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The maximum value of  \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cos \theta \end{array}\right| is 1/2.

Answers (1)

\\ \begin{array}{l} \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cos \theta \end{array}\right| \\ \text { Apply, } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \\ \Rightarrow \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & \sin \theta & 0 \\ 0 & 0 & \cos \theta \end{array}\right| \end{array}

 

\\ \vspace{\baselineskip}= cos \theta . sin \theta \\ \vspace{\baselineskip}\text{Multiply and divide by 2},\\ \vspace{\baselineskip} = 1/2 (2sin \theta cos \theta )\\ \vspace{\baselineskip} \text{We know}, 2 sin \theta cos \theta = sin 2 \theta \\ \\ \vspace{\baselineskip}= 1/2 (sin 2 \theta )\\ \\

 

Since the maximum value of  sin 2 \theta \ is \ 1, \theta = 45 ^{\circ} .\\

\\ \\ \vspace{\baselineskip} \therefore \Delta = 1/2 (sin 2(45 ^{\circ} ))\\ \\ \vspace{\baselineskip}= 1/2 sin 90 ^{\circ} \\ \\ \vspace{\baselineskip}= 1/2 (1)\\ \\ \vspace{\baselineskip} \therefore \Delta = 1/2\\ \vspace{\baselineskip}
 

Thus the given statement is true.

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