#CSIRNET #MATHEMATICS FEB 2022 | Linear Algebra | Row space, Column space and Null space of Matrix

Hello friends, this is Mahesh Yuvlekar and you are watching NextGenMaths. If you have not subscribed my channel, then please subscribe the channel and press the bell icon. In this video, we are going to discuss the problem of CSI NET 522, problem of Linear Algebra. Question. Consider the two statements given below. First statement, there exists matrix N belongs to M4R such that 1, 1, 1, minus 1, 1, minus 1, 1, 1 is basis for row space of N and 1, 2, 1, 4 belongs to null space of N. And second statement, there exists matrix M belongs to M4R such that 1, 1, 1, 0, 1, 1, 0, 1, 1 is basis for the column space of M and 1, 1, 1, 1 and 1, 0, 1, 0 belongs to null space of M. Which of the following statements is true? We have again four options. First option, first statement is false, second is true, second option, first is true, second is false, third option, both are false and the fourth option, both are true. Before going into detailed solution, I have good news for the candidate of CSI NET December 2022. IFS, the leading institute for CSR, UGC NET, Mathematical Science launched special series on YouTube absolutely free for all the candidates of CSI NET from 4th July 2022. It includes 30 units in 60 days with 2000 plus practice questions which will teach by India's base educator. Motto of this series is Aapki Jeet Pakki. You will be Ash of CSI NET with IFAS. So my friend, please visit YouTube channel of IFAS which is IFAS Maths. In the description, link is available. Now we back to the solution. We will check both the statements one by one. First statement, it says that there exists matrix N such that 1, 1, 1, minus 1, 1, minus 1, 1, 1 is basis for row space of N. So if we construct N with this condition, then N equal to 1, 1, 1, minus 1, first row, 1, minus 1, 1, 1, second row and we don't know about the last two rows. Then we have given 1, 2, 1, 4 belongs to null space of N. Suppose V equal to 1, 2, 1, 4 belongs to null space of N which is a solution of NX equal to 0. So if V is an element of null space of N, then it satisfy NX equal to 0. Therefore, we have NV equal to 0. So N multiply with V, 1, 2, 1, 4, then we have two equations, 1 plus 2 plus 1 minus 4 equal to 0 and the second equation 1 minus 2, 1 plus 4 equal to 0. So first equation gives 0 equal to 0 but the second equation gives 4 equal to 0 which is not possible. Therefore, matrix N does not exist with this condition. So the first statement is a false statement. Now statement number 2 which is there exists matrix M such that 1, 1, 1, 0, 1, 0, 1, 1 is basis of column space of M and 1, 1, 1, 1 and 1, 0, 1, 0 belongs to null space of M. So with this condition, matrix M with the first column 1, 1, 1, 0 and the second column 1, 0, 1, 1, we don't know about the last two columns so it is A, B, C, D and the fourth column X, Y, Z, W. Another condition we have again 1, 1, 1, 1, 0, 1, 0 belongs to null space of M. So V equal to 1, 1, 1, 1 and U equal to 1, 0, 1, 0 belongs to null space of M which is X such that MX equal to 0. V and U are the elements of M therefore it satisfy the condition MX equal to 0. So we have first condition MV equal to 0 so M multiply with V 1, 1, 1, 1 then we have four equation 1 plus 1 plus A plus X equal to 0 that means 2 plus A plus X equal to 0. Second condition 1 plus B plus Y equal to 0. Third condition 1 plus 1 plus C plus Z equal to 0 that means 2 plus C plus Z equal to 0 and the fourth condition 1 plus D plus W equal to 0. Then the second equation MU equal to 0. So M multiply with U 1, 0, 1, 0 we have four equation 1 plus A equal to 0, 1 plus B equal to 0, 1 plus C equal to 0 and D equal to 0. So this four equation will give the value of A, B, C, D. A equal to minus 1, B equal to minus 1, C equal to minus 1 and D equal to 0. Using first four condition we can find the value of X, Y, Z by substituting the value of A, B, C, D. So if you put A equal to minus 1 then we have X equal to minus 1, B equal to minus 1 then Y equal to 0, C equal to minus 1 then Z equal to minus 1 and D equal to 0 then we have W equal to minus 1. Therefore we have A equal to minus 1, B equal to minus 1, C equal to minus 1, D equal to 0 and X equal to minus 1, Y equal to 0, Z equal to minus 1, W equal to minus 1. If we substitute this value in M then we are able to find matrix M which satisfy this two condition therefore matrix M is exist with this given condition. So the statement 2 is true, statement 1 is false and statement 2 is true which is option number 1 so option number 1 is the correct answer. Thanks for watching. For more videos like, share, subscribe, press the bell icon of NextGenMaths.