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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Prove the following identities:
$\begin{vmatrix}2\text{y}&\text{y}-\text{z}-\text{x}&2\text{y}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})^3$
✓

Answer

$\text{L.H.S}=\begin{vmatrix}2\text{y}&\text{y}-\text{z}-\text{x}&2\text{y}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$ $[R_1 = R_1 + R_2 + R_3]$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&1&1\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&0&0\\2\text{z}&0&-\text{x}-\text{y}-\text{z}\\\text{x}-\text{y}-\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\end{vmatrix}$ $[C_2 = C_2 - C_1, C_3 = C_3 - C_1]$
$=(\text{x}+\text{y}+\text{z})\big[1\{0+(\text{x}+\text{y}+\text{z})(\text{x}+\text{y}+\text{z})\}\big]$
$=(\text{x}+\text{y}+\text{z})^3$
$=\text{R.H.S}$
Hence proved.

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