Question
Prove the following identities:
$\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$

Answer

$\text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 \rightarrow R_1 - R_2$
$=\begin{vmatrix}\text{y}&-\text{x}&\text{y}-\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 \rightarrow R_1 - R_3​​​​​​​$​​​​​​​
$=\begin{vmatrix}0&-2\text{x}&-2\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$=2\text{x}[\text{z}(\text{x}+\text{y})-\text{xy}]-2\text{x}[\text{zx}-\text{y}(\text{z}+\text{x})]$
$=2\text{x}[\text{zx}+\text{zy}-\text{xy}-\text{zx}+\text{yz}+\text{yx}]$
$=4\text{xyz}$

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