Question

Find the value of the determinant \(\begin{vmatrix}x&a&x+a\\y&b&y+b\\z&c&z+c\end{vmatrix}\).

Hint:

Column 3 is exactly column 1 + column 2 in every row, so the columns are linearly dependent.

Answer 1

Step 1: Compare the third column to the first two:
In every row, the third entry is the sum of the first two: \(x+a\), \(y+b\), \(z+c\). So column 3 = column 1 + column 2.

Step 2: Apply the column operation \(C_3\to C_3-C_1-C_2\):
This is a legal determinant operation (adding a multiple of other columns doesn't change the value) and turns column 3 entirely to zeros.

Step 3: A determinant with a zero column is zero:
Expanding along the (now all-zero) third column gives \(0\).

Final Answer:
\[ \boxed{0} \]