Q : 1        Using the property of determinants and without expanding, prove that 

                  \begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}=0

Answers (1)

We can split it in manner like;

 \begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}= \begin{vmatrix} x &a &x \\ y & b &y \\ z &c &z \end{vmatrix} + \begin{vmatrix} x &a & a\\ y &b &b \\ z&c & c \end{vmatrix}

So, we know the identity that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Clearly, expanded determinants have identical columns.

\therefore 0 + 0 = 0

Hence the sum is zero.

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