If = 1&x&x^2 1&y&y^2 1&z&z^2 , _1= 1&1&1 &zx&xy &y&z , then prove that + _1=0

+ _1= 1&x&x^2 1&y&y^2 1&z&z^2 + 1&1&1 &zx&xy &y&z = 1&x&x^2 1&y&y^2 1&z&z^2 + 1&yz&x 1&zx&y 1&xy&z [Interchanging rows and coloumns in _1] = 1&x&x^2 1&y&y^2 1&z&z^2 - 1&x&yz 1&y&zx 1&z&xy [Applying C_2 _3 in _1] = 1&x&x^2 0&y-x&y^2-x^2 0&z-x&z^2-x^2 - 1&x&yz 0&y-x&zx-yz 0&z-x&xy-yz [Applying R_2 R_2-R_1 and R_3 R_3-R_1 ] =(y-x)(z-x) 1&x&x^2 0&1&y+x 0&1&z+x -(y-x)(z-x) 1&x&yz 0&1&-z 0&1&-y [Taking (y - x) common from R_2 and (z - x) common from R_3] =(y-x)(z-x)(z+x-y-x)-(y-x)(z-x)(-y+z) [Expanding along first column] =(y-x)(z-x)(z-y)(1-1) =0 + _1=0

If = 1&x&x^2 1&y&y^2 1&z&z^2 , _1= 1&1&1 &zx&xy &y&z , then prove…

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Question

If $\triangle=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix},$ $\triangle_1=\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix},$ then prove that $\triangle+\triangle_1=0$

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Answer

$\triangle+\triangle_1=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&\text{yz}&\text{x}\\1&\text{zx}&\text{y}\\1&\text{xy}&\text{z} \end{vmatrix}$
$[$Interchanging rows and coloumns in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\1&\text{y}&\text{zx}\\1&\text{z}&\text{xy} \end{vmatrix}$
$[$Applying $\text{C}_2\leftrightarrow\text{C}_3$ in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{y}-\text{x}&\text{y}^2-\text{x}^2\\0&\text{z}-\text{x}&\text{z}^2-\text{x}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\0&\text{y}-\text{x}&\text{zx}-\text{yz}\\0&\text{z}-\text{x}&\text{xy}-\text{yz}\end{vmatrix}$
[Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$ ]
$=(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{y}+\text{x}\\0&1&\text{z}+\text{x}\end{vmatrix}-(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{yz}\\0&1&-\text{z}\\0&1&-\text{y}\end{vmatrix}$
[Taking (y - x) common from $R_2$ and (z - x) common from $R_3$]
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}+\text{x}-\text{y}-\text{x})-(\text{y}-\text{x})(\text{z}-\text{x})(-\text{y}+\text{z})$
[Expanding along first column]
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}-\text{y})(1-1)$
$=0$
$\therefore\ \triangle+\triangle_1=0$