SOLVING DETERMINANTS USING PROPERTIES

Question 1 :

Without expanding the determinant, prove that

Solution :

To prove  the given determinant as zero, we may try to show any two rows or columns as identical.

For that we use elementary operations.

Now we may factor out "s" and  "(a² + b² + c²)" from the 1^st and and 3^rd column respectively.

So, we get

If any two columns are identical, then the value of determinant will become zero.

  =  s(a² + b² + c²) (0)

  =  0

Hence proved.

Question 2 :

Show that 

Solution :

In order to show any two rows or columns are same, let us multiply "a", "b" and "c" by the 1^st, 2^nd and 3^rd row respectively.

Now we may factor abc from 2^nd and 3^rd column respectively.

Since column 1 and 2 are identical, the value of determinant will become 0.

So, we get (abc)² (ab + bc + ca) (0).

Hence the answer is 0.

Question 3 :

Prove that

Solution :

First let us factor out a from a² + ab, b from b² + bc and c from c² + ac respectively.

Factor out a, b and c from column 1, 2 and 3 respectively.

So, we get

Now we may factor out 2 from the first row.So, we get

Now let us subtract column 3 from column 2. So we get

  =  2abc [ c(ab + b² - b(b - a)) ]

  =  2abc [ c(ab + b² - b² + ab) ]

  =  2abc [ 2abc ]

  =  4a²b²c²

Hence proved.

Question 4 :

Prove that

Solution :

First let us subtract second row from 1^st row.

Since we cannot simplify this hereafter, let us find the expansion.

  =  a [(1+b)(1+c) - 1] + b[(1+c) - 1]

  =  a [(1 + c + b + bc) - 1] + b[1+c-1]

  =  a [c + b + bc] + b[c]

  =  ac + ab + abc + bc

  =  abc +ab + bc + ca

  =  abc (1 + 1/a  + 1/b  +  1/c)

Hence proved.

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