MATRICES-(E)-PART-II-05-THEORY

Properties of adjoint matrix : If A, B are square matrices of order n and is corresponding unit matrix, then (i) (Thus A (adj A) is always a scalar matrix) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) A is symmetric  adj A is also symmetric. (xii) A is diagonal  adj A is also diagonal. (xiii) A is triangular  adj A is also triangular. (xiv) A is singular  |adj A|= 0 8.2.14 Inverse of a Matrix. A non-singular square matrix of order n is invertible if there exists a square matrix B of the same order such that . In such a case, we say that the inverse of A is B and we write The inverse of A is given by The necessary and sufficient condition for the existence of the inverse of a square matrix A is that Properties of inverse matrix: If A and B are invertible matrices of the same order, then (i) (ii) (iii) (iv) [In particular (v) (vi) (vii) A = diag (viii) A is symmetric  is also symmetric. (ix) A is diagonal, is also diagonal. (x) A is scalar matrix  is also scalar matrix. (xi) A is triangular,  is also triangular. (xii) Every invertible matrix possesses a unique inverse. Note :  (Cancellation law with respect to multiplication) If A is a non singular matrix i.e., if ,then exists and    Example: 24 If , then is equal to (a) 16 (b) 10 (c) 6 (d) None of these Solution: (b) Example: 25 If 3, – 2 are the Eigen values of non-singular matrix A and |A| = 4 . Then Eigen values of adj (A) are (a) 3/4, –1/2 (b) 4/3, –2 (c) 12, –8 (d) –12, 8 Solution: (b) Since and if is Eigen value of A then is Eigen value of , thus for corresponding to Eigen value is = 4/3 and for is = = –2 Example: 26 If matrix and , then K is (a) 7 (b) –7 (c) 1/7 (d) 11 Solution: (d) We know that . We have i.e. and Example: 27 The inverse of matrix is (a) (b) (c) (d) Solution: (b) Here , . Hence Example: 28 Let and . If B is the inverse of matrix A, then is (a) 5 (b) –1 (c) 2 (d) –2 Solution: (a) We have, ,  and Then According to question, B is the inverse of matrix A. Hence Example: 29 Matrix is invertible for (a) (b) (c) (d) All real K Solution: (d) For invertible, i.e.,   , which is true for all real K . Example: 30 Let be a non-singular matrix, (0 denotes the null matrix), then = [] (a) (b) (c) (d) None of these Solution: (a) We have, Multiplying both sides by ,  . Example: 31 Let where ,then is equal to (a) (b) (c) (d) None Solution: (a) , adj of ......(i) and ......(ii) From (i) and (ii), Example: 32 If I is a unit matrix of order 10, then the determinant of I is equal to (a) 10 (b) 1 (c) 1/10 (d) 9 Solution: (b) Determinant of unit matrix of any order =1. Example: 33 If and then = (a) (b) (c) (d) 0 Solution: (a) and   Example: 34 If |A| denotes the value of the determinant of the square matrix A of order 3, then (a) (b) (c) (d) None of these Solution: (a) We know that, det. , where n is order of square matrix If A is square matrix of order 3, Then . Hence . Example: 35 For how many value (s) of x in the closed interval [–4, –1] is the matrix singular (a) 2 (b) 0 (c) 3 (d) 1 Solution: (d) ,   Hence only one value of x in closed interval [–4,–1] i.e. Example: 36 Inverse of diagonal matrix (if it exists) is a (a) Skew-symmetric matrix (b) Diagonal matrix (c) Non invertible matrix (d) None of these Solution: (b) Let As A is invertible, therefore   for i = 1, 2, 3…..n Here, cofactor of each non diagonal entry is 0 and cofactor of which is a diagonal matrix 8.2.15 Elementary Transformations or Elementary Operations of a Matrix. The following three operations applied on the rows (columns) of a matrix are called elementary row (column) transformations (1) Interchange of any two rows (columns) If ith row (column) of a matrix is interchanged with the ith row (column), it will be denoted by Example : , then by applying , we get (2) Multiplying all elements of a row (column) of a matrix by a non-zero scalar If the elements of ith row (column) are multiplied by a non-zero scalar k, it will be denoted by , or If , then by applying we obtain (3) Adding to the elements of a row (column), the corresponding elements of any other row (column) multiplied by any scalar k. If k times the elements of jth row (column) are added to the corresponding elements of the ith row (column), it will be denoted by If then the application of elementary operation gives the matrix If a matrix B is obtained from a matrix A by one or more elementary transformations, then A and B are equivalent matrices and we write Let then applying , applying An elementary transformation is called a row transformation or a column transformation according as it is applied to rows or columns. 8.2.16 Elementary Martix. A matrix obtained from an identity matrix by a single elementary operation (transformation) is called an elementary matrix. Example : are elementary matrices obtained from by subjecting it to the elementary transformations and respectively. Theorem 1 : Every elementary row (column) transformation of an m×n matrix (not identity matrix) can be obtained by pre-multiplication (post- multiplication) with the corresponding elementary matrix obtained from the identity matrix by subjecting it to the same elementary row (column) transformation. Theorem 2 : Let be a product of two matrices. Any elementary row (column) transformation of AB can be obtained by subjecting the pre-factor A (post factor B) to the same elementary row (column) transformation. Method of finding the inverse of a matrix by elementary transformations : Let A be a non singular matrix of order n. Then A can be reduced to the identity matrix by a finite sequence of elementary transformation only. As we have discussed every elementary row transformation of a matrix is equivalent to pre-multiplication by the corresponding elementary matrix. Therefore there exist elementary matrices such that (post multiplying by ) Algorithm for finding the inverse of a non singular matrix by elementary row transformations Let A be non- singular matrix of order n Step I : Write Step II : Perform a sequence of elementary row operations successively on A on the LHS and the pre factor on the RHS till we obtain the result Step III : Write Note :  The following steps will be helpful to find the inverse of a square matrix of order 3 by using elementary row transformations. Step I : Introduce unity at the intersection of first row and first column either by interchanging two rows or by adding a constant multiple of elements of some other row to first row. Step II : After introducing unity at (1,1) place introduce zeros at all other places in first column. Step III Introduce unity at the intersection of 2nd row and 2nd column with the help of 2nd and 3rd row. Step IV : Introduce zeros at all other places in the second column except at the intersection of 2nd row and 2nd column. Step V : Introduce unity at the intersection of 3rd row and third column. Step VI : Finally introduce zeros at all other places in the third column except at the intersection of third row and third column. Example: 37 Using elementary row transformation find the inverse of the matrix [] (a) (b) (c) (d) None of these Solution: (a) We have A=IA  Applying Applying and , Applying Applying and , Applying , Applying , 8.2.17 Rank of Matrix . Definition : Let A be a m×n matrix. If we retain any r rows and r columns of A we shall have a square sub-matrix of order r. The determinant of the square sub-matrix of order r is called a minor of A order r. Consider any matrix A which is of the order of 3×4 say, . It is 3×4 matrix so we can have minors of order 3, 2 or 1. Taking any three rows and three columns minor of order three. Hence minor of order Making two zeros and expanding above minor is zero. Similarly we can consider any other minor of order 3 and it can be shown to be zero. Minor of order 2 is obtained by taking any two rows and any two columns. Minor of order . Minor of order 1 is every element of the matrix. Rank of a matrix: The rank of a given matrix A is said to be r if (1) Every minor of A of order r+1 is zero (2) There is at least one minor of A of order r which does not vanish Note :  If a minor of A is zero the corresponding submatrix is singular and if a minor of A is not zero then corresponding submatrix is non-singular. Here we can also say that the rank of a matrix A is said to be r if (i) Every square submatrix of order r+1 is singular. (ii) There is at least one square submatrix of order r which is non-singular. The rank r of matrix A is written as Working rule : Calculate the minors of highest possible order of a given matrix A. If it is not zero, then the order of the minor is the rank. If it is zero and all other minors of the same order be also zero, then calculate minor of next lower order and if at least one of them is not zero then this next lower order will be the rank. If, however, all the minors of next lower orders are zero, then calculate minors of still next lower order and so on. Note :  The rank of the null matrix is not defined and the rank of every non-null matrix is greater than or equal to 1.  The rank of a singular square matrix of order n cannot be n. 8.2.18 Echelon form of a Matrix. A matrix A is said to be in Echelon form if either A is the null matrix or A satisfies the following conditions: (1) Every non- zero row in A precedes every zero row. (2) The number of zeros before the first non-zero element in a row is less than the number of such zeros in the next row. It can be easily proved that the rank of a matrix in Echelon form is equal to the number of non-zero row of the matrix. Example : The rank of the matrix is 2 because it is in Echelon form and it has two non-zero rows. The matrix is not in Echelon form, because the number of zeros in second row is not less than the number of zeros in the third row. To reduce A in the echelon form, we apply some elementary row transformations on it. Applying we obtain which is in Echelon form and contains 2 non zero rows. Hence, Rank of a matrix in Echelon form : The rank of a matrix in Echelon form is equal to. the number of non-zero rows in that matrix. Algorithm for finding the rank of a matrix : Let be an m×n matrix. Step I : Using elementary row transformations make Step II : Make all zeros by using elementary transformations, Step III : Make by using elementary row transformations. Step IV : Make all zeros by using The process used in steps III and IV is repeated upto row. Finally we obtain a matrix in Echelon form, which is equivalent to the matrix A. The rank of A will be equal to the number of non-zeros rows in it. Example: 38 The rank of the matrix is [ (a) 2 (b) 3 (c) 1 (d) Indeterminate Solution: (a) We have , Considering 3×3 minor its determinant is 0. Similarly considering , , and , their determinant is 0 each rank can not be 3 Then again considering a 2×2 minor, , which is non zero. Thus, rank =2 Example: 39 The rank of the matrix is (a) 1 if a = 6 (b) 2 if a =1 (c) 3 if a = 2 (d) 1 if a = – 6 Solution: (b,d) Let When , , When , , When , , When , , Example: 40 The value of x so that the matrix has rank 3 is (a) (b) (c) and (d) Solution: (c) Since rank is 3, , , Applying  ,  8.2.19 System of Simultaneous Linear Equations . Consider the following system of m linear equations in n unknowns ... ... ... ... ... ... ... ... ... ... The system of equations can be written in matrix form as or Where , The m×n matrix A is called the coefficient matrix of the system of linear equations. (1) Solution : A set of values of the variables which simultaneously satisfy all the equations is called a solution of the system of equations. Example : is a solution of the system of linear equations , because 3(2) +(–3)=3 and 2(2)+(–3)=1 (2) Consistent system : If the system of equations has one or more solutions, then it is said to be a consistent system of equations, otherwise it is an inconsistent system of equations. Example : the system of linear equation is consistent, because x=1, y =1 and are solutions of it. However, the system of linear equations is inconsistent, because there is no set of values of x, y which satisfy the two equations simultaneously. (3) Homogeneous and non-homogeneous system of linear equations: A system of equations AX=B is called a homogeneous system if . Otherwise, it is called a non-homogeneous system of equations. Example : The system of equations, is a homogeneous system of linear equations whereas the system of equations given by is a non homogeneous system of linear equations. 8.2.20 Solution of a Non Homogeneous System of Linear Equations. There are three methods of solving a non homogeneous system of simultaneous linear equations. (1) Determinant Method (Cramer's Rule) (2) Matrix method (3) Rank method We have already discussed the determinant method (Cramer's rule) in chapter determinants. (1) Matrix method : Let be a system of n linear equations with n unknowns. If A is non-singular, then exists. , [pre-multiplying by ] , [by associativity] . Thus, the system of equations has a solution given by . Now, let and be two solutions of . then, and    . Hence, the given system has a unique solution. Thus, if A is a non-singular matrix, then the system of equations given by has a unique solution given by . If A is a singular matrix, then the system of equations given by AX=B may be consistent with infinitely many solutions or it may be inconsistent also. Criterion of consistency : Let be a system of n-linear equations in n unknowns. (i) If then the system is consistent and has a unique solution given by (ii) If and (adj A) , then the system is consistent and has infinitely many solutions. (iii) If and , then the system is inconsistent Algorithm for solving a non-homogeneous system of linear equations : We shall give the algorithm for three equations in three unknowns. But it can be generalized to any number of equations. Let be a non-homogenous system of 3 linear equations in 3 unknowns. To solve this system of equations we proceed as follows Step I : Write the given system of equations in matrix form, and obtain A, B. Step II : Find Step III : If , then write "the system is consistent with unique solution". obtain the unique solution by the following procedure. Find by using obtain the unique solution given by Step IV : If then write "the system is either consistent with infinitely many solutions or it is inconsistent. To distinguish these two, proceed as follows: Find (adj A) B. If , then write "the system is inconsistent”. If , then the system is consistent with infinitely many solution. To find these solutions proceed as follows. Put (any real number) and take any two equations out of three equations. Solve these equations for x and y. Let the values of x and y be and respectively. Then , is the required solution, where any two of are functions of the third. (2) Rank method : Consider a system of m simultaneous linear equations in n unknowns , given by This system of equations can be written in matrix form as or ,where and The matrix A is called the coefficient matrix and the matrix is called the augmented matrix of the given system of equations. This matrix is obtained by adding column to A. The elements of this column are For example, the augmented matrix of the system of equation is A non-homogeneous system of linear equations may have a unique solution, or many solutions or no solution at all. If it has a solution (whether unique or not) the system is said to be consistent. Otherwise it is called an inconsistent system. The following theorems tell us about the condition for consistency of a system of linear equations Theorem 1 : The system of linear equations is consistent iff the rank of the augmented matrix is equal to the rank of the coefficient matrix A. Theorem 2 : Let be a system of m simultaneous linear equations in n unknowns. Case I : If , then (i) if then system of linear equations has a unique solution. (ii) if then system of linear equations is consistent and has infinite number of solutions. In fact, in this case variables can be assigned arbitrary values. (iii) if then the system of linear equations is inconsistent i.e. it has no solution. Case II : If and then and so from (ii) in case I, there are infinite number of solutions. Thus, when the number of equations is less than the number of unknowns and the system is consistent, then the system of equations will always have an infinite number of solutions. Algorithm for solving a non-homogeneous system AX=B of linear equations by rank method Step I: Obtain A, B. Step II : Write the Augmented matrix [A : B]. Step III : Reduce the augmented matrix to Echelon form by applying a sequence of elementary row-operations. Step IV : Determine the number of non-zero rows in A and [A : B] to determine the ranks of A and [A :B] respectively. Step V: If then write "the system is inconsistent" STOP else write "the system is consistent", go to Step VI Step VI : If = number of unknowns, then the system has a unique solution which can be obtained by back substitution. If < number of unknowns, then the system has an infinite number of solutions which can also be obtained by back substitution. Example: 41 If then values of are (a) 2, 2, 3, 4 (b) 2, 3, 1,2 (c) 3, 3, 0, 1 (d) None of these Solution: (a) We have , and  and Example: 42 The system of linear equation , has unique solution if (a) (b) (c) (d) Solution: (a) The given system of equation has a unique solution if 8.2.21 Cayley-Hamilton Theorem . Every matrix satisfies its characteristic equation e.g. let A be a square matrix then is the characteristics equation of A. If is the characteristic equation for A, then Roots of characteristic equation for A are called Eigen values of A or characteristic roots of A or latent roots of A. If is characteristic root of A, then is characteristic root of . 8.2.22 Geometrical Transformations . (1) Reflexion in the x-axis: If is the reflexion of the point on the x-axis, then the matrix describes the reflexion of a point in the x-axis. (2) Reflexion in the y-axis : Here the matrix is (3) Reflexion through the origin : Here the matrix is (4) Reflexion in the line y = x : Here the matrix is (5) Reflexion in the line y = – x : Here the matrix is (6) Reflexion in y = x tan  : Here matrix is (7) Rotation through an angle : Here matrix is 8.2.23 Matrices of Rotation of Axes . We know that if x and y axis are rotated through an angle about the origin the new coordinates are given by and is the matrix of rotation through an angle . Example: 43 Characteristic equation of the matrix (a) (b) (c) (d) None of these Solution: (a) The characteristic equation is . So, i.e. By cayley-Hamilton theorem , Example: 44 The transformation due to the reflection of through the origin is described by the matrix (a) (b) (c) (d) Solution: (b) If is the new position   Transformation matrix is Example: 45 The rotation through is identical to (a) The reflection in x -axis (b) The reflection in y-axis (c) A point reflection (d) Identity transformation Solution: (c) Rotation through gives . Hence this a point reflection.