(ii) Prove that: \( \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|=x y \) (iii) Find the value of \( \int \log x \cdot d x \)

To prove the determinant, we start with the matrix: \[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{pmatrix} \] We will compute the determinant \( \det(A) \). Using the determinant formula for a \( 3 \times 3 \) matrix, we have: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] where the matrix is represented as: \[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \] 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 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: \[ 1(1+y) - 1 = y \] 3. The third term: \[ 1 - (1+x) = -x \] Putting it all together: \[ \det(A) = (x + y + xy) - y - x = xy \] Thus, we have shown that: \[ \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right| = xy \] Now, for part (iii), we need to find the integral: \[ \int \log x \, dx \] We can use integration by parts. Let: - \( u = \log x \) so that \( du = \frac{1}{x} \, dx \) - \( dv = dx \) so that \( v = x \) Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \): \[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx \] This simplifies to: \[ \int \log x \, dx = x \log x - \int 1 \, dx = x \log x - x + C \] where \( C \) is the constant of integration. Therefore, the final result is: \[ \int \log x \, dx = x \log x - x + C \]