CSIR NET MATHEMATICS | Linear Algebra | Best Solution Determinant of Matrix June 2019

In this video, we are going to discuss the problem of Linear Algebra from CSIR NET June 2019. Example number 70, Booklet C, marks 4.74. Let A equal to aij be a 3 cross 3 complex matrix. Identify the correct statement from the following. First option, determinant minus 1 power i plus j aij is equal to determinant a or it has value minus determinant a. Third option square root minus 1 power i plus j a i j has value determinant a or in the fourth option it has value minus determinant a. We have given a equal to a i j is a 3 cross 3 complex matrix so we can write down in the form a 1 1 a 1 2 a 1 3 similarly a 3 1 a 3 2 a 3 3. Now corresponding determinant value now we required in the first option minus 1 power i plus j a i j determinant. So every term is multiplied by minus 1 power i plus j. Just for example a11 is multiplied by minus 1 power 1 plus 1 which is 2. a12 is multiplied by minus 1 power 1 plus 2 is 3. a13 multiplied by minus 1 power 1 plus 3 is 4. a21 is multiplied by minus 1 power 2 plus 1 is 3 and similarly a33 multiplied by minus 1 power 6. Now we can see in the first row minus 1 power 2 is a common factor. In the second row minus 1 power 3 is a common factor and in the third row minus 1 power 4 is a common factor. So using the property of determinant we can write down minus 1 power 2 minus 1 power 3 minus 1 power 4 into the remaining part. Now in the remaining part second column has common factor minus 1 and the third column has common factor minus 1 power 2. So which is also take out from the determinant then we have minus 1 power 2 minus 1 power 3 minus 1 power 4 minus 1 and minus 1 power 2 and the remaining part is a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 which is nothing but the determinant a now the total minus 1 power is 2 plus 3 plus 4 plus 1 plus 2 which is 12 and into determinant so minus 1 power 12 is 1 so it has value determinant a so first option is correct second option say it is minus determinant a which is wrong now for the third option we have to find the determinant value square root minus 1 power i plus j a i j but the square root minus 1 is represented by the complex number iota i power i plus j with the property i square equal to minus 1. So, the corresponding determinant every value a i j multiply by i power i plus j. So, a 1 1 is multiplied by i power 1 plus 1 2 a 1 2 multiply by i power 1 plus 2 3 a 1 3 multiply by i power 1 plus 3 which is 4 and similarly a 3 3 multiply by i power 3 plus 3 6 now we can see first row as common factor i square second row as common factor i cube and third row as common factor i power 4 so now we can see in the second column i is a common factor and in the third column i square is common factor so take out from the determinant then we have i square i cube i power 4 i into i square and the remaining part is determinant a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 which is nothing but the determinant a so it is i power 2 plus 3 plus 4 plus 1 plus 2 which is i power 12 into determinant a i power 4 has value 1 so 1 power 3 is 1 so remaining part is determinant a so it has value determinant a so third option says that has value determinant a which is correct and the fourth option says is determinant minus 1 which is wrong thanks for watching for more videos like share subscribe press the bell icon of NextGenMaths