(ii) 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. Let \[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{pmatrix}. \] We can compute the determinant of \( A \) using the formula for the determinant of 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 \( A \): - \( a = 1, b = 1, c = 1 \) - \( d = 1, e = 1+x, f = 1 \) - \( g = 1, h = 1, i = 1+y \) 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: \[ 1 - (1+x) = 1 - 1 - x = -x. \] Putting it all together, we have: \[ \det(A) = 1 \cdot (x + y + xy) - 1 \cdot y + 1 \cdot (-x). \] This simplifies to: \[ \det(A) = x + y + xy - y - x = xy. \] Thus, we have shown that: \[ \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right| = xy. \] Therefore, the statement is proven: \[ \boxed{xy}. \]