum of two determinants, i.e.

a1x b1y c1z

(9)

c1

D = a1 b1 c 1 a 3 b3 c 3

i.e.

(8)

= 0.

a1

b1

c1

x

y

z

a2

b2

c2

 a2

b2

c 2  a2

b2

c2

a3

b3

c3

a3

b3

c3

b3

c3

a3

The value of a determinant is not altered by adding to the elements of any row (or column) a constant multiple of the corresponding elements of any other row (or column),

a1

b1

c1

D1 = a 2 a3

b2

c2

b3

c3

and D2 =

a1  ma 2

b1  mb 2

c 1  mc 2

a2

b2

c2

a 3  na1

b3  nb1

c 3  nc 1

. Then D2= D1

Let A = [aij]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).

a b c Example # 4

Simplify

b c a c a b

Solution :

Let



R1  R1 + R2 + R3

abc abc abc b c a c

a

b


9

MATHS 1 1 1 b c a = (a + b + c) c a b C 1 C 1 – C 2, C 2  C 2 – C 3

Apply

0

0

1

= (a + b + c) b  c c  a a c a a b b = (a + b + c) ((b – c) (a – b) – (c – a) 2) = (a + b + c) (ab + bc – ca – b2 – c 2 + 2ca – a2) = (a + b + c) (ab + bc + ca – a2 – b2 – c 2)  3abc – a3 – b3 – c 3

Example # 5

a a2

Simplify

b b2

c c2

bc ca ab Solution :

Given detereminant is equal to

=

1 abc

a2

b2

c2

a3

b3

c3

abc abc abc

abc = abc

a2

b2

c2

3

3

c3

1

1

a

1

b

Apply C1  C1 – C2,

=

C2  C2 – C3

a2  b2

b2  c 2

c2

a b

b c

c3

3

0

3

3

0

3

1 ab

2 2 = (a – b) (b – c) a  ab  b 0

bc

c2

b 2  bc  c 2

c3

0

1

= (a – b) (b – c) [ab2 + abc + ac 2 + b3 + b2C + bc 2 – a2b – a2c – ab2 – abc – b3 – b2c] = (a – b) (b – c) [c(ab + bc + ca) – a(ab + bc + ca)] = (a – b) (b – c) (c – a) (ab + bc + ca)

Factor Theorem : 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.


10

MATHS

Example # 6

a b c 2 2 2 Prove that a b c bc ca ab

Solution :

Let a = b



a b c 2 2 2 D= a b c bc ac ab

= (a – b) (b – c) (c – a) (ab + bc + ca) by using factor theorem.

=0

Hence (a – b) is a factor of determinant Similarly, let b = c, D = 0 c = a, D = 0 Hence, (a – b) (b – c) (c – a) is factor of determinant. But the given determinant is of fifth order so

a a2

b b2

c c2

= (a – b) (b – c) (c – a) { (a2 + b2 + c 2) + µ (ab + bc + ca)}

bc ca ab Since this is an identity so in order to find the values of  and µ. Let a = 0, b = 1, c = – 1 – 2 = (2) (2 – µ) (2 – µ) = – 1. ........(i) Let a = 1, b = 2, c = 0

1 2 0 1 4 0

= (–1) 2 (– 1) (5 + 2µ)

0 0 2  5 + 2µ = 2 .......(ii) from (i) and (ii)  = 0 and µ = 1

a

b

2

2

Hence a

b

c c2

= (a – b) (b – c) (c – a) (ab + bc + ca).

bc ca ab

Self practice problems

(5)

0 ba c a 0 c b . Find the value of  = a  b ac bc 0


11

MATHS b 2  ab b  c bc  ac 2 2 Simplify ab  a a  b b  ab . bc  ac c  a ab  a 2

(6)

(7)

Prove that

abc 2a 2b bc a 2c

c ab

2c

1 a bc Show that 1 b ca

(8)

2a 2b

= (a + b + c)3.

= (a – b) (b – c) (c – a) by using factor theorem .

1 c ab Answers :

(5)

0

(6)

0

Multiplication of two determinants : If A and B are two square matrices of same order, then |AB| = |A| |B|.

a1

b1

a2

b2



1

m1

2

m2



a1 1b1 2

a1m1b1m 2

a 2  1b 2  2

a 2 m1b 2 m 2

a1 b1 c 1 a1 1  b1 2  c 1 3  1 m1 n1 a 2 b 2 c 2 ×  2 m 2 n 2 = a 2 1  b 2 2  c 2  3 a 3 b3 c 3 a 3  1  b 3 2  c 3 3  3 m 3 n3

Note : As |A| = |A|, we have

Example # 7

Solution.

Find the value of 1

2

1 3

×

3

a 3m1  b 3m 2  c 3m3

a 3n1  b 3n 2  c 3n3

|A| |B| = |AB| (row - row method) |A| |B| = |AB| (column - column method) |A| |B| = |AB| (column - row method) 2

1 3

×

3

0

1 4

and prove that it is equal to

1

8

 6 12

.

0

1 0  2  4

 1 3  3  ( 1)  1 0  3  4 1

=

a1n1  b1n 2  c 1n 3 a 2n1  b 2n 2  c 2n3

1 4

1 3  2  1

=

1

a1m1  b1m 2  c 1m 3 a 2m1  b 2m 2  c 2m3

8

 6 12

= 60

a1x1  b1y1 Example # 8

a1x 2  b1y 2

Prove that a 2 x1  b 2 y1 a 2 x 2  b 2 y 2 a 3 x1  b 3 y 1 a 3 x 2  b 3 y 2

a1x 3  b1y 3 a2 x 3  b 2 y 3 = 0 a3 x 3  b 3 y 3


12

MATHS Solution.

Given determinant can be splitted into product of two determinants

a1x1  b1y1

a1x 2  b1y 2

a1x 3  b1y 3

a 2 x1  b 2 y 1 a 2 x 2  b 2 y 2

i.e.

a 3 x1  b 3 y 1 a 3 x 2  b 3 y 2

Example # 9

b1

c1

a2 x 3  b 2 y 3 = a 2 b2 a3 x 3  b 3 y 3 a 3 b3

c2

(a1  b1 )2

(a1  b 2 )2

(a1  b 3 )2

Prove that (a 2  b1 )2

(a 2  b 2 ) 2

(a 2  b 3 )2

(a3  b1 )2

(a 3  b 2 ) 2

(a 3  b 3 )2

a1

c3

x1 x 2 × y1 y 2 0 0

x3 y3 = 0 0

= 2(a1 – a2) (a2 – a3) (a3 – a1) (b1 – b2) (b2 – b3) (b3 – b1).

Solution.

(a1  b1 )2

(a1  b 2 )2

(a1  b 3 )2

(a 2  b1 )2

(a 2  b 2 ) 2

(a 2  b 3 )2

(a3  b1 )2

(a 3  b 2 ) 2

(a 3  b 3 )2

=

2

2

2

2

a1  b3  2a1b 3 2 2 a 2  b3  2a 2b 3

2

2

2

2

a 3  b3  2a 3b 3

1

1

a1  b1  2a1b1 a1  b 2  2a1b 2 2 2 2 2 a2  b1  2a 2b1 a 2  b 2  2a 2b 2 a3  b1  2a 3b1 a3  b 2  2a 3b 2 2

a1

1  2a1

= a 2 1  2a 2 2 a3 1  2a 3 2

×

2 b1

b2

b1

2

b2

2

2

2

2

2

1 b3

2

b3

1 a1

a1

2

= 2 1 a2 2 1 a3

a2 a3

2

b1

2

b2

1 b1

× 1 b2 2 1 b3

b3

= 2(a1 – a2) (a2 – a3) (a3 – a1) (b1 – b2) (b2 – b3) (b3 – b1) Note : The above problem can also be solved using factor theorem method. Self practice problems (9)

Find the value of 

2bc  a 2 c

2

b2

c2 2ca  b a2

b2 2

a2 2ab  c 2

1 (10)

cos(B  A ) cos(C  A )

1 cos(C  B) . If A, B, C are real numbers then find the value of  = cos( A  B) cos( A  C) cos(B  C) 1

Answers :

(9)

(3abc – a3 – b3 – c 3)2

(10)

0

f(r) g(r ) h(r ) Summation of determinants : Let (r) = a1 a 2 a3 b1 b 2 b 3

where a1, a2, a3, b1, b2, b3 are constants

indepedent of r, then


13

MATHS n



n



f (r )

r 1

n

 (r ) = r 1

n

g(r )

r 1

 h(r ) r 1

a1

a2

a3

b1

b2

b3

Here the functions 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.

2r  1

n

Example # 10 Evaluate



n

r

n2 2

n

n

2

Cr

r 1

r

r 1

n



n

cos 2 

y

n2

2n – 1

2n 1 – 2

n

(2r  1)



x n2

cos 2  y n n1 2 1 2  2

r 1

x

=

r 1

n Cr

2

r

r 1

2n  1 2n 1  2 cos 2 

x n

n

r 1

r 1

=

y

 (2r  1) 

n

D

cos 

2n – 1 2n1 – 2

n2

n

Solution :

2r

2

x

r 1

Cr

y

2 1 2

n 1

n 2

n2

n

2

=0

Example # 11 : Dr =

Cr  2

Cr 1

n2

Cr

3

1

1

2

1

0

n

evaluate

D

r

r 2

n 2 n

n

Solution :

D

r

r 2

n2

=

=

 r 2

Cr  2

n2

Cr 1

n2

Cr

3

1

1

2

1

0

C0  n  2 C1  ....  n 2 Cn 2

n2

C1  n 2 C2  ....  n 2 Cn 2

n2

C2  n 2 C3  ....  n 2 Cn 2

3

1

1

2

1

0


14

MATHS 2 n 2 =

2n  2  1 2n  2  1  n

3

1

1

2

1

0

C1  C1 – 2 × C2

2n 2  2n1  2 2n2  1 2n 2  1  n =

= (–1)

1

1

1

0

1

0

2n2  2n1  2 2n2  1  n 1

1

= 2n – 1 – n – 3

r 1

1

0

n

3  r , find Example # 12 If r = r 1 1  2 2

Solution.

r



r

r 1

On expansion of determinent, we get Dr = (r –1) (3 – r) + 7 + r2 + 4r = 8r + 4 n





= 4n (n + 2)

r

r 1

Self Practice Problem n

(11)

Evaluate

D

r

r 1

Answer :

(11)

r 1

x

6

(r  1)

2

y

4n  2

(r  1)

3

z 3n2  3n

0

Differentiation of determinant :

f1( x )

f1( x ) f2 ( x ) f3 ( x ) g Let (x) = 1( x ) g2 ( x ) g3 ( x ) h1( x ) h 2 ( x ) h3 ( x )

f2 ( x )

f3 ( x )

f1( x ) f2 ( x ) f3 ( x ) f1( x ) f2 ( x ) f3 ( x ) then (x) = g1( x ) g2 ( x ) g3 ( x ) + g1 ( x ) g2 ( x ) g3 ( x ) + g1( x ) g2 ( x ) g3 ( x ) h1( x ) h 2 ( x ) h3 ( x ) h1 ( x ) h2 ( x ) h3 ( x ) h1( x ) h2 ( x ) h3 ( x ) Note : We can differentiate a determinant columnwise also.


15

MATHS 3 Example # 13 If f(x) = 6x 1

2 2

2x a

3 Solution.

1 3

2

f(x) = 12x 6 x 1 a

x 4 , then find the value of f(a). a2 1

2

4x 3 a2

3 2 1 12 12 x 12 x2 f(x) = 1 a a2 3 2 1 2 f(a) = 12 1 a a = 0. 1 a a2



Example # 14 Let  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

A( x ) A( )

B( x ) B( )

C( x ) C( )

divisible by f(x).

A( ) B( ) C( )

Solution.

Let

g(x) =

A( x )

B( x )

C( x )

A ( )

B( )

C( )

A( ) B( ) C( ) A( x ) B( x ) C( x ) 

A ( )

g(x) =

B( )

C( )

A( ) B( ) C( ) Since 

g() = g() = 0 g(x) = (x – )2 h(x) i.e.  is the repeated root of g(x) and h(x) is any polynomial expression of degree 3. Also f(x) = 0 have repeated root . So g(x) is divisible by f(x).

Example # 15 Prove that F depends only on x 1, x 2 and x 3

1

1

x1  a1

F=

x12

 b1x1  b 2

1

x 2  a1 x 22

 b1x 2  b 2

x 3  a1 x 32

 b1x 3  b 2

and simplify F.

Solution :

dF = da1

0

0

x1  a1 x12

x 2  a1

 b1x1  b 2

+

0

x 22

x 3  a1

 b1x 2  b 2

x 32

 b1x 3  b 2

1

1

1

1

1

1

x12  b1x1  b 2

x 22  b1x 2  b 2

x 32  b1x 3  b 2

1

1

1

+ x1  a1 x 2  a1 x 3  a1 = 0 0 0 0

Hence F is independent of a1.


16

MATHS dF dF = = 0. db 2 db1

Similarly

Hence F is independent of b1 and b2 also. So F is dependent only on x 1, x 2, x 3 Put a1 = 0, b1 = 0, b2 = 0

1 F = x1 x12



1 x2

1 x3

x 22

x 32

= (x 1 – x 2) (x 2 – x 3) (x 3 – x 1).

ex

Example # 16 If Solution :

sin x

= A + Bx + Cx 2 + ....., then find the value of A and B.

cos x n(1  x )

Put x = 0 in

ex

sin x

= A + Bx + Cx 2 + .......

cos x n(1  x ) 1 0



1 0

=A

A = 0. Differentiating the given determinant w.r.t x, we get

ex

sin x 1 +  sin x cos x n(1  x ) 1 x Put x = 0, we get

ex

cos x

1 1 1 0



= B + 2 C x + ......

1 0

+

0 1

=0

 B = –1 + 1 = 0 A = 0, B = 0

Self practice problem

(12)

If (i)

Answers :

x x 1 x  2x x  1 1 x 1

1

= ax 3 + bx 2 + cx + d. Find

x

d

(ii)

a+b+c+d

(12)

(i)

–1

Integration of a determinant :

Let

(iii)

b

(iii)

–4

(ii)

–5

(x) =

f ( x ) g( x ) h( x ) a1 b1 c1 a2

b2

c2

where a1, b1, c 1, a2, b2, c 2 are constants independent of x. Hence


17

MATHS b



b



b

f ( x ) dx

a

( x ) dx =

 a

a1 a2

a

b

g( x ) dx

 h( x) dx a

b1 b2

c1 c2

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

cos x 1 0 1 2 cos x 1 Example # 17 If f(x) = , then find 0 1 2 cos x

Solution.

 f ( x ) dx 0

Here f(x) = cos x (4 cos 2x – 1) –2 cos x = 4 cos 3x – 3 cos x = cos 3x /2

/2

so



sin 3 x  3  0

cos 3 x dx =

0

1 3

=–

 2  1 2  2  2  3 Example # 18 If  =

6

4

3

x

x2

x3

2  1

1

Solution.

/2

 (x ) dx 0

=

1

, then find

0

2  2

6

2  3

4

1

3

1

 x dx  x 0

 (x ) dx

1

2

dx

0

x

3

dx

0

 2  1 2  2  2  3 =

6 1 2

4 1 3

3 1 4

 2  1 2  2  2  3 1 12

6

4

3

6

4

3

a 2x + b 2y + c 2 = 0

then:

=

=0

Cramer's Rule: System of linear equations : (i)

Two variables Let a1x + b1y + c 1 = 0

&

a1 b c  1  1  a2 b2 c2

Given equations are inconsistent

a1 b c  1  1  a2 b2 c2

Given equations are consistent


&

18

MATHS (ii)

Three variables consider the system

Then,

a 1x + b 1y + c 1z = d 1 a 2x + b 2y + c 2z = d 2 a 3x + b 3y + c 3z = d 3

D.x = D1, D.y = D2, D.z = D3

Where D =

a1 b1 c 1 a 2 b2 c 2 a3

b3

c3

; D1 =

d1 b1 c 1 d2 b 2 c 2 d3

b3

c3

; D2 =

a1 a2

d1 c 1 d2 c 2

a3

d3

c3

& D3 =

a1 b1 a2 b2

d1 d2

a3

d3

b3

Consistency of a system of equations (a)

If D  0 and alteast one of D1, D2, 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. (For 2 × 2 system, D = 0 = D1 = D2  system has infinitely many solutions).

(d)

If D = 0 but atleast one of D1, D2, D3 is not zero then the equations are inconsistent and have no solution.

Homogeneous system : a1x + b1y + c1z = 0 a 2x + b 2y + c 2z = 0 a 3x + b 3y + c 3z = 0 (x, y, z) = (0, 0, 0) is always a solution of this system. This solution is called as the trivial solution (or zero solution) of this system. D0  this system has only the trivial solution. D=0  this system has nontrivial solutions (infinitely many solutions).

Three equation in two variables : If x and y are not zero, then condition for a1x + b1y + c 1 = 0 ; a2x + b2y + c 2 = 0 &

a1 b1 c 1 a3x + b3y + c 3 = 0 to be consistent in x and y is a 2 b 2 c 2 = 0. a 3 b3 c 3

Example # 19 : Find the nature of solution for the given system of equations. x + 2y + 3z = 1 2x + 3y + 4z = 3 3x + 4y + 5z = 0


19

MATHS

Solution :

1 2 3 Let D = 2 3 4 3 4 5 apply C1  C1 – C2 , C2  C2 – C3

1 1 3 D=

1 1 4

=0

D=0

1 1 5 1 2 3 Now, D1 = 3 3 4 0 4 5 C3  C3 – C2

1 2 1 D1 = 3 3 1 0 4 1 R1  R1 – R2 , R2  R2 – R3

 2 1 0 D1 =

3

1 0

0

4

=5

1

D = 0 But D1  0

Hence no solution

Example # 20 : Solve the following system of equations x+y+z=1 2x + 2y + 2z = 3 3x + 3y + 3z = 4

1 1 1 Solution :



D= 2 2 2 =0 3 3 3

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 # 21: Solve the following system of equations x+y+z=2 2x + 2y + 2z = 4 3x + 3y + 3z = 6

Solution :



D=

1 1 1 2 2 2

=0

3 3 3 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.


20

MATHS Let z = t1, y = t2 where t1, t2  R.



x = 2 – t1 – t2

Example # 22 Consider the following system of equations x+y+z=6 x + 2y + 3z = 10 x + 2y + z =  Find values of  and  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 + z = 

D=

1 1 1 1 2 3 1 2 

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

6 D1 =

D2 =

D3 =

1 1

10 2 3 

2 

1 6 1 1 10 3 1 



1 1 6 1 2 10 1 2



If  = 3 then D1 = D2 = D3 = 0 for µ = 10 (i)

For unique solution D  0 i.e. 3

(ii)

For infinite solutions D=0  =3 D1 = D2 = D3 = 0  µ = 10. For no solution D=0  =3 Atleast one of D1, D2 or D3 is non zero

(iii)



µ  10.

Self practice problems (*13)

Solve the following system of equations x + 2y + 3z = 1 2x + 3y + 4z = 2 3x + 4y + 5z = 3

(14)

Solve the following system of equations


21

MATHS x + 2y + 3z = 0 2x + 3y + 4z = 0 x–y–z=0 (15)

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.

(16)

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

Answers :

(13) (14)

x=1+t y = –2t x = 0, y = 0, z = 0

(15)

x =

(16)

t=7

z=t

where t  R

ac ba c b ,y= ,z= a bc a bc a bc

Application of determinants : Following examples of short hand writing large expressions are: (i)

Area of a triangle whose vertices are (x r, yr); r = 1, 2, 3 is:

1 D= 2

x1 x2

y1 1 y2 1

x3

y3 1

If D = 0 then the three points are collinear.

x (ii)

Equation of a straight line passing through (x 1, y1) & (x 2, y2) is x1 x2

(iii)

The lines:

1

y1 1 = 0 y2 1

a1x + b1y + c 1 = 0........ (1) a2x + b2y + c 2 = 0........ (2) a3x + b3y + c 3 = 0........ (3)

are concurrent if,

(iv)

y

a1 a2 a3

b1 b2 b3

c1 c 2 = 0. c3

Condition for the consistency of three simultaneous linear equations in 2 variables. ax² + 2 hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines if:

a h g abc + 2 fgh  af²  bg²  ch² = 0 = h b f g f c

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.

Cofactor matrix & adjoint matrix :

Let A = [aij] 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.


22

MATHS i.e. if A = [aij]n then cofactor A = [c ij]n when c ij is the cofactor of aij  i & j. Adj A = [dij]n where dij = c ji  i & j.

Properties of cofactor A and adj A: (a) (b) (c) (d)

A . adj A = |A| n = (adj A) A where A = [aij]n. |adj A| = |A|n – 1, where n is order of A. In particular, for 3 × 3 matrix, |adj A| = |A|2 If A is a symmetric matrix, then adj A are also symmetric matrices. If A is singular, then adj A is also singular.

Example # 23 : For a 3×3 skew-symmetric matrix A, show that adj A is a symmetric matrix.

Solution :

 c 2  bc ca    2  ab  cof A =  bc b  ca  ab a 2   

a b 0    a 0 c A=    b  c 0

 c 2  bc ca    2  ab  which is symmetric. adj A = (cof A) =  bc b  ca  ab a 2   

Inverse of a matrix (reciprocal matrix) : 1 adj A is the |A| multiplicative inverse of A (we call it inverse of A) and is denoted by A–1. We have A (adj A) = |A| n = (adj A) A Let A be a non-singular matrix. Then the matrix



 1  adj A  = n = A  | A | 



A–1 =

 1   adj A  A, for A is non-singular | A | 

1 adj A. |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, a22, ....., ann) where aii  0  i, then A–1 = diag (a11– 1, a22–1, ...., ann–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 =

7.

1 |A–1| = | A | for |A|  0.


1 –1 A . k

23

MATHS 8.

Let A be a non-singular matrix. Then AB = AC  B = C &

BA = CA  B= C.

9.

A is non-singular and symmetric  A–1 is symmetric.

10.

(AB)–1 = B–1 A–1 if A and B are non- singular.

11.

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.

Example # 24 : For two non-singular matrices A & B, show that adj (AB) = (adj B) (adj A) Solution : We have (AB) (adj (AB)) = |AB| n = |A| |B| n A–1 (AB)(adj (AB)) = |A| |B| A–1 

B adj (AB) = |B| adj A

 

B–1 B adj (AB) = |B| B–1 adj A adj (AB) = (adjB) (adj A)

(

1 A–1 = | A | adj A)

Self practice problems : (17)

If A is non-singular, show that adj (adj A) = |A|n – 2 A.

(18)

Prove that adj (A–1) = (adj A)–1.

(19)

( n 1) For any square matrix A, show that |adj (adj A) | = | A | .

(20)

If A and B are non-singular matrices, show that (AB) –1 = B–1 A–1.

2

System of linear equations & matrices : Consider the system a11 x 1 + a12x 2 + .......... + a1nx n = b1 a21x 1 + a22 x 2 + ..........+ a2n x n = b2 ................................................. am1x 1 + am2x 2 + ..........+ amnx n = bn.

Let

 a11 a12  a 21 a 22 A =  ..... .....  a m1 a m2

 b1  a1n   x1       b 2  .......... a 2n  x2   ,X= & B =  ...  . ....  .......... .....        ...  .......... a mn  xn  b   n ..........

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.

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. Results :

(1) (2)

If A is non-singular, 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 infinitely many solutions.


24

MATHS (3)

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

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 = ....... bn = 0), where A is a square matrix.

Results :

(1) (2)

If A is non-singular, 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.

Rank of a matrix : Let A = [aij]m×n. A natural number  is said to be the rank of A if A has a non-singular submatrix of order  and it has no non-singular submatrix of order more than . Rank of zero matrix is regarded to be zero.

eg.

3  1 2 5   A = 0 0 2 0 0 0 5 0

3 2 we have   as a non-singular submatrix. 0 2

The square matrices of order 3 are

3  1 2 3  1 5 3 2 5       0 0 2 , 0 0 0 , 0 2 0 , 0 0 5 0 0 0 0 5 0    

 1 2 5    0 2 0  0 5 0

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

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. Remarks : 1. 2.

Elementary transformation of a matrix does not affect its rank. Two matrices A & B are said to be equivalent if one is obtained from other using elementary transformations. We write A  B.

Echelon form of a matrix : (a) (b)

A matric is said to be in Echelon form if it satisfy the followings: 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. The number of zeroes before the first non zero element in any non zero row is not more than the number of such zeroes in succeeding rows.


25

MATHS 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 :

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.

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 + 1th column & first n columns are that of A). (A) denote rank of A and ([AB]) denote rank of the augmented matrix. Clearly (A) ([AB]). Results :

(1) (2) (3)

If (A) < ([AB]) then the system has no solution (i.e. system is inconsistent). If (A) = ([AB]) = number of unknowns, then the system has unique solution. (and hence is consistent) If (A) = ([AB]) < number of unknowns, then the systems has infinitely many solutions (and so is consistent).

Homogeneous system & rank of matrix : Let the homogenous system be AX = 0, m equations in 'n' unknowns. In this case B = 0 and so (A) = ([AB]). Hence if (A) = n, then the system has only the trivial solution. If (A) < n, then the system has infinitely many solutions.

xyz6 Example # 25 : Solve the system x  y  z  2 using matrix inverse. 2x  y  z  1

Solution :

6 1 1 1  x       Let A = 1  1 1  , X =  y  & B = 2 .  1 2 1  1  z  Then the system is AX = B. |A| = 6. Hence A is non singular.

3 0 3   2  3 1 Cofactor A =  2 0  2 2 0 2   adj A = 3  3 0  3 1  2 A–1

2 1/ 3 1/ 3  0 2  0 1 1     3  3 0 1 / 2  1 / 2 0  = | A | adj A =  =  6  3 1  2 1/ 2 1/ 6  1/ 3


26

MATHS

X= 

A–1

1/ 3 1/ 3   0   0  B = 1/ 2  1/ 2 1/ 2 1/ 6  1/ 3

6    2  1

x = 1, y = 2, z = 3.

Example # 26 : Test the consistency of the system

Solution :

i.e.

x 1      y  = 2  z  3

1  1  1 1 A =  1 3  2 4

x – y  2z  1 x y z3 x – 3y  3z  – 1 2x  4y  z  8.

. Also find the solution, if any..

1 2 x    1   3 y X = , B =    1 3  z     1 8

1  1  1 1 [AB] =  1  3  2 4

1  3 3  1  1 8 2 1

1 1  1 2   R 2  R 2  R1 0 2  1 2  R 3  R 3  R1   0  2 1  2   R 4  R 4  2R1 0 6  3 6 

2 1  1  0 1  1/ 2   0 1  1/ 2  0 1  1/ 2

1  0   0  0

1 1 R 2  R 2 2  1 1 R3   R3 2 1  1 1 R 4  R 4 6

2  R1  R1  R 2 1  1/ 2 1 R3  R3  R2 0 0 0  R4  R4  R2 0 0 0 0

3/2

This is in Echelon form. (AB) = 2 = (A) < number of unknowns Hence there are infinitely many solutions n –  = 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. Then x – y = 1 – 2r x+y=3–r 4  3r 2r  x= &y= 2 2  4  3r 2  r  , ,r  .  Solutions are (x, y, z) =  2  2 


27

MATHS Self practice problems:

(21)

(22)

(23)

0 1 2   A =  1 2 3 . Find the inverse of A using |A| and adj A. 3 1 1 Find real values of  and µ so that the following systems has (i) unique solution (ii) infinitely many solutions x+y+z=6 x + 2y + 3z = 1 x + 2y + z = µ

(iii)

No solution.

Find  so that the following homogeneous system have a non zero solution x + 2y + 3z = x 3x + y + 2z = y 2x + 3y + z = z

Answers : (21)

1 5 4 2 2 1 3  3  2 2 1 1 1 2 2

(22) (i)  3, µ  R (ii)  = 3, µ = 1 (iii)  = 3, µ  1 (23)  = 6

Characteristic polynomial & characteristic equation : Let A be a square matrix. Then the polynomial | A – x| is called as characteristic polynomial of A & the equation | A – x| = 0 is called as characteristic equation of A.

Cayley - Hamilton theorem : i.e.

Every square matrix A satisfies its characteristic equation a0 x n + a, x n – 1 + ........ + an – 1x + an = 0 is the characteristic equation of A, then a0An + a1An – 1 + ......... + an – 1 A + an  = 0

1 2 0    Example # 27 : If A = 2  1 0  , show that 5A–1 = A2 + A – 5. 0 0  1 Solution : We have the characteristic equation of A. | A – x| = 0 2 0  1  x    1 x 0  =0 i.e.  2  0 0  1  x  i.e. x 3 + x 2 – 5x – 5 = 0. Using Cayley - Hamilton theorem. A3 + A2 – 5A – 5 = 0 Multiplying by A–1, we get 5A–1 = A2 + A – 5



5 = A3 + A2 – 5A

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 Ap = O.


28

MATHS Idempotent matrix : A square matrix A is said to be idempotent if, A2 = A.  1 0 e.g.   is an idempotent matrix. 0 1

Involutory matrix :

A square matrix A is said to be involutory if A2 = , being the identity matrix.  1 0 e.g. A =   is an involutory matrix. 0 1

Orthogonal matrix : A square matrix A is said to be an orthogonal matrix if, A A =  = AA.

Example # 28 : Show that a square matrix A is involutory, iff ( – A) ( + A) = 0 Solution :

Let A be involutory Then A2 =  ( – A) ( + A) =  + A – A – A2 =  + A – A – A2 =  – A2 =0 Conversly, let ( – A) ( + A) = 0  + A – A – A2 = 0   + A – A – A2 = 0   – A2 = 0  A is involutory

Self practice problems (24)

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

(25)

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

(26)

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

c 0  (27) Let A =  c 0  b  a Answer (4) a2

 b  a  . If A3 + A = 0, find . 0  + b 2 + c 2.


29

Matrices and Determinants Theory_E

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