Use Properties of determinants, prove that:
$open vertical bar table row cell 1 plus straight a end cell cell space 1 end cell cell space 1 end cell row 1 cell 1 plus straight b end cell 1 row 1 1 cell 1 plus straight c end cell end table close vertical bar space equals space abc plus bc plus ca plus ab$

Solution

Consider the determinant
$increment space equals open vertical bar table row cell 1 plus straight a end cell 1 1 row 1 cell 1 plus straight b end cell 1 row 1 1 cell 1 plus straight c end cell end table close vertical bar$
Taking abc common outside, we have
$increment equals abc space open vertical bar table row cell 1 over straight a plus 1 end cell cell 1 over straight b end cell cell space 1 over straight c end cell row cell 1 over straight a end cell cell 1 over straight b plus 1 end cell cell 1 over straight c end cell row cell 1 over straight a end cell cell 1 over straight b end cell cell space 1 over straight c plus 1 end cell end table close vertical bar$
Applying the transformation, $straight C subscript 1 rightwards arrow straight C subscript 1 plus straight C subscript 2 plus straight C subscript 3 comma$
$increment equals space abc open vertical bar table row cell 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c space space space end cell cell 1 over straight b end cell cell 1 over straight c end cell row cell 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c end cell cell space 1 over straight b plus 1 end cell cell 1 over straight c end cell row cell 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c end cell cell 1 over straight b end cell cell space space 1 over straight c plus 1 end cell end table close vertical bar$
$rightwards double arrow increment equals abc open parentheses 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses space open vertical bar table row 1 cell 1 over straight b end cell cell 1 over straight c end cell row cell 1 space end cell cell 1 over straight b plus 1 end cell cell 1 over straight c end cell row 1 cell 1 over straight b end cell cell 1 over straight c plus 1 end cell end table close vertical bar$
Apply the transformations, $straight R subscript 2 rightwards arrow straight R subscript 1 space and space straight R subscript 3 rightwards arrow straight R subscript 3 minus straight R subscript 1$
$increment equals abc open parentheses 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses space open vertical bar table row 1 cell space 1 over straight b end cell cell 1 over straight c end cell row 1 1 0 row 1 0 1 end table close vertical bar$
Expanding along $straight C subscript 1$ , we have
$increment equals abc open parentheses 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses cross times 1 cross times open vertical bar table row 1 0 row 0 1 end table close vertical bar rightwards double arrow space increment equals abc open parentheses 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses equals abc plus ab plus bc plus ca$

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