CSIR NET MATHEMATICS | Linear Algebra | Simplest Solution Similar Matrices June 2019

Hello friends, in this video we are going to discuss the problem of linear algebra from CSIR NET June 2019. Example number 75 from booklet C marks 4.75. Consider the matrix A equal to 0, 1, 1, 0 over the field Q of rational numbers. Which of the following matrices are of the form P transpose AP for the suitable 2 cross 2 invertible P over Q? Here P transpose denote the transpose of P. First option 2 0 0 -2, second option 2 0 0 2, third option 1 0 0 -1 and the fourth option 3 4 4 5. Now matrix A is given 0 1 1 0. We want to find matrix P which is invertible such that any one of the option can be written in the form of P transpose AP. P equal to abcd is invertible 2 cross 2 matrix then for invertibility determinant ad minus bc should not be 0. Now P transpose AP is equal to ac bd 0 1 1 0 abcd. So after the multiplication we have 2ac, ad plus bc, ad plus bc and 2bd. Now we are going to check one by one option. So now P transpose AP is equal to 2 0 0 minus 2 option number 1. Now P transpose AP equal to 2 AC AD plus BC AD plus BC 2 BD is equal to 2 0 0 minus 2. By comparing the components of matrix A we have 2 AC equal to 2, AD plus BC equal to 0, 2 BD equal to minus 2. Therefore AC equal to 1, AD plus BC equal to 0, BD equal to minus 1. For the invertibility of P, determinant AD minus BC should not be 0. So, we are going to find the value of AD minus BC. So, AD minus BC square equal to AD plus BC whole square minus 4 AD BC. So, we have the value of AD plus BC, AC and BD. So, after the substituting the values, we have AD minus BC square is equal to 4, which is possible for some P. Therefore, P's exist. And therefore, the option number 1 is correct. Now, we repeat the process for the option 2. P transpose AP equal to 2, 0, 0, 2. Then, 2AC, AD plus BC, AD plus BC, 2BD equal to 2, 0, 0, 2. By comparing the elements of matrix A, we have AC equal to 1, AD plus BC equal to 0 and BD equal to 1. Now find the value of AD minus BC whole square, so AD plus BC whole square minus 4AD BC, so we have AD plus BC equal to 0, AC equal to 1, BD equal to 1, so 4 into 1, 1, so therefore AD minus BC whole square equal to minus 4, so square of any number should not be negative for the real values, so which is not possible, therefore P does not exist and hence the option Option number two is wrong. Now for the option number 3, P transpose AP equal to 1, 0, 0, minus 1, then repeating the process, we have AC equal to 1 by 2, AD plus BC equal to 0 and BD equal to minus 1 by 2 and value of AD minus BC whole square equal to 1 and therefore P can be exist for some values of ABCD and hence the option number 3 is also correct. Now, option number 4. Where P transpose AP equal to 3, 4, 4, 5. So, 2AC, AD plus BC, AD plus BC, 2BD equal to 3, 4, 4, 5. By comparing the component, 2AC equal to 3, AD plus BC equal to 4 and 2BD equal to 5. Therefore, AD minus BC whole square is equal to 4 square minus 4 into 3 by 2 into 5 by 2. And we have AD minus BC whole square equal to 1. So, P can be exist. Therefore, Option number 4 is correct. Hence, this question has 3 correct answers 1, 3 and 4. Thanks for watching. For more videos, like, share, subscribe, press the bell icon of NextGen Maths.