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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Using properties of determinants, prove that:
$\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\\text{y}&\text{y}^2&1+\text{py}^3\\\text{z}&\text{z}^2&1+\text{pz}^3\end{vmatrix}=(1+\text{pxyz})(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x}),$ where p is any scalar.
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Answer

$\triangle=\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\\text{y}&\text{y}^2&1+\text{py}^3\\\text{z}&\text{z}^2&1+\text{pz}^3\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$, we have:
$\triangle=\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\\text{y}-\text{x}&\text{y}^2-\text{x}^2&\text{p}(\text{y}^3-\text{x}^3)\\\text{z}-\text{x}&\text{z}^2-\text{x}^2&\text{p}(\text{z}^3-\text{x}^3)\end{vmatrix}$
$=(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\1&\text{y}+\text{x}&\text{p}(\text{y}^2+\text{x}^2+\text{xy})\\1&\text{z+x}&\text{p}(\text{z}^2+\text{x}^2+\text{xz})\end{vmatrix}$
Applying $R_3 \rightarrow R_3 - R_2​​​​​​​$, we have:
$\triangle=(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\1&\text{y}+\text{x}&\text{p}(\text{y}^2+\text{x}^2+\text{xy})\\0&\text{z}-\text{y}&\text{p}(\text{z}-\text{y})(\text{x+y+z})\end{vmatrix}$
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}-\text{y})\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\1&\text{y}+\text{x}&\text{p}(\text{y}^2+\text{x}^2+\text{xy})\\0&1&\text{p}(\text{x+y+z})\end{vmatrix}$
Expanding along $R_3​​​​​​​$, we have:
$\triangle = (x - y) (y - z) (z - x) [(-1) (p) (xy^2 + x^3 + x^2y) + 1 + px^3 + p(x + y + z) (xy)]$
$= (x - y) (y - z) (z - x) [-pxy^2 - px^3 - px^2y + 1 + px^3 + px^2y + pxy^2 + pxyz]$
$= (x - y) (y - z) (z - x) (1 + pxyz)$
Hence, the given result is proved.

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