# True or False: The maximum value of $\begin{vmatrix}1 & 1 & 1\\1 &(1-\sin x) & 1\\1 & 1 & 1-cos x\end{vmatrix}\;is\;\Large \frac{1}{2}$.

$\begin{array}{1 1} True \\ False \end{array}$

## 1 Answer

Toolbox:
• Every square matrix can be associated to an expression or a number which is known as determinant.
• Determinant of matrix $A=\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}$
• $|A|=a_1\begin{vmatrix}b_2& c_2\\b_3 & c_3\end{vmatrix}-b_1\begin{vmatrix}a_2& c_2\\a_3 & c_3\end{vmatrix}+C_1\begin{vmatrix}a_2& b_2\\a_3 & b_3\end{vmatrix}$
Given $A=\begin{vmatrix}1 & 1 & 1\\1 &(1-sin x) & 1\\1 & 1 &1-cos x\end{vmatrix}$
Now let us apply $C_2\rightarrow C_2-C_3$ and
$A=\begin{vmatrix}1 & 0 & 0\\1&-sin x& 0\\1 & 0 &-cos x\end{vmatrix}$
Now expanding along $R_1$ we get,
$\;\;\;=1(sin x cos x-0)-0+0.$
The value of the determinant is
$|A|=sin x cos x$
But we know $sin 2x=2sin x cos x$
Therefore $sin x cos x=\frac{1}{2}sin 2x$
$sin x cos x$ can be written as $\frac{1}{2}sin 2x$
$|A|=\frac{1}{2}sin 2x$
When x=0,sin 0$\Rightarrow |A|=0$
When $x=\frac{\pi}{2};sin\frac{\pi}{2}=sin \pi$ again $sin\pi=0.$
$|A|=0.$
When $x=\frac{\pi}{4};sin2\frac{\pi}{4}=sin\frac{\pi}{2}=1$
|A|=$\frac{1}{2}.sin\big(\frac{\pi}{2}\big)=\frac{1}{2}\times 1=\frac{1}{2}$
Hence the maximum value it can take is $\frac{1}{2}$
So the statement is True.

answered Mar 27, 2013

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