 Welcome to the session. In this session we discuss the following question that says using properties of determinants show the following. Determinant with elements in the first row as b plus c whole square a b c a. In the second row the elements are a b a plus c whole square b c. In the third row the elements are a c b c a plus c whole square is equal to 2 a b c into a plus b plus c whole cube. Let's move on to the solution now. First of all we assume the given determinant as delta. So this is the given determinant taken as delta. Now in the next step we multiply the first row of the determinant that is r1 by a the second row that is r2 by b and the third row that is r3 by c. So we get delta is equal to the determinant with elements in the first row as a into b plus c whole square that is we multiply the first row by a. So we multiply a b by a we get a square b. Multiplying c a by a we get c a square. We now multiply the second row by b. So multiply a b by b we get a b square then we have a plus c b whole square into b then b square c. In the same way we multiply the third row by c. So we have a c square then b c square then c into a plus b b whole square. The whole determinant delta is equal to 1 upon a b c that is we are dividing the whole determinant by a b c into the determinant with elements in the first row as a into b plus c b whole square a square b. In the second row the elements are a b square c whole square into b b square c. In the third row the elements are a c square b c square c into a plus b whole square. Now in the next step we have b c common respectively we get is equal to column c1. So we write here a taking b common from the column c2 we write here b. I am taking c common from the column c3 we write here c delta is equal to a b c upon a b c into determinant with elements in the first row as b plus c whole square then a square then in the second row the elements would be b square a plus c whole square b square and the third row the elements would be c square c square a plus c whole square. These are the elements that we obtain after taking out a b and c common from the column c1 c2 c3 respectively. Now this a b c cancels with a b c delta as the determinant with elements in the first row as b plus c whole square a square a square in the second row the elements are b square a plus c whole square b square in the third row the elements are c square c square a plus b whole square. Next we apply our c3 elements of the first column obtain by subtracting the elements of the first and the third columns. So here we have b plus c whole square minus a square is the first element of the column 1. Now second element of the column 1 is obtained by subtracting b square and b square so we have here 0. Third element of the column c1 is obtained by subtracting these two so we have c square a plus b b whole square column c2 and c3 elements as it is so here we have a square then a plus c b whole square c square column c3 also known as it is so here we have a square b square a plus b b whole square. Next we apply c2 minus c3 and when applying this we get delta is equal to with the determinants where the elements of the column c2 are obtained by subtracting the elements of the column c2 and c3 so this means column c1 will remain as it is so here we have b plus c whole square minus a square 0 and c square minus a plus b whole square. Now the elements of the column c2 will change and it would be obtained by subtracting elements of column c2 and c3 so a square minus a square is 0 then here we have a plus c whole square minus b square and here we have c square minus a plus b whole square and elements of column c3 also remains as it is that is we have a square b square a plus b whole square. Now further let's expand this term that is b plus c whole square minus a square can be written as b plus c plus a this whole into b plus c minus a the whole then consider this element which is c square minus a plus b b whole square this is equal to c plus a plus b b whole into c minus a minus b the whole. Now we consider this element that is a plus c b whole square minus b square this can be written as a plus c plus b the whole into a plus c minus b the whole. We have delta is equal to determinant with elements in the first row as this element can be written as this that is a plus b plus c b whole into b plus c minus a b whole then the second element of the first row is 0 the third element of the first row is a square. Now consider the second row its first element is 0 second element would be written as this that is a plus c plus b b whole into a plus c minus b b whole and where we have b square then the elements of the third row are written as increase of c square minus a plus b b whole square we write this that is we have c plus a plus b the whole into c minus a minus b the whole. The second element of the third row would also be the same that is c plus a plus b the whole into c minus a minus b the whole third element would be written as this that is a plus b the whole square. As you can see a plus b plus c is common in the columns c1 and c2 so taking a plus b plus c common from the columns c1 delta is equal to a plus b plus c the whole square into determinant with elements in the first column as b plus c minus a 0 c minus a minus b with elements in the second column as 0 a plus c minus b c minus a minus b in the third column the elements are a square b square and a plus b the whole square. We apply r3 minus r1 and in applying this we get delta is equal to a plus b plus c the whole square into determinant with elements of the third row determined by r3 minus r1 and 2 as it is that is in the first row we have b plus c minus a 0 a square in the second row we have 0 a plus c minus b b square. Now the first element of the third row that is c minus a minus b is equal to r3 that is c minus a minus b minus r1 which is b plus c minus a so here we have minus b minus c plus a minus r2 which is 0 this c minus c cancels a minus a cancels and we are left with minus 2b so we write here minus 2b for the second element of the third row that is c minus a minus b we write this as r3 which is b minus a minus b minus r1 which in this case is 0 minus r2 which is a plus c minus b so here we have minus a minus c plus b now here c minus c cancels b minus b cancels and we have minus 2a so we write here minus 2 now consider the third element of the third row which is a plus b the whole square this is equal to r3 that is a plus b the whole square minus r1 which is a square minus r2 which is b square so this is equal to a square plus b square plus 2ab minus a square minus b square a square minus a square cancels b square minus b square cancels and this is equal to 2ab so we write here 2ab next we apply the operation which is c1 as c1 plus 1 upon a c3 and on applying this we get delta is equal to a plus b plus cg whole square into determinant may be elements of the first column are determined by c1 plus 1 upon a c3 consider the first element of the first column which is b plus c minus a this is equal to b plus c minus a that is c1 plus 1 upon a into the element of the column c3 which is a square so we get b plus c minus a plus a where this a minus a cancels and this is written as b plus c so we write here b plus c as the first element plus the second element of the first column which is 0 this is equal to 0 plus 1 upon a into the corresponding element of the third column which is b square and so this is equal to b square upon a so we write here b square upon a next is the third element of the first column which is minus 2b this is equal to minus 2b plus 1 upon a into the corresponding element of the third column which is 2ab where this a and a cancels and we have minus 2b plus 2b which is equal to 0 so we write here 0 now the elements of the second and the third column remain as it is so here we have 0 a plus c minus b minus 2a a square b square 2ab now next applying upon b c3 we get theta is equal to a plus b plus cb whole square into determinant where the elements of the second column are obtained as c2 plus 1 upon b c3 so this means the elements of the first and the third column remains as it is so first of all we write the elements of the first column b plus c b square upon a 0 elements of the third column as a square b square 2ab determine the elements of the second column the first element of the second column which is 0 this is equal to c2 that is 0 plus 1 upon b into c3 that is the corresponding element of the third column which is a square in this case a square upon b we write the first element of the second column as square upon b so there the second element of the second column which is a plus c minus b this is equal to c2 that is a plus c minus b plus 1 upon b into the corresponding element of the third column which is b square so this is equal to a plus c minus b plus b b minus b cancels and we have a plus c here so the second element of the second column is consider the third element of the second column which is minus 2a this is equal to minus 2a plus 1 upon b into the corresponding element of the third column which is 2ab and is 0 so we write here 0 the third element of the third row that is r3 we get delta is equal to a plus b plus c the whole square now as you can see in the third row the first two elements are 0 so we will only consider the third element of delta equal to a plus b plus c whole square into the determinant with b plus c square upon b b square upon a further we get delta is equal to a plus b plus c the whole square into 2ab into b plus c the whole into whole minus 1b into b square delta is equal to a plus b plus c the whole square into 2ab into ab plus bc plus ac cancels a and a cancels so we have minus ab the whole this ab minus ab cancels so we get delta is equal to a plus b plus c the whole square into 2ab now we take c common from these three terms so we have 2abc b plus a plus this with us delta equal to 2abc into a plus b plus c the whole tube the given determinant with elements in the first row as b plus c whole square ab c a in the second row the elements are ab a plus c whole square bc in the third row the elements are ac bc 12 plus b whole square and this is equal to 2abc into a plus b plus c the whole tube and this is what we were supposed to prove so I should hope you understood the solution of this question