#CSIRNET CSIR NET MATHEMATICS | 30th Nov 2020 | Complete Solution of Linear Algebra | Section B

Hello friends, this is Mahesh Yolekar and you are watching NextGenMaths. If you have not subscribed my channel, then please subscribe the channel and press the bell icon of NextGenMaths. In this video, we are going to discuss the problem of CSI NET November 2020, Problem of Linear Algebra. Example, let S equal to U1, U2, UK be a subset of non-zero vectors from Rn. Now consider two statements given below. First statement, if S is linearly dependent set in Rn, then UK is a linear combination of U1, U2, UK-1. And the statement number two, if S is linearly independent set in Rn, then K less than N, which of the following statements is true? So we have four options. Statement one is false and statement two is true. Second option, statement one is true and statement two is false. Third option, both statement one and two are false. And the fourth option, both statement one and two are true. If you want notes of this solution, then please visit our website www.nextgenmaths.com. In the description, link is available. It is a problem of linearly independent and dependent set. So first, we recall the definition, a set S equal to U1, U2, UK is a linearly independent set. If vector equation alpha on V1 plus alpha to V2 plus alpha K VK equal to zero has only trivial solution. That means all alpha is a zero, otherwise S is linearly dependent set. That means this vector equation has trivial solution. That means all alpha is zero, then it is a linearly independent set, otherwise it is a linearly dependent set. Now here we will use the counter example for this statement one and statement two. So first counter example, if we take S equal to 1 0 0, 2 0 0, 0 1 0. It is a subset of R3. Now consider the vector equation alpha on U1 plus alpha to U2 plus alpha 3 U3 equal to zero. Here U1, U2, U3 are the 1 0 0, 2 0 0 and 0 1 0. Now substitute that value. Then we have alpha 1 plus 2 alpha 2 equal to zero and alpha three equal to zero. That means alpha 1 equal to minus 2 alpha 2 and alpha 3 equal to zero. Here alpha 3 equal to zero, but we have infinitely many values of alpha 1 and alpha 2. Like if we take alpha to equal to 1, then alpha 1 equal to minus two. If we take alpha 2 equal to two, then alpha 1 equal to minus four. So we are infinitely non-zero values. So, this vector equation has non-trivial solution, hence S is linearly dependent set. But you can see U3 equal to 0, 1, 0 cannot be expressed as a linear combination of 1, 0, 0, 2, 0, 0. Therefore, the U3 cannot express a linear combination of U1 and U2. So, first statement says that if S is linearly dependent set, then UK is a linear combination of U1, U2, UK minus 1, that is not the correct statement. So, statement 1 is a false statement. Now counter example 2, if we take X equal to 1, 0, 0, 0, 1, 0, 0, 0, 1. These are the standard basis for R3 and they are the linearly independent vector. So, S is a linearly independent set in R3. S contain three vectors, so K equal to 3 and they are the elements of R3, so n equal to 3. Here K equal to n, K is not less than n. So, statement 2 says that S is linearly independent, then K must be less than n. Here we have example where S is linearly independent, but K equal to, so statement 2 is also a false statement. Therefore, the option number 3 which says that both the statements are false is the correct option. Example, let M be 5 cross 5 matrix with real entries such that rank of M equal to 3. Consider the linear system Mx equal to b, let the row reduce echelon form of the augmented matrix Mb be R and let Ri colon denote the ith row of R. Suppose that the system admits a solution, then which of the following statement is necessarily true? If you want notes of this solution, please visit our website www.nextgenmath.com, in the description link is available. Here we are given that matrix is of order 5 cross 5, rank of M is 3, therefore row reduce echelon form of M will be 1 0 0 star star, star represent any number, 0 1 0 star star, 0 0 1 star and star and the last 2 row must be 0 because rank of M is 3. Now if the row reduce echelon form of augmented matrix Mb of system Mx equal to b is R and the linear system admits the solution that is given in the data. So R represent the augmented matrix Mb and system admits the solution, therefore rank of M equal to rank of Mb equal to rank of R is 3, therefore row reduce echelon form of Mb that means R is in the form 1 0 0 star star star, 0 1 0 then the remaining values 0 0 1 and any values and last 2 row must be 0 because rank of R is 3. The options is about the rows of matrix R, the first option is about the third row of matrix R which is 0 0 1 and star star star, here third row is in the option is 0 1 0 which is not equal to the third row of R, so first option is wrong. The second option is about the fifth row, so in the second option fifth row is non-zero but we have fifth row is 0, so second option is also wrong. Now we have third option which is about the fourth row, so fourth row is a 0 row but the third option is non-zero row, therefore third option is wrong, here fourth option shows that fourth row is 0, therefore the fourth option is the correct answer. Example, let A be n cross n matrix of rank 1, let alpha equal to determine I plus A where I is the identity matrix and let B equal to trace of A, which of the following is true? We have again four option, first option beta minus alpha equal to 1, second option alpha minus beta equal to 1, third option alpha less than beta plus 1 and fourth option alpha greater than beta plus 1. If you want notes of this solution, please visit our website www.nextgenmaths.com, in the description link is available. We have given A is a n cross n matrix and rank of A equal to 1, if we take n equal to 1, then A equal to k and k not equal to 0, because rank of A equal to 1, so k should not be 0, if k equal to 0, then rank of A equal to 0, so that's why k not equal to 0 and I is identity matrix of order 1 cross 1, so it is 1. Now beta is a trace of A, so trace of A is a sum of diagonal, so a diagonal only one entry a11, so its value is k, alpha equal to determinant i plus a and which has value k plus 1, but k is a trace of A, so we can write down k equal to beta, so alpha equal to beta plus 1 and in other words alpha minus beta equal to 1, so which is a option number 2. So, result of n equal to 1 is enough to discard other three option 1, 3 and 4, but we will check for n greater than 1, for n greater than 1, rank of A equal to 1 that is given, by the definition all the minor of order greater than 1 are 0, because rank of A is 1, therefore determinant A equal to 0. Now if we take n equal to 2, A equal to ABCD and its generalized matrix, I is identity matrix of order 2, 1 0 0 1, alpha equal to determinant i plus a, so we have determinant A plus 1 BCD plus 1 and if we evaluate then we have A plus 1 into D plus 1 minus BC, if we open the brackets then AD plus A plus D plus 1 minus BC, if we rewrite then A plus D plus 1 AD minus BC, so A plus D is nothing but the trace of A which is sum of its diagonal plus 1 and AD minus BC is nothing but the determinant of A, so alpha equal to trace of A plus 1 plus determinant A, but as we know rank of A equal to 1 and so the determinant A equal to 0, so we have alpha equal to trace of A plus 1, but trace of A is nothing but the beta, so alpha equal to beta plus 1 that means alpha minus beta equal to 1, so we will get this similar relation for n equal to 2. Now for n equal to 3, alpha equal to determinant i plus A equal to trace of A plus 1 plus determinant A plus m11 plus m22 plus m33 where m11, m22, m33 are the minors of a11, a22, a33 respectively, as we know A is 0 and all the minors of order greater than 1 are 0, so m11 0, m22 0, m33 0 and we have trace of A plus 1 and the other entry will be 0, so trace of A equal to beta, so beta plus 1, that's why alpha equal to beta plus 1, in other words alpha minus beta equal to 1 which is a option number 2 and option number 2 is the correct answer. Example let A, B and C be distinct integers, let A be the matrix, A equal to A square B square C square, A power 5, B power 5, C power 5, A power 11, B power 11, C power 11 which among the following is the set of all possible ranks of A, we have again 4 option, first option set contain only one element 3, second option 2, 3, third option 1, 2, 3 and the fourth option 0, 1, 2, 3, if you want notes of this solution then please visit our website www.nextsenmas.com, in the description link is available, so it is a problem of rank of A, so first we recall the definition of rank of A, the order of non-zero minor of highest order obtained from the matrix is called the rank of A matrix, that means order of highest non-zero determinant obtained from the matrix will be the rank of A matrix, now case 1, we will check all the ranks 0, 1, 2, 3, 1 by 1, so first we assume rank of A equal to 0, rank of A will be 0 only when all the minors obtained from the matrix are 0, that means all the order minors are 0 only when matrix A is a 0 matrix, it is only possible when A equal to 0, B equal to 0 and C equal to 0, but A, B, C are the distinct integers that is given in the data, so our assumption at rank of A equal to 0 which is not possible and so the option number 4 which contain 0 as a rank is not possible, so option number 4 is a wrong option, case 2, suppose rank of A equal to 1 and we also assume A and B are the non-zero distinct integer, as we have given that A, B, C are the distinct integer, so we can assume that A and B are the non-zero integers, C may be 0, assume that rank of A equal to 1, therefore all the minors of order 2 and order 3 are 0, so we consider minor of M33 will be 0, but minor of M33 is the determinant A square B square A power 5 B power 5 equal to 0, if we evaluate the determinant then we have A square B power 5 minus A power 5 B square equal to 0, you can write down A square into B square B cube minus A cube equal to 0, but here A and B are the non-zero integers, therefore A square into B square cannot be 0 and we have B cube minus A cube equal to 0, so B cube equal to A cube and B equal to 0, but A and B are the distinct integers, so B cannot equal to A, so our assumption that rank of A equal to 1 which is not possible, so option number 3 which contain 1 as a rank of matrix A which is not possible, now case 3, if we take A equal to minus 1, B equal to 1 and C equal to 0, therefore A equal to 1 1 0, minus 1 1 0, minus 1 1 0, determinant A equal to 0, we can easily say that because last column is 0, therefore rank of A is less than the order of A which is 3, now if we are able to find at least one minor of order 2 which is non-zero, then rank of A is 2, so we consider the minor of m33 which is 1 1 minus 1 1 and its value is 2 which is non-zero, therefore rank of A equal to order of m33 which is 2, so rank of A equal to 2, that means we are able to find one example which gives the rank of A equal to 2, now case 4, if we take A equal to 1, B equal to 1 and C equal to 2, therefore A is equal to 1 1 2, minus 1 1 32 and minus 1 1 2048 and its determinant value is 4032 which is non-zero, therefore rank of A equal to order of matrix A which is 3, so rank of A equal to 3, so we are able to find the examples where matrix has rank 2 and 3, so these are the possible ranks of matrix A and which is option number 2, so option number 2 is the correct, example which of the following is an inner product on vector space of all real-valued continuous function on closed interval 0 1, we are given four option, first option inner product f with g equal to modulus integration 0 to 1 f t g t dt, second option integration 0 to 1 modulus f t g t dt, third option f 0 g 0 plus f 1 g 1 and the fourth option integration 0 to 1 f t g t dt, if you want notes of this solution then please visit our website www.nextgenmatch.com, in the description link is available, so it is a problem of inner product, so first we recall the definition of inner product, a function or a operation will be the inner product if it satisfy the following condition, so first condition inner product u with v equal to inner product v with u, second condition inner product u plus w with v equal to inner product u with v plus w with v, third condition inner product alpha u with v equal to alpha times inner product u with v for any scalar alpha, fourth condition inner product u with u is always greater than equal to 0 and inner product u with u equal to 0 if and only if v equal to 0, so any operation which satisfy this four condition will be the inner product, now we are given the four different inner products, so we check all the four inner product one by one, so for the option one inner product f with g equal to modulus integration 0 to 1 f t g t dt, taking alpha equal to minus 1, consider the condition number 3, so inner product alpha f with g equal to inner product minus f with g because alpha equal to minus 1, so we have modulus integration 0 to 1 f replaced by minus f into g t dt, but this modulus we have modulus integration 0 to 1 f t g t, now if we consider right hand side of this condition alpha inner product f with g equal to minus times inner product f with g will give the minus modulus integration 0 to 1 f t g t dt, therefore these two are not the same inner product alpha f with g not equal to alpha times inner product f with g, so it's not satisfy the third condition and hence the given operation is not an inner product, so option number one is the wrong answer, now option number two inner product f with g equal to integration 0 to 1 modulus f t g t dt, it is similar to the option number one, so we consider the same way alpha equal to minus 1 and repeat the option number one procedure, consider the condition number three, then we have inner product alpha f with g equal to inner product minus f with g equal to integration 0 to 1 modulus minus f t g t dt, so this modulus will give the positive values integration 0 to 1 modulus f t g t dt, and the right hand side of that condition alpha times inner product f with g equal to minus times inner product f with g will give the minus times integration 0 to 1 modulus f t g t dt, so these two are not the same and it does not satisfy the condition number three, therefore the given operation is not an inner product, so option number two is also the wrong option, option number three where inner product is in the form inner product f with g equal to f of 0 g of 0 plus f 1 into g 1, here f and g are the real value continuous function, so we consider the function f x equal to x into 1 minus x in the closed interval 0 1, it is a real value continuous function, now we consider the fourth condition of the inner product, inner product f with f equal to f of 0 f of 0 plus f of 1 into g of 1, here the function f x equal to x into 1 minus x, so if you find the f of 0 will give the 0 value and f of 1 will also be 0, therefore inner product f with f is 0, if we evaluate f at 1 by 2, then we have f of 1 by 2 equal to 1 by 2 into 1 minus 1 by 2, so 1 by 2 into 1 by 2 will be the 1 by 4 which is non-zero, therefore inner product f with f is 0, but f is not 0, so it does not satisfy the second part of the fourth condition, therefore it is not an inner product, so option number three is also the wrong option, now option number four, it is an integration 0 to 1 f t g t dt, so it is a simple integration, so it satisfy first three condition, because first three condition are also the property of integration, now we consider the fourth condition, so inner product f with f equal to integration 0 to 1 f t into f t dt, so integration 0 to 1 f t square dt is always greater than equal to 0, because f t square is a non-negative continuous real valued function for all x belongs to closed interval 0 1, and with the similar argument inner product f with f integration 0 to 1 f t square dt equal to 0, if and only if f equal to 0, so the operation in the option number four will satisfy all the four condition of inner product, so it will be the inner product, so option number four is the correct answer. Example, let's define a belongs to M and R to be positive, if every column vectors v belongs to R n, we have inner product A v with v greater than equal to 0, where this operation is the inner product on R n, let A alpha beta equal to matrix alpha 1 1 1 0 0 1 0 beta, let S is equal to alpha beta belongs to R 2, A alpha beta is positive, which among the following statement is true, so we have four option, first option S equal to empty, second option alpha beta belongs to S, if and only if alpha beta greater than 0, alpha beta belongs to S, if and only if alpha plus beta plus 4 greater than 0, and fourth option S is equal to R 2, if you want notes of this solution, then please visit our website www.nextgenmath.com, in the description link is available. Here, we are given A alpha beta is 3 cross 3 matrix, therefore A v is possible, if v belongs to, now if we consider v equal to v 1, v 2, v 3 belongs to R 3, here we have to find the value of alpha beta, such that A alpha beta is a positive, A alpha beta is positive, if inner product A v is always greater than equal to 0, so for that we need A into v, so we are going to find A alpha beta with v, if we take this matrix multiplication, then we have alpha v 1 plus v 2 plus v 3, v 1 and v 1 plus beta v 3. Now, we are going to find the inner product A alpha beta v with v, it is a standard inner product, so it is a dot product, so simple dot product will give the alpha v 1 square plus v 1 v 2 plus v 1 v 3 plus v 1 v 2 plus v 1 v 3 plus beta v 3 square. And if we simplify, then we will have alpha v 1 square plus 2 times v 1 v 2 plus 2 times v 1 v 3 plus beta times v 3 square, we have given that A alpha beta is positive, if inner product alpha beta v with v is always greater than equal to 0, for all v belongs to R3. And to find the elements of S, we need to find the value of alpha and beta, if we take example v equal to 1 0 0 and alpha equal to minus 1 beta equal to minus 1. Now, note that alpha beta belongs to R2, because alpha equal to minus 1 beta equal to minus 1, so minus 1 minus 1 belongs to R2, alpha into beta that means minus 1 into minus 1 will be the plus 1 greater than 0, and alpha plus beta plus 4 that means minus 1 minus 1 minus 2 plus 4 equal to 2 which greater than 0. Now, we are going to find the value of inner product alpha beta v with v, for that we substitute the value of v 1 v 2 v 3 and alpha beta in this expression, here v 1 equal to 1, v 2 equal to 0, v 3 equal to 0. Therefore, last three terms are 0, and for the first term v 1 equal to 1 and alpha equal to minus 1, so we will get the minus 1, which is less than 0, that means this particular v and alpha beta we have inner product negative, therefore a alpha beta is not positive, therefore alpha beta greater than 0, but a is not positive, so alpha beta does not belongs to S. So, second option says that alpha beta belongs to S, if and only if alpha beta greater than 0, we have counter example which valid the condition of second option, so option number 2 is the wrong option. Now, alpha plus beta plus 4 greater than 0, but a is not positive, the same example shows that option number 3 is a wrong option, and alpha beta belongs to R2, but a is not positive, because minus 1 minus 1 belongs to R2, but a is not positive, so alpha beta does not belongs to S, so S not equal to R2, so option number 4 is also wrong option. Therefore, the option number 1 is the correct answer. 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