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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
✓

Answer

$\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y})\\\sin\text{x}\sin\text{y}&\cos\text{x}\sin\text{y}&\sin^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
[Applying $R_2 \rightarrow$ siny $R_2$ and $R_3 \rightarrow$ cosy $R_3$ ]
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}\sin\text{y}-\cos\text{x}\cos\text{y}&\cos\text{x}\sin\text{y}+\sin\text{x}\sin\text{y}&\sin^2\text{y}-\cos^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
$[$Applying $R_2→ R_2+ R_3]$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
$=0$

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