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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
If $\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0,$ then using properties of determinants, find the value of $\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}},$ where $\text{x},\text{y},\text{z}\neq0.$
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Answer

$\Rightarrow\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
$R_1 → R_1 - R_2$
$\Rightarrow\begin{vmatrix}\text{x}&-\text{y}&0\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
$R_2→ R_2 - R_3$​​​​​​​
$\Rightarrow\begin{vmatrix}\text{x}&-\text{y}&0\\0&\text{y}&-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
Expanding along first row, we get
$\text{x}(\text{yc}+\text{zb}-\text{zy})+\text{y}(0-\text{za}+\text{zx})=0$
$\Rightarrow\text{xyc}+\text{xzb}-2\text{xyz}+\text{zya}=0$
$\Rightarrow\text{xyc}+\text{xzb}-2\text{xyz}+\text{zya}=0$
Dividing by $xyz$, we get
$\frac{\text{c}}{\text{z}}+\frac{\text{b}}{\text{y}}-2+\frac{\text{a}}{\text{x}}=0$
$\therefore\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}}=2$

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