Question

Let the value of the determinant $\left| {\begin{array}{*{20}{c}}
  1&{\sin \alpha }&1 \\
  { - \sin \alpha }&1&{\sin \alpha } \\
  { - 1}&{ - \sin \alpha }&1
\end{array}} \right|$ be $\Delta $, then show that $\Delta $lies in the interval (2, 4).

Answer 1

Hint – In this question first find the value of determinant by applying basic row transformations like ${R_3} \to {R_3} + {R_1}$, then determinant will be obtained in terms of $\sin \alpha $. Use the concept that $\sin \alpha $ always lies between (-1, 1), so manipulate the determinant accordingly to get the required range of the determinant.

Complete step-by-step answer:
Given determinant
$\Delta = \left| {\begin{array}{*{20}{c}}
  1&{\sin \alpha }&1 \\
  { - \sin \alpha }&1&{\sin \alpha } \\
  { - 1}&{ - \sin \alpha }&1
\end{array}} \right|$
Now apply determinant rule i.e.
$ \Rightarrow {R_3} \to {R_3} + {R_1}$
$ \Rightarrow \Delta = \left| {\begin{array}{*{20}{c}}
  1&{\sin \alpha }&1 \\
  { - \sin \alpha }&1&{\sin \alpha } \\
  { - 1 + 1}&{ - \sin \alpha + \sin \alpha }&{1 + 1}
\end{array}} \right|$
$ \Rightarrow \Delta = \left| {\begin{array}{*{20}{c}}
  1&{\sin \alpha }&1 \\
  { - \sin \alpha }&1&{\sin \alpha } \\
  0&0&2
\end{array}} \right|$
Now expand the determinant we have,
$ \Rightarrow \Delta = 1\left| {\begin{array}{*{20}{c}}
  1&{\sin \alpha } \\
  0&2
\end{array}} \right| - \sin \alpha \left| {\begin{array}{*{20}{c}}
  { - \sin \alpha }&{\sin \alpha } \\
  0&2
\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}
  { - \sin \alpha }&1 \\
  0&0
\end{array}} \right|$
$ \Rightarrow \Delta = 1\left( {2 - 0} \right) - \sin \alpha \left( { - 2\sin \alpha - 0} \right) + 1\left( {0 - 0} \right)$
\[ \Rightarrow \Delta = 2 + 2{\sin ^2}\alpha \]................ (1)
Now as we know that $\sin \alpha $ always lie between (-1, 1)
Therefore, $ - 1 < \sin \alpha < 1$
So ${\sin ^2}\alpha $lie between (0, 1) because after squaring all negative quantities becomes positive,
$ \Rightarrow 0 < {\sin ^2}\alpha < 1$
Now multiply by 2 we have,
$ \Rightarrow 0 < 2{\sin ^2}\alpha < 2$
Now add by 2 we have,
$ \Rightarrow 2 < 2 + 2{\sin ^2}\alpha < 4$
Now from equation (1) we have,
$ \Rightarrow 2 < \Delta < 4$
Therefore lies in the interval (2, 4)
Hence proved.

Note – The determinant in the initial stages of the solution could have been solved directly even but however the main aim of applying the row transformation or even the column transformations in some of the problems of this kind is to get maximum number of zeroes inside the determinant as it helps simplification of the determinant.