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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 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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