CSIR NET Mathematics Exam DEC 2018 [ Linear Algebra ] Section C Ex 70 (Set C)

Welcome friends. In this video, we are going to discuss example number 17 from that exam December 2018, which is a problem of linear algebra. Example, consider the quadratic forms Q1(x,y) equal to xy, Q2(x,y) is equal to x square plus 2xy plus y square and Q3(x,y) is equal to x square plus 3xy plus 2y square. On R2, choose the correct statement from the following. The first statement Q1 and Q2 are equivalent. Q1 and Q3 are equivalent. Q2 and Q3 are equivalent. All are equivalent. To solve this example, we are required to understand the equivalence of quadratic forms. So, we begin with the definition of equivalent quadratic form. Two quadratic forms in x bar, that is binary quadratic forms are said to be equivalent if there exist integer a, b, c, d such that f of x bar is equal to G of AX plus BY, CX plus DY and the determinant value of ABCD, AD minus BC should be 1, now we are given the quadratic form Q1(x,y) is equal to xy, Q2(x,y) is equal to x square plus 2xy plus y square which can be written in the form of x plus y whole square and Q3(x,y) is equal to x square plus 3xy plus 2y square which is factorized into x plus 2y, x plus y. Now, first we are going to check the equivalence of Q1 and Q2. Q1(x,y) equal to Q2(AX plus BY, CX plus DY). If we are able to find the integer value of ABCD, then we can say that Q1 and Q2 are equivalent. So xy, now the Q2(x,y) is equal to x plus y whole square. So here we replace x by AX plus BY and y replaced by CX plus DY. So XY is equal to AX plus BY plus CX plus DY whole square. Now, gathering the terms of x and y, then we have xy is equal to a plus c x plus b plus d y whole square. Now, opening the brackets, xy is equal to a plus c whole square x square plus a plus c b plus d xy plus b plus d whole square y square. Now, comparing the coefficient of x square, xy and y square, we have a plus c whole square equal to 0, b plus d whole square equal to 0, a plus c into b plus d equal to 1. That means a plus c equal to 0, b plus d equal to 0 and a plus c into b plus d equal to 1 which is not possible because a plus c equal to 0, b plus d equal to 0 then its multiplication would not be 1. So which is not possible therefore the q1 is not equivalent to the q2. The first option is wrong and the also fourth option is wrong which is all are equivalent. Now we are going to check the equivalence of q2 and q3. So, now we assume q2 x ax plus by cx plus dy equal to q3 xy. Now, ax plus by plus cx plus dy whole square is equal to x square plus 3xy plus 2y square. Now, opening the brackets, a plus c whole square x square plus a plus c into b plus d xy plus b plus d whole square y square equal to x square plus 3xy plus 2y square. By comparing the coefficient of x square, xy and y square, we have This relation a plus c whole square equal to 1, b plus d whole square equal to 2 and a plus c into b plus d is equal to 3. But b plus d whole square is equal to 2 which is not possible because a, b, c, d are integers. So square of any integer should not be 2. So therefore this equivalence is also not possible. So option number 3 is also wrong. Now we are going to check the equivalence of q1 and q3. Q1 AX plus BY and CX plus DY equal to Q3 XY. AX plus BY into CX plus DY as a definition of Q1 XY, X into Y. So, X replaced by AX plus BY and Y replaced by CX plus DY. So, we have AX plus BY into CX plus DY is equal to X square plus 3XY plus 2Y square. Now, factorize the right hand side term. So, we have X plus Y into X plus 2Y. By natural comparison, we can say that one of the value of ABCD is A equal to 1, B equal to 1, C equal to 1 and D equal to 2. AD minus BC is equal to 1. So, ABCD are integer and AD minus BC determinant value is 1. Therefore, we are able to find ABCD such that Q1 of AX plus BY, CX plus DY equal to Q3XY. Therefore, Q1 equivalent to Q3. So, this is the correct answer 2. Thanks for watching. For more videos, like, subscribe, press the bell icon of Next Gen.