Prove that: \( \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|=x y \)

To prove that \[ \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|=x y, \] we will compute the determinant of the given matrix. The matrix is \[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{pmatrix}. \] We can use the determinant formula for a \(3 \times 3\) matrix: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg), \] where the matrix is \[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}. \] For our matrix, we have: - \(a = 1\), \(b = 1\), \(c = 1\) - \(d = 1\), \(e = 1+x\), \(f = 1\) - \(g = 1\), \(h = 1\), \(i = 1+y\) Now we can compute the determinant: \[ \det(A) = 1 \cdot ((1+x)(1+y) - 1 \cdot 1) - 1 \cdot (1 \cdot (1+y) - 1 \cdot 1) + 1 \cdot (1 \cdot 1 - 1 \cdot (1+x)). \] Calculating each term: 1. The first term: \[ (1+x)(1+y) - 1 = 1 + x + y + xy - 1 = x + y + xy. \] 2. The second term: \[ 1 \cdot (1+y) - 1 = 1 + y - 1 = y. \] 3. The third term: \[ 1 \cdot 1 - 1 \cdot (1+x) = 1 - (1+x) = -x. \] Putting it all together, we have: \[ \det(A) = 1 \cdot (x + y + xy) - 1 \cdot y + 1 \cdot (-x). \] This simplifies to: \[ x + y + xy - y - x = xy. \] Thus, we conclude that: \[ \det(A) = xy. \] Therefore, we have shown that \[ \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right| = xy. \] The final result is \[ \boxed{xy}. \]