AP DSC School Assistant Math Model Test 2

Concept:The determinant of a matrix is zero if its rows (or columns) are linearly dependent. Here we expand the given determinant and factorise to find the condition that makes the product equal zero.Explanation:Let the determinant be:$$\Delta = \begin{vmatrix} xp+y & x & y \\ yp+z & y & z \\ 0 & xp+y & yp+z \end{vmatrix}$$Expanding along the first row:$$\Delta = (xp+y)\bigl(y(yp+z) - z(xp+y)\bigr) - x\bigl((yp+z)(yp+z) - z\cdot 0\bigr) + y\bigl((yp+z)(xp+y) - y\cdot 0\bigr)$$Simplify each term:

• First term: $(xp+y)(y^2p + yz - xzp - yz) = (xp+y)(y^2p - xzp) = (xp+y)\,p\,(y^2 - xz)$
• Second term: $-x\,(yp+z)^2$
• Third term: $y\,(yp+z)(xp+y)$

So,$$\Delta = p(xp+y)(y^2 - xz) - x(yp+z)^2 + y(yp+z)(xp+y)$$Group the terms with $(yp+z)$:$$\Delta = p(xp+y)(y^2 - xz) + (yp+z)\bigl[y(xp+y) - x(yp+z)\bigr]$$Simplify the bracket: $y(xp+y) - x(yp+z) = xyp + y^2 - xyp - xz = y^2 - xz$.Thus,$$\Delta = (y^2 - xz)\bigl[p(xp+y) + (yp+z)\bigr] = (y^2 - xz)(xp^2 + 2yp + z)$$Hence $\Delta = 0$ if either $y^2 - xz = 0$ or $xp^2 + 2yp + z = 0$. The condition $y^2 = xz$ means that $x, y, z$ are in geometric progression (G.P.).Answer:Option B: x, y, z are in G.P.