SYMMETRY IN CHEMISTRY

SYMMETRY IN CHEMISTRY Professor MANOJ K. MISHRA CHEMISTRY DEPARTMENT IIT BOMBAY ACKNOWLEGDEMENT: Professor David A. Micha Professor F. A. Cotton

WHY SYMMETRY? An introduction to symmetry analysis For H atom: Hψ = Eψ Each member of the CSCO labels, For molecules: Symmetry operation R CSCO H is INVARIANT under R ( by definition too) 2

An introduction to symmetry analysis Hψ = Eψ gives NH 3 normal modes = NH 3 rotation or translation MUST be A, A 2 or E! NO ESCAPING SYMMETRY! 3

Molecular Symmetry (Ref.: Inorganic chemistry by Shirver, Atkins & Longford, ELBS) One aspect of the shape of a molecule is its symmetry (we define technical meaning of this term in a moment) and the systematic treatment and symmetry uses group theory. This is a rich and powerful subject, by will confine our use of it at this stage to classifying molecules and draw some general conclusions about their properties. An introduction to symmetry analysis An introduction to symmetry analysis Our initial aim is to define the symmetry of molecules much more precisely than we have done so far, and to provide a notational scheme that confirms their symmetry. In subsequent chapters we extend the material present here to applications in bonding and spectroscopy, and it will become that symmetry analysis is one of the most pervasive techniques in inorganic chemistry. Symmetry operations and elements A fundamental concept of group theory is the symmetry operation. It is an action, such as a rotation through a certain angle, that leave molecules apparently unchanged. An example is the rotation of H2O molecule by 80 (but not any smaller angle) around the bisector of HOH angle. Associated with each symmetry operation there is a symmetry element; this is a point, a line, or a plane with respect to which symmetry operation is performed. The most important symmetry operation and their corresponding elements are listed in table 2.6. All operations leaves at least one point of the molecule unmoved, just as rotation of a sphere leaves its center unmoved. Hence they are operations of point-group symmetry. The identity operation E leaves whole molecule unchanged. Fig.2.5. An H2O molecule may be rotated through any angle about the bisector of the HOH bond angle, but only a rotation of 80, C2, leaves it apparently unchanged. 4

The rotation of an H2O molecule by 80 (n = 2) around a line bisect the HOH angle is a symmetry operation, so the H2O molecule possess two-fold rotation axis C2 (fig. 2.5). In general, an n-fold rotation symmetry operation if the molecule appears unchanged after rotation 360 /n. The corresponding symmetry element is a line, the n-fold rotational axis Cn, about which the rotation is performed. The triangular pyramid NH3 molecule has a threefold rotation axes, denoted C3, but there are operations associated with it, one a rotation by 20 and the other a rotation through twice this angle (Fig. 2.6). The two operations are denoted C3 and. We do not need to consider, a rotation through 3x 20 = 360, since it is equivalent to the identity. Table 2.6. Important symmetry operations and symmetry elements Symmetry element Symmetry operation Symbol Identity N-fold symmetry axis Rotation by 2π/n C n Mirror plane Reflection σ Center of inversion Inversion i An introduction to symmetry analysis E N-fold axis of improper rotation Rotation by 2π/n followed by reflection perpendicular to rotation axis The symmetry element can be thought of as the molecule as a whole. Note the equivalences S = σ and S 2 = i. S n 5

An introduction to symmetry analysis Fig. 2.6. A three-fold rotation and the corresponding C 3 axis in NH 3. There are two rotations associated with this axis. One through 20 (C 3 ) and the other through 240 (C 32 ). The reflection of an H 2 O molecule in either of the two planes shown in Fig. 2.7 is a symmetry operation: corresponding symmetry element is a mirror plane σ. The H 2 O molecule has two mirror planes which intersect at the bisector of the HOH angle. Because the planes are vertical (in the sense of being parallel to the rotational axis of the molecule), they are labeled σ v and σ v. The C 6 H 6 molecule has a mirror plane σh in the plane of the molecule. Fig. 2.7. The two vertical mirror planes σ v and σ v in H 2 O and the corresponding operations. Both planes cut through the C 2 axis. The h signifies that the plane is horizontal in the sense that the principal rotational axis of the molecule is perpendicular to it. This molecule also has two more sets of three mirror planes that intersect the six-fold axis (Fig. 2.8). In such cases, the members of one set are called vertical and the members of the other are called dihedral. The symmetry elements (and the associated operations) are denoted σ v and σ d respectively. 6

An introduction to symmetry analysis Fig. 2.8. Some of the symmetry elements of the benzene ring. There is one horizontal reflection plane (σh) and two sets of vertical reflection plane (σv and σd); one example of each is shown. Fig. 2.9. The inversion operation and the center of inversion (i) in SF 6 The inversion operation consists of imagining that each point of the molecules is taken through a single center and projected an equal distance on the other side (Fig. 2.9). The symmetry element is the center of inversion (i), the point through which projection is made. An N 2 molecule has a center of inversion midway between the two nitrogen nuclei. The H 2 O molecule does not possess this element, but a benzene molecule and a sulfur hexafluoride molecule both do. In due coarse we shall see the importance of recognizing that an octahedron has a center of inversion but that a tetrahedron does not. Fig. 2.0 A four fold axis of improper rotation S4 in the CH4 molecule A=, B=2, C=3, &D=4. The operation of improper rotation is composite: it consist of a rotation followed by a reflection in the plane perpendicular to the axis of rotation. Fig. 2.0 shows a fourfold improper rotation of a tetrahedral CH 4 molecule: in this case, the operation consists of a 90ᴼ rotation about an axis bisecting two HCH bond angles followed by a reflection through a plane perpendicular to the rotation axis. 7

An introduction to symmetry analysis Neither the operation C 4 nor the reflection σ h alone is a symmetry operation for CH 4, but their products C 4 x σ h is a symmetry operation, the improper rotation S 4. The symmetry element, the improper rotation axis S n (S 4 in the example), is the corresponding combination of an n-fold rotational axis and a perpendicular to the mirror plane. The only Sn axes that exist are those for n = and n = 2, 4, 6,. Moreover, S is equivalent to a horizontal reflection σ h and S 2 is equivalent to the inversion (i) (Fig. 2.). Fig. 2. (a) An S axis is equivalent to a mirror plane and (b) an S 2 axis is equivalent to a center of inversion. Example 2.6: Identifying a symmetry element Q. Which conformation of CH 3 CH 3 molecule has an S 6 axis? Answer: We need to find a conformation that leaves the molecule looking the same after a 60 % rotation followed by a reflection in a plane perpendicular to that axis. The conformation and axis are shown in (5); this is the staggered conformation of the molecule, and also the one of lowest energy. Exercise: Identity a C 3 axis of an NH 4 + ion. How many of these axes are there in the ion? 8

The point groups of molecules The symmetry elements possessed by molecules determine the point group to which they belong. Assigning a molecule to a particular group depends on making a list of symmetry elements it possesses and then comparing it with the list that is characteristic of each point group. For example, if a molecule has only the identity element (CHBrClF is an example), we list its elements as E and look for the group that has only this one. In fact, the group called C is the group with only the element E, so the CHBrClF (6) molecule belongs to that group. The molecule CH 2 BrCl belongs to a slightly richer group: it has the elements E (all groups have that element) and a mirror plane. The group of elements E, σ is called C S, so the CH 2 BrCl molecule belongs to that group. This process can be continued, with molecules being assigned to the group that matches the symmetry elements they possess. Some of the more common groups and their names are listed in Table 2.7. Assigning a molecule to its group depends on listing the symmetry elements it possesses and then referring to the table. However, it is often easier to work through the tree in Fig. 2.2 and to arrive at the correct point group by answering the questions at each decision point on the correct point on the chart. (Note that we need to be careful to distinguish the names of the groups, C 2, and so on. From the symbols for the symmetry elements, such as C 2, and the corresponding operations, also C 2. The context will always make it clear what interpretation is intended. 9

Table 2.7: The composition of some common groups Point Symmetry elements Examples group C E SiBrClFl C 2 E, C 2 H 2 O 2 C 8 E, NHF 2 C 2v E, C 2, 2 v H 2 O, SO 2 Cl 2 C 3v E, C 2, 3 v NH 3, PCl 3, POCl 3 C v E, C CO, HCl, OCS D 2h E, 3C 2, 2 v, h, i N 2 O 4, B 2 H 6 D 3h E, C 3, 3C 2, 3 v, h BF 3, PCl 5 D 4h E, C 4, C 2, i, S 4, h,.. XeF 4, trans-ma 4 B 2 D h E, C, v,.. H 2, CO 2, C 2 H 2 T d E, 3C 2, 4C 3, 6, 3S 4 CH 4, SiCl 4 O h E, 6C 2, 4C 3, 4S 6, 3S 4, i,.. SF 6 0

(a) (b) Fig. 2.2: The decision tree for identifying a molecular point group. After passing through part (a), go to part (b) if necessary.

Example 2.7: Identifying the point group of a molecule To what point groups do H 2 O and NH 3 belong? Answer: The symmetry elements are shown in Fig. 2.3. (a) H 2 O possesses the identity (E), a two fold rotation axis (C 2 ), and two vertical mirror planes (2σ V ). The set of elements (E, C2, 2σ V ) corresponds to the group C 2V. (b) NH 3 possesses the identity (E), a three fold rotation axes (C 3 ), and three vertical mirror planes (3σ V ). The set of elements (E, C 3, 3σ V ) corresponds to the group C 3V. Exercise: Identity the point groups of (a) BF 3, a trigonal planar molecule and D 3h (b) the tetrahedral SO 4 2 ion (T d ). Fig. 2.3: The symmetry elements of (a) H 2 O and (b) NH 3. The diagrams on the right are views from above and summarize the diagrams to their left. 2

Linear molecules with a center of symmetry (H 2, CO 2, C 2 H 2,) (7) belong to the point group D h. A molecule that is linear but has no center of symmetry (HCl, OCS, NNO, ) (8) belongs to the point group C v. (9) (0) () Fig. 2.4: shapes with the symmetries of the groups (a) Td (b) Oh and (c) Ih. They are all closely related to the symmetries of a cube. Tetrahedral (T d ) and octahedral (O h ) molecules (Fig. 2.4), which are of great importance in coordination chemistry, have more than one principal axis of symmetry: a tetrahedral CH 4 molecule, for instance, has four C 3 axes, one along each C-H bond. A closely related group, the icosahedral group (I h ) characteristic of the icosahedron (Fig. 2.4c), is important for boron compounds. The three groups are quite easy to recognize: a regular tetrahedron has four equilateral triangles for its faces, an octahedron has eight, and an icosahedrons has 20. 3

(2) (3) For our present purposes, the most important groups are C 2v and C 3v and the tetrahedral and octahedral groups. We shall also sometimes encounter C v and D h for linear molecules, D 3h for trigonal planar molecules (such as BF 3, 9) and trigonal bipyramidal molecules such as PCl 5, 0), and D 4h for square-planar molecules () and octahedral molecules with two substituents opposite to each other, as in (2). This last example shows that the point group classification of molecule is more precise than the casual use of the term octahedral or tetrahedral. For instance, a molecule may loosely be called octahedral even if it has six different groups attached to the central atom. However, it only belongs to the octahedral point group O h, if all six groups are identical (3) 4

Order: The member of elements in a group (G) is called the ORDER of the group. e.g., if S is a Group, its order h = 6 Subgroup: If a smaller sub set of elements of a Group fulfills all the group requirement (-4 above), then, this subset is called a SUBGROUP of that Group. 5

GROUP MULTIPLICATION TABLE Operation A (ii) 4 classes (iii) 4 irreps (iv) Subgroups {EC 2 } {E, v } &{E, v } 6

Each irreps furnishes vectors 7

ABELIAN GROUP If А, B G : AB = BA then G is Abelian. For an Abelian group EACH element is in a CLASS by ITSELF. Proof: X - AX = X - XA = A A is in a class by itself Illustration : C 2v {E, C 2, v (xz), v (y, z)} E (x, y, z) = (x, y, z) C 2 (x, y, z) = (-x, y, z) v (x, y, z) = (x, -y, z) v (x, y, z) = (-x, y, z) C 2 v (x, y, z) = C 2 { v (x, y, z)} = C 2 (x, -y, z) = (-x, y, z) = v (x, y, z) C 2 v = v 8

Similarly, V C 2 (x, y, z) = v (-x, -y, z) = (-x, y, z) = v (x, y, z) or V C 2 = V = C 2 v V C 2 = C 2 v Similarly, C 2 v = v C 2 = v and v v = v v = C 2 (SHOW) C 2v is ABELIAN. Is every element in a class by itself? C 2 2 (x, y, z) = C 2 C 2 (x, y, z) = C 2 (-x, -y, z) = (x, y, z) = E (x, y, z) C 2 C 2 = E C 2 - = C 2 SIMILARLY v - = v & ( v ) - = v (SHOW) 9

Symmetry Element H 2 O : C 2v { E, C 2, σ v (xz), σ v (yz) } Symmetry Point Group Symmetry Operation 20

NH 3 : C 3v { E, C 3,C 3 2, σ, σ 2, σ 3 } 2

C 3 = 2 C 3 = 3 C 3v is not abelian. 22

Representation: Consider the following set of simultaneous equation. C = a b + a 2 b 2 + a 3 b 3 +. () C 2 = a 2 b + a 22 b 2 + a 23 b 3 +. (2) C 3 = a 3 b + a 32 b 2 + a 33 b 3 +. (3) Using matrix notation C a a 2 a 3 b C 2 = a 2 a 22 a 23 b 2 C 3 a 3 a 32 a 33 b 3 or C = A b or in operative form C = A b 23

Or the matrix representation of any operator A can be constructed by examining its effect on a vector and using equation () (3) e.g., C 2 (x, y, z) = (-x, -y, +z) -x = (-)x + 0 y + 0 z -y = 0 x + (-)y + 0 z +z = 0 x + 0 y + z - 0 0 or C 2 = 0-0 0 0 - -x x -y = C 2 y & Tr (C 2 ) = (-) + (-) + = - +z z 24

Representation of Rotation Operation x = r Cos ( + ) = r Cos Cos - r Sin Sin x or x = Cos x Sin y, Similarly, y y = r Sin ( + ) = r Sin Cos + r Cos Sin y x or, y = Sin x + Cos y 25

or, X = Cos - Sin x y Sin Cos y & Since x = R 2 () x y y R 2 () = Cos -Sin Sin Cos C 2 = R z (80) = Cos 80 -Sin 80 = 0 Sin 80 Cos 80 0 C 3 = R 2 (20) = -Y 3 2 2 3 -Y 2 2 26

GROUP REPRESENTATION C 2v E C 2 v v A z, x 2, y 2, z 2 A 2 - - R 2, xy B - - X, R y, xz B 2 - - y, R x, yz C 3 = Ȓz(20) = Cos 20 -Sin 20 Sin 20 Cos 20 C 3 t = Cos 20 Sin 20 C 3 C 3 t = C 3 t C 3 = 0 -Sin 20 Cos 20 0 27

C 3v E C 3 C 2 3 2 3 A A 2 - - - E 0 0 2 3 2 3 2 2 C 3 t 0 0 - C 3 C 32 (E) 2 - - 0 0 0 REPRESENTATIONS ARE NOT UNIQUE If Γ (R) is a representation R G, so is Γ (R) = - Γ(R) R G since Γ (R) Γ (S) = Γ(R) - Γ (S) = - Γ (R) Γ(S) = - Γ (RS) = Γ (RS) Representations are NOT UNIQUE indeed! VARI! With!! 28

However, Trace or character of a matrix A, is invariant i.e. i.e. CHARACTER of representations is invariant and we should be decline with character tables and not Representation Tables. 29

Let us collect the character for each representation in a six dimensional (component) vector for C 3v i.e. ( ) = (,,,,,, ) ( 2 ) = (,,, -, -, -) & ( ) = (2, -, -, 0, 0, 0) Notice that E (C 3 ) = E (C 32 ) same for & A 2. i.e. i (C 3 ) = i ( - C 3 ) i.e. character is a Class property. Similarly, i ( ) = i ( 2 ) = i ( 3 ) for all imeps, can be easily verified that where i & j are irrep labels and h is the order of the group for C 3v, h = 6. 30

Since reduction of a matrix is through similarity transform which presents character 3

32 2 2 2 2 2 2, ˆ ˆ ˆ ˆ B R Similarly B R R R B orb R R R C A R R R A ora R R R C x y y y v y y y z z z v z z z Symmetry type of R x, R y, R z σ v (yz)

Labels for IRREPS:. An One-dimensional irrep is given the symbol A or B. 2. A if (C X ) = + B if (C X ) = - 3. Subscripts may be applied as follows: (i) There are v or x plane then if () = + 2 if () = - (ii) If there is i then g if (i) = + u if (i) = - 4. Superscripts when h symmetry point group (i) if ( h ) = + (ii) if ( h ) = - 5. Two-dimensional irrep is given the symbol E. 6. Three-dimensional irrep is given the symbol T, etc. 33

f i f j d 0 unless A or A Where & are the irrep labels for f i & f j respectively. Proof: Let & be irreps of dimension i.e. (R) = R G & (R) = R G S ij f i i.e. R S ij = + S ij R G but, f j d a number Totally symmetric R Sij R fi Rf j d ( R) ( R) S S for some R unless ij ij 34

i.e. if R S ij = -S ij = S ij = 0 for S ij 0 = & = A is required. In general, ˆ d i O j O Iff A i o j i.e. if o = A then, A I j 35

Vibrational Modes of H 2 O C 2v E C 2 v (xz) v (yz) A Z A 2 - - R z B - - x, R y B 2 - - y, R x CONSTRUCT A REPRESENTATION IN THE BASIS OF CARTESIAN DISPLACEMENTS ( ) FOR EACH ATOM. 36

(E) = 9 (C 2 ) = - ( xz ) = ( yz ) = 3 a a a A A i h 4 RC ( R) ( R) 2v 9 3 3, a 2, 2 B a B 2 3 Total Modes (T + R + V) = 3A + 3B 2 + 2B + A 2 Translational Modes : xb, yb 2, za Rotational Modes : R x B 2, R y B, R z A 2 Vibrational Modes = Total Rot Trans = (3A + 3B 2 + 2B + A 2 ) (2B + 2B 2 + A + A 2 ) = B 2 + 2A i.e. H 2 O has 2a, type and b 2 type NORMAL MODES OF VIBRATION. 37

EX = X = X + 0Y + 0Z + 0X 2 + 0Y 2 + 0Z 2 + 0X 3 + 0Y 3 + 0Z 3 EY = Y = 0X + Y + 0Z + 0X 2 + 0Y 2 + 0Z 2 + 0X 3 + 0Y 3 + 0Z 3 EZ = Z = 0X + 0Y + Z + 0X 2 + 0Y 2 + 0Z 2 + 0X 3 + 0Y 3 + 0Z 3 OR E X 0 0 0 0 0 0 0 0 X Y 0 0 0 0 0 0 0 0 Y Z 0 0 0 0 0 0 0 0 Z X 2 0 0 0 0 0 0 0 0 X 2 Y 2 = 0 0 0 0 0 0 0 0 Y 2 Z 2 0 0 0 0 0 0 0 0 Z 2 X 3 0 0 0 0 0 0 0 0 X 3 Y 3 0 0 0 0 0 0 0 0 Y 3 Z 3 0 0 0 0 0 0 0 0 Z 3 E i ( E) 9 4 I 99 38

C 2 X = -X 2 = 0 X + 0 Y + 0 Z - X 2 + 0 Y 2 + 0 Z 2 + 0 X 3 + 0 Y 3 + 0 Z 3 C 2 Y = -Y 2 = 0 X + 0 Y + 0 Z + 0 X 2 - Y 2 + 0 Z 2 + 0 X 3 + 0 Y 3 + 0 Z 3 C 2 Z = Z 2 = 0 X + 0 Y + 0 Z + 0 X 2 + 0 Y 2 + Z 2 + 0 X 3 + 0 Y 3 + 0 Z 3 C 2 X 2 = -X = - X + 0 Y + 0 Z + 0 X 2 + 0 Y 2 + 0 Z 2 + 0 X 3 + 0 Y 3 + 0 Z 3 C 2 Y 2 = -Y = 0 X - Y + 0 Z + 0 X 2 + 0 Y 2 + 0 Z 2 + 0 X 3 + 0 Y 3 + 0 Z 3 C 2 Z 2 = Z = 0 X + 0 Y + Z + 0 X 2 + 0 Y 2 + 0 Z 2 + 0 X 3 + 0 Y 3 + 0 Z 3 C 2 X 3 = -X 3 = 0 X + 0 Y + 0 Z + 0 X 2 + 0 Y 2 + 0 Z 2 - X 3 + 0 Y 3 + 0 Z 3 C 2 Y 3 = -Y 3 = 0 X + 0 Y + 0 Z + 0 X 2 + 0 Y 2 + 0 Z 2 + 0 X 3 - Y 3 + 0 Z 3 C 2 Z 3 = +Z 3 = 0 X + 0 Y + 0 Z + 0 X 2 + 0 Y 2 + 0 Z 2 + 0 X 3 + 0 Y 3 + Z 3 Or, 0 0 0-0 0 0 0 0 0 0 0 0-0 0 0 0 0 0 0 0 0 + 0 0 0-0 0 0 0 0 0 0 0 C 2 = 0-0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0-0 0 0 0 0 0 0 0 0-0 0 0 0 0 0 0 0 0 + (C 2 ) = - + (-) + = - 39

WHAT KIND OF DISPLACEMENT PATTERNS DO THESE VIBRATIONAL MODES REPRESENT? Θ Θ ϵ A representation using r and r 2, (E) = 2 a A = (r + r 2 ) (C 2 ) = 0 a A2 = 0 ( v ) = 0 a B = 0 ( v ) = 2 a B2 = (r - r 2 ) Symmetric (r + r 2 ) STRETCH Antisymmetric (r - r 2 ) STRETCH and, Symmetric () BEND 40

ν s B > ν s 2 A > ν A b 4

WHICH OF THESE MODES ARE IR/RAMAN ACTIVE? Symmetry type of q j mode A IR : ( q j ) x ( or y or z) ( q j symmetry type of ) d 0 q j mod e Raman : A 2 2 2 ( q ) xy ( or other tensor products xz, yz, x, y, z etc.) ( q ) d j j 0 ALL THREE MODES ARE IR ACTIVE and, also ALL THREE MODES ARE RAMAN ACTIVE. 42

43 H 2 O: Is from a b 2 allowed? WHAT is the of the light? polarization emitted transition 0 0 0 : : : 2 2 2 2 y B y B x A z d y d x d z x z b a y b a x b a z TRANSITION a b 2 is and will be in Y

SYMMETRY IN QUANTUM CHEMISTRY H E ; N i Leads to C = $ C Є e.g. HF or Huckel etc. C i LCAO i MO N X N gets block diagonalised e.g., for C 0 H 8. Naphthalene: D 2h {E, C 2 (x), C 2 (y), C 2 (z), i, (xy), (xz), (yz) } 0 P & $ are 0 x 0 UNDOABLE by HAND. 44

, 4, 5, 8 2, 3, 6, 7 9, 0 A u, B u, B 2g, B 3g A u, B u, B 2g, B 3g B u, B 3g SALC: A u 2 x 2 B u 3 x 3 B 2g 2 x 2 B 3g 3 x 3 EACH BLOCK CAN BE SOLVED SEPARATELY BY HAND. 45

QUALITATIVE MO THEORY BENZENE D 6h Pˆ l h RD on 6 p 6h ( R) Rˆ AOs as basis A 2 B 2 E E 2 Set up Huckel Hamiltonian Matrix x + 2 x - 2 x + () () x + = 0, x - x x E 46

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SYMMETRY OF ENERGY LEVELS IN HMO THEORY C 6 H 6, C 6 H 6-, C 6 H 6 + C 5 H 5, C 5 H 5-, C 5 H 5 + C 4 H 4, C 4 H 4-, C 4 H 4 + 48

CONSERVATION OF ORBITAL SYMMETRY WOODWARD-HOFFMANN RULES APPLICATION TO CYCLIZATION OF -3 CIS BUTADIENE TO CYCLOBUTENE C 2, V containing C 2 v (plane of molecule) Butadiene Same symmetry elements as for Butadiene Cyclobutene

H H 2 R H con-rotatory H 3 4 R H H C 2 survives σ v & σ v don t H H R R

H H 2 R H dis-rotatory H 3 4 R H H σ v survives C 2 & σ v don t H R R H

Cyclobutene dis Butadiene con Cyclobutene σ v C 2 σ* A A π 4 S A σ* π* A S π 3 A S π* π S A π 2 S A π σ S Photo chemically Allowed S π A S Thermally Allowed σ Thermally disallowed Why not (S) with (S) cyclo Because of Non-crossing Rule

MATRICES = M M 2 ----- M n M 2 M 22 ------ M 2n... M n M n2 ------ M nn Square M contains n 2 elements M ij i th row j th column + = P ij = M ij + N ij = P ij = k M ik N kj 54

Let M = 2 and, N = 5 6 3 4 7 8 M + N = 6 8 0 2 MN = P = 9 22 43 50 NM MN NM = 23 34 3 46 Matrices are the operators. 55

SPECIAL NAME MATRICES (i) DIAGONAL MATRIX: - M ij = 0 if i j i.e. M = 0 is diagonal but 0 is not. 0 2 2 0 (ii) THE UNIT MATRIX IS A DIAGONAL MATRIX = 0 0 0 0 0 0 3 dimensional unit matrix 56

(iii) THE TRANSPOSE OF A MATRIX Thus if = 2 ; t = 3 3 4 2 4 If is symmetric. (iv) THE INVERSE OF A MATRIX IS WRITTEN AS 57

THANK YOU FOR YOUR PATIENCE 58