3. Find value of determinant using properties of determinants ...
`[[100,205,105],[200,408,207],[300,608,310]]`

Solution:

`A=`	100   205   105   200   408   207   300   608   310

Now, `C_2=C_2 - 2 xx C_1`

`=`	100   5   105   200   8   207   300   8   310

Now, `C_3=C_3 - C_1`

`=`	100   5   5   200   8   7   300   8   10

take 100 as a comman factor from `C_1`

`=100 xx `

1   5   5   2   8   7   3   8   10

`=100 xx [1 xx (8 × 10 - 7 × 8) -5 xx (2 × 10 - 7 × 3) +5 xx (2 × 8 - 8 × 3)]`

`=100 xx [1 xx (80 -56) -5 xx (20 -21) +5 xx (16 -24)]`

`=100 xx [1 xx (24) -5 xx (-1) +5 xx (-8)]`

`=100 xx [24 +5 -40]`

`=100 xx [-11]`

`=-1100`

Method-2: Determinant by expanding cofactors

`|A|`

=

`100`   `205`   `105`   `200`   `408`   `207`   `300`   `608`   `310`

=

`100`   ×    `408`   `207`   `608`   `310`

`-205`   ×    `200`   `207`   `300`   `310`

`+105`   ×    `200`   `408`   `300`   `608`

`=100 xx (408 × 310 - 207 × 608) -205 xx (200 × 310 - 207 × 300) +105 xx (200 × 608 - 408 × 300)`

`=100 xx (126480 -125856) -205 xx (62000 -62100) +105 xx (121600 -122400)`

`=100 xx (624) -205 xx (-100) +105 xx (-800)`

`= 62400 +20500 -84000`

`=-1100`

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4. Find value of determinant using properties of determinants ...
`[[6,3,9],[1,0,2],[40,50,20]]`

Solution:

`A=`	6   3   9   1   0   2   40   50   20

take 3 as a comman factor from `R_1`

`=3 xx `

2   1   3   1   0   2   40   50   20

take 10 as a comman factor from `R_3`

`=30 xx `

2   1   3   1   0   2   4   5   2

`=30 xx [2 xx (0 × 2 - 2 × 5) -1 xx (1 × 2 - 2 × 4) +3 xx (1 × 5 - 0 × 4)]`

`=30 xx [2 xx (0 -10) -1 xx (2 -8) +3 xx (5 +0)]`

`=30 xx [2 xx (-10) -1 xx (-6) +3 xx (5)]`

`=30 xx [-20 +6 +15]`

`=30 xx [1]`

`=30`

Method-2: Determinant by expanding cofactors

`|A|`

=

`6`   `3`   `9`   `1`   `0`   `2`   `40`   `50`   `20`

=

`6`   ×    `0`   `2`   `50`   `20`

`-3`   ×    `1`   `2`   `40`   `20`

`+9`   ×    `1`   `0`   `40`   `50`

`=6 xx (0 × 20 - 2 × 50) -3 xx (1 × 20 - 2 × 40) +9 xx (1 × 50 - 0 × 40)`

`=6 xx (0 -100) -3 xx (20 -80) +9 xx (50 +0)`

`=6 xx (-100) -3 xx (-60) +9 xx (50)`

`= -600 +180 +450`

`=30`

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