If x& & - &-x&1 &1&x =8, write the value of x. - Vidyadip

Question

If $\begin{vmatrix}\text{x}&\sin\theta&\cos\theta\\-\sin\theta&-\text{x}&1\\\cos\theta&1&\text{x} \end{vmatrix}=8,$ write the value of $x$.

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Answer

$\begin{vmatrix}\text{x}&\sin\theta&\cos\theta\\-\sin\theta&-\text{x}&1\\\cos\theta&1&\text{x} \end{vmatrix}=8$
Expanding along $R_1$ we get
$\text{x}(-\text{x}^2-1)-\sin\theta(-\text{x}\sin\theta-\cos\theta)\\+\cos\theta(-\sin\theta+\text{x}\cos\theta)=8$
$\Rightarrow-\text{x}^3-\text{x}+\text{x}\sin^2\theta+\sin\theta\cos\theta\\-\sin\theta\cos\theta+\text{x}\cos^2\theta=8$
$\Rightarrow-\text{x}^3-\text{x}+\text{x}(\sin^2\theta+\cos^2\theta)=8$
$\Rightarrow-\text{x}^3-\text{x}+\text{x}=8$
$\Rightarrow\text{x}^3+8=0$
$\Rightarrow(\text{x}+2)(\text{x}^2-2\text{x}+4)=0$
$\Rightarrow\text{x}+2=0$ $[\because\text{x}^2-2\text{x}+4>0\ \forall\text{ x}]$
$\Rightarrow\text{x}=-2$

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