Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Determinants
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
  • 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 $sin\pi=0.$
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 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App