Question and Answers Forum

All Questions      Topic List

Matrices and Determinants Questions

Previous in All Question      Next in All Question      

Previous in Matrices and Determinants      Next in Matrices and Determinants      

Question Number 10245 by j.masanja06@gmail.com last updated on 31/Jan/17

prove that determinant ((1,1,1),(x,y,z),(x^2 ,y^2 ,z^2 ))=(x−y)(y−z)(z−y)

$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{x}}&{\mathrm{y}}&{\mathrm{z}}\\{\mathrm{x}^{\mathrm{2}} }&{\mathrm{y}^{\mathrm{2}} }&{\mathrm{z}^{\mathrm{2}} }\end{vmatrix}=\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{y}−\mathrm{z}\right)\left(\mathrm{z}−\mathrm{y}\right) \\ $$

Answered by prakash jain last updated on 31/Jan/17

determinant ((1,1,1),(x,y,z),(x^2 ,y^2 ,z^2 )) C1−C2→C1, C2−C3→C2 determinant ((0,0,1),((x−y),(y−z),z),((x^2 −y^2 ),(y^2 −z^2 ),z^2 )) determinant ((0,0,1),((x−y),(y−z),z),(((x−y)(x+y)),((y−z)(y+z)),z^2 )) (x−y)(y−z) determinant ((0,0,1),(1,1,z),(((x+y)),((y+z)),z^2 )) Expanding wrt row 1 (x−y)(y−z)((y+z)−(x+y)) =(x−y)(y−z)(z−x)

$$\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{x}}&{\mathrm{y}}&{\mathrm{z}}\\{\mathrm{x}^{\mathrm{2}} }&{\mathrm{y}^{\mathrm{2}} }&{\mathrm{z}^{\mathrm{2}} }\end{vmatrix}\:\: \\ $$$$\mathrm{C1}−\mathrm{C2}\rightarrow\mathrm{C1},\:\mathrm{C2}−\mathrm{C3}\rightarrow\mathrm{C2} \\ $$$$\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{x}−\mathrm{y}}&{\mathrm{y}−\mathrm{z}}&{\mathrm{z}}\\{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }&{\mathrm{y}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} }&{\mathrm{z}^{\mathrm{2}} }\end{vmatrix}\:\: \\ $$$$\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{x}−\mathrm{y}}&{\mathrm{y}−\mathrm{z}}&{\mathrm{z}}\\{\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}+\mathrm{y}\right)}&{\left(\mathrm{y}−\mathrm{z}\right)\left(\mathrm{y}+\mathrm{z}\right)}&{\mathrm{z}^{\mathrm{2}} }\end{vmatrix}\:\: \\ $$$$\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{y}−\mathrm{z}\right)\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{z}}\\{\left(\mathrm{x}+\mathrm{y}\right)}&{\left(\mathrm{y}+\mathrm{z}\right)}&{\mathrm{z}^{\mathrm{2}} }\end{vmatrix}\:\: \\ $$$$\mathrm{Expanding}\:\mathrm{wrt}\:\mathrm{row}\:\mathrm{1} \\ $$$$\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{y}−\mathrm{z}\right)\left(\left(\mathrm{y}+\mathrm{z}\right)−\left(\mathrm{x}+\mathrm{y}\right)\right) \\ $$$$=\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{y}−\mathrm{z}\right)\left(\mathrm{z}−\mathrm{x}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com