Show that `|[a, b, c],[ a+2x, b+2y, c+2z],[ x, y ,z]|=0`

Explanation

Solving L.H.S.

`|[a, b, c],[ a+2x, b+2y, c+2z],[ x, y ,z]|`

Expressing elements of `2^(nd)` row as sum of two elements.

` " " " " =|[a, b, c],[ a, b, c],[ x, y ,z]|+|[a, b, c],[ 2x,2y, 2z],[ x, y ,z]|`

`R_1 & R_2` are identical in `1^(st)` matrix,

`therefore` determinant of `1^(st)` matrix is zero.

` " " " " =0+|[a, b, c],[ 2x,2y, 2z],[ x, y ,z]|`

` " " " " =|[a, b, c],[ 2x,2y, 2z],[ x, y ,z]|`

Since, all elements of `2^(nd)` row are multiplied by 2,

We can take 2 outside the determinant.

` " " " " =2|[a, b, c],[ x,y, z],[ x, y ,z]|`

`R_1 & R_3` are identical

` " " " " =2xx0`

` " " " " =`R.H.S.

Thus, L.H.S. = R.H.S.

Hence proved.