# SOLVING DETERMINANTS USING PROPERTIES

## About "Solving Determinants Using Properties"

Solving Determinants Using Properties :

Here we are going to see some example problems to understand solving determinants using properties.

To know properties of determinants, please visit the page "Properties of determinants".

## Solving Determinants Using Properties - Questions

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 1st and and 3rd 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 1st, 2nd and 3rd row respectively.

Now we may factor abc from 2nd and 3rd 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).

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 1st 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.

After having gone through the stuff given above, we hope that the students would have understood, "Solving Determinants Using Properties".

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