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The value of determinant $A$, where $A$$=$\[\left( \begin{matrix}
   1 & \cos \theta & 0 \\
   -\cos \theta & 1 & \cos \theta \\
   -1 & -\cos \theta & 1 \\
\end{matrix} \right)\] lies:
(a) in the close interval $\left[ 1,2 \right]$
(b) in the close interval $\left[ 0,1 \right]$
(c) in the open interval $\left( 0,1 \right)$
(d) in the open interval $\left( 1,2 \right)$

Answer Verified Verified
Hint: Find the value of determinant $A$ by expanding the given determinant as per rule. The value will be obtained as a function of $\cos \theta $. Now, find the range of that function of $\theta $. The obtained range will be the answer.

Complete step-by-step answer:
In linear algebra, the determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformation described by the matrix. The determinant of a matrix $A$ is denoted as $\det \left( A \right)$, $\det A$ or $\left| A \right|$. Determinants can only be found if the given matrix is a square matrix. Square matrix is a type of matrix in which there are equal numbers of rows and columns. We already know how to expand a determinant. Consider a matrix of $3\times 3$ dimension: $\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$. Now, ${{a}_{11}},{{a}_{12}},{{a}_{13}},{{a}_{14}}......$ are its entries. To find its determinant value, we use the formula,$\det $$\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$$=$ \[{{a}_{11}}\times \det \left( \begin{matrix}
   {{a}_{22}} & {{a}_{23}} \\
   {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)-{{a}_{12}}\times \det \left( \begin{matrix}
   {{a}_{21}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{33}} \\
\end{matrix} \right)+{{a}_{13}}\times \det \left( \begin{matrix}
   {{a}_{21}} & {{a}_{22}} \\
   {{a}_{31}} & {{a}_{32}} \\
\end{matrix} \right)\]
Now, \[\det \left( \begin{matrix}
   {{a}_{22}} & {{a}_{23}} \\
   {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)={{a}_{22}}\times {{a}_{23}}-{{a}_{32}}\times {{a}_{23}}\].
Now, $\det $$A$$=$$\det $\[\left( \begin{matrix}
   1 & \cos \theta & 0 \\
   -\cos \theta & 1 & \cos \theta \\
   -1 & -\cos \theta & 1 \\
\end{matrix} \right)\]
                      \[\begin{align}
  & =1\times \det \left( \begin{matrix}
   1 & \cos \theta \\
   -\cos \theta & 1 \\
\end{matrix} \right)-\cos \theta \times \det \left( \begin{matrix}
   -\cos \theta & \cos \theta \\
   -1 & 1 \\
\end{matrix} \right)+0\times \det \left( \begin{matrix}
   -\cos \theta & 1 \\
   -1 & -\cos \theta \\
\end{matrix} \right) \\
 & =1\times \left\{ (1\times 1)-\cos \theta \times (-\cos \theta ) \right\}-\cos \theta \times \left\{ (-\cos \theta \times 1)-(-1\times \cos \theta ) \right\}+0\times \left\{ (-\cos \theta )\times (-\cos \theta )-(-1\times 1) \right\} \\
\end{align}\]$\begin{align}
  & =1+{{\cos }^{2}}\theta -\cos \theta (-\cos \theta +\cos \theta )+0 \\
 & =1+{{\cos }^{2}}\theta \\
\end{align}$
Now, we know that ${{\cos }^{2}}\theta $ has range from 0 to 1 in closed interval and therefore, $1+{{\cos }^{2}}\theta $ will have a range from 1 to 2 in closed interval.
Hence, option (a) is the correct answer.

Note: Since, the value of $\cos \theta $ lies in a closed interval of -1 and 1, therefore, value of ${{\cos }^{2}}\theta $ will have a maximum of 1 and minimum of 0 and when 1 is added to the minimum and maximum value of ${{\cos }^{2}}\theta $ the new minima and maxima of the function becomes 1 and 2 respectively. We can even convert the final answer into sine function and do the same application for range.

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