Pttempt any five :- Prove that:- \( \left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2}\end{array}\right|=(x-y)(y-z)(z-x) \)

To prove the determinant identity \[ \left|\begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array}\right| = (x-y)(y-z)(z-x), \] we will compute the determinant of the matrix on the left-hand side. The determinant of a \(3 \times 3\) matrix \[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \] is given by the formula: \[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg). \] For our specific matrix, we have: \[ \begin{pmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{pmatrix}. \] Let's denote the rows as \(R_1\), \(R_2\), and \(R_3\): - \(R_1 = (1, x, x^2)\) - \(R_2 = (1, y, y^2)\) - \(R_3 = (1, z, z^2)\) Now, we can compute the determinant: \[ \text{det} = 1 \cdot (y \cdot z^2 - z \cdot y^2) - x \cdot (1 \cdot z^2 - 1 \cdot y^2) + x^2 \cdot (1 \cdot y - 1 \cdot z). \] Calculating each term: 1. The first term is: \[ y z^2 - z y^2 = y z^2 - z y^2 = y z^2 - y^2 z = y(z^2 - yz). \] 2. The second term is: \[ -x(z^2 - y^2) = -x(z - y)(z + y). \] 3. The third term is: \[ x^2(y - z). \] Putting it all together, we have: \[ \text{det} = y(z^2 - yz) - x(z^2 - y^2) + x^2(y - z). \] Now, we can factor this expression. Notice that the determinant is a polynomial in \(x\), \(y\), and \(z\) that is symmetric in \(x\), \(y\), and \(z\) and has degree 3. The roots of this polynomial are clearly \(x = y\), \(y = z\), and \(z = x\), which suggests that the determinant can be expressed as: \[ k(x-y)(y-z)(z-x) \] for some constant \(k\). To find \(k\), we can evaluate the determinant at specific values. Let's set \(x = 0\), \(y = 1\), and \(z = 2\): \[ \left|\begin{array}{ccc} 1 & 0 & 0^2 \\ 1 & 1 & 1^2 \\ 1 & 2 & 2^2 \end{array}\right| = \left|\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{array}\right|. \] Calculating this determinant: \[ = 1(1 \cdot 4 - 1 \cdot 2) - 0 + 0 = 1(4 - 2) = 2. \] Now, substituting \(x = 0\), \(y = 1\), and \(z = 2\) into \((x-y)(y-z)(z-x)\): \[ (0-1)(1-2)(2-0) = (-1)(-1)(2) = 2. \] Since both sides equal 2, we conclude that \(k = 1\). Thus, we have shown that: \[ \left|\begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array}\right| = (x-y)(y-z)(z-x). \] This completes the proof.