Question

Let $M=\left[ \begin{matrix} {{\sin }^{4}}\theta & -1-{{\sin }^{2}}\theta \\\ 1+{{\cos }^{2}}\theta & {{\cos }^{4}}\theta \\\ \end{matrix} \right]=\alpha I+\beta {{M}^{-1}}$ where $\alpha =\alpha \left( \theta \right)$ and $\beta =\beta \left( \theta \right)$ are real numbers and I is the $2\times 2$ identity matrix. If ${{\alpha }^{*}}=$ is the minimum of the set \left\\{ \alpha \left( \theta \right):\theta \in \left[ 0,2\pi \right] \right\\} and ${{\beta }^{*}}=$ is the minimum of the set \left\\{ \beta \left( \theta \right):\theta \in \left[ 0,2\pi \right] \right\\}
Then the value of ${{\alpha }^{*}}+{{\beta }^{*}}$ is:
(a) $-\dfrac{37}{16}$
(b) $-\dfrac{29}{16}$
(c) $-\dfrac{31}{16}$
(d) $-\dfrac{17}{16}$

Answer 1

We have given M in such a way that $M=\left[ \begin{matrix} {{\sin }^{4}}\theta & -1-{{\sin }^{2}}\theta \\\ 1+{{\cos }^{2}}\theta & {{\cos }^{4}}\theta \\\ \end{matrix} \right]=\alpha I+\beta {{M}^{-1}}$ and we have to find the value of $\alpha \And \beta$. For that, we are going to find the inverse of M. We know that inverse of a matrix M is equal to $\dfrac{adj\left( M \right)}{\left| M \right|}$. In this formula, adj (M) is the adjoint of M which is calculated by taking the cofactor of matrix M then take the transpose of the matrix and $\left| M \right|$ is the determinant of M. After calculating the inverse of M, substitute in the matrix given and also substitute the identity matrix and then play with the equation to see, what values of $\alpha \And \beta$ will satisfy the equation. After that take the minimum value of $\alpha \And \beta$ individually and then add them to get the required answer.

Complete step by step answer:
We have given the matrix M as follows:
$M=\left[ \begin{matrix} {{\sin }^{4}}\theta & -1-{{\sin }^{2}}\theta \\\ 1+{{\cos }^{2}}\theta & {{\cos }^{4}}\theta \\\ \end{matrix} \right]=\alpha I+\beta {{M}^{-1}}$
To solve the above equation, we need the value of inverse of matrix M i.e. ${{M}^{-1}}$ so we are going to find the inverse of the matrix M.
We know the inverse of a matrix M is equal to:
$\dfrac{adj\left( M \right)}{\left| M \right|}$
Now, we are going to find the adj (M) or adjoint of the matrix M by first of all taking the co – factor of M and then take the transpose of that matrix.
Finding the co – factor of matrix M as follows:
The cofactor of matrix M contains cofactor of first row and first column, first row and second column likewise you can write all the position of the elements. In the below, we are showing the cofactor at the position of first row and first column.
The cofactor of the element lies at ${{i}^{th}}$ row and ${{j}^{th}}$ column is given below:
${{C}_{ij}}={{\left( -1 \right)}^{i+j}}\left( {{M}_{ij}} \right)$
In the above equation, ${{M}_{ij}}$ represents the minor corresponding to the position of i and j of the matrix.
The matrix M given as:
$\left[ \begin{matrix} {{\sin }^{4}}\theta & -1-{{\sin }^{2}}\theta \\\ 1+{{\cos }^{2}}\theta & {{\cos }^{4}}\theta \\\ \end{matrix} \right]$
Finding the cofactor of the position where ${{\sin }^{4}}\theta$ is placed so value of i and j are both equal to 1 we get,
$\begin{aligned} & {{C}_{11}}={{\left( -1 \right)}^{1+1}}\left( {{M}_{11}} \right) \\\ & \Rightarrow {{C}_{11}}={{\left( -1 \right)}^{2}}\left( {{M}_{11}} \right) \\\ & \Rightarrow {{C}_{11}}=1\left( {{M}_{11}} \right) \\\ \end{aligned}$
Now, we have to find the minor of the position (i, j) is equal to 1 which we have shown below:
Hide the elements of the first row and first column which we have shown below by red color and then the element remaining is the minor corresponding to the first row and first column.
$\left[ \begin{matrix} \color {red} {{\sin }^{4}} \color {red} \theta & \color {red} -\color {red} 1\color {red} -\color {red} {{\sin }^{2}} \color {red} \theta \\\ \color {red} 1\color {red} +\color {red} {{\cos }^{2}} \color {red} \theta & {{\cos }^{4}}\theta \\\ \end{matrix} \right]$
Hence, the value of ${{M}_{11}}={{\cos }^{4}}\theta$
Similarly, we can find the other minors also so all the minors are:
$\begin{aligned} & {{M}_{12}}=1+{{\cos }^{2}}\theta \\\ & {{M}_{21}}=-1-{{\sin }^{2}}\theta \\\ & {{M}_{22}}={{\sin }^{4}}\theta \\\ \end{aligned}$
Now, using these minors, we can find the co factors as follows:
$\begin{aligned} & {{C}_{11}}={{M}_{11}} \\\ & \Rightarrow {{C}_{11}}={{\cos }^{4}}\theta \\\ \end{aligned}$
$\begin{aligned} & {{C}_{12}}={{\left( -1 \right)}^{1+2}}\left( {{M}_{12}} \right) \\\ & \Rightarrow {{C}_{12}}=-1\left( 1+{{\cos }^{2}}\theta \right) \\\ \end{aligned}$
$\begin{aligned} & {{C}_{21}}={{\left( -1 \right)}^{2+1}}\left( {{M}_{21}} \right) \\\ & \Rightarrow {{C}_{21}}=-1\left( -1-{{\sin }^{2}}\theta \right) \\\ \end{aligned}$
$\begin{aligned} & {{C}_{22}}={{\left( -1 \right)}^{2+2}}\left( {{M}_{22}} \right) \\\ & \Rightarrow {{C}_{22}}=1\left( {{\sin }^{4}}\theta \right) \\\ \end{aligned}$
Now, writing these cofactors in the matrix we get,
$\begin{aligned} & \left[ \begin{matrix} {{C}_{11}} & {{C}_{12}} \\\ {{C}_{21}} & {{C}_{22}} \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} {{\cos }^{4}}\theta & -1-{{\cos }^{2}}\theta \\\ 1+{{\sin }^{2}}\theta & {{\sin }^{4}}\theta \\\ \end{matrix} \right] \\\ \end{aligned}$
Taking the transpose of the above matrix by interchanging rows with columns we get,
$\left[ \begin{matrix} {{\cos }^{4}}\theta & 1+{{\sin }^{2}}\theta \\\ -1-{{\cos }^{2}}\theta & {{\sin }^{4}}\theta \\\ \end{matrix} \right]$
Hence, we got the adjoint of the matrix M as:
$adj\left( M \right)=\left[ \begin{matrix} {{\cos }^{4}}\theta & 1+{{\sin }^{2}}\theta \\\ -1-{{\cos }^{2}}\theta & {{\sin }^{4}}\theta \\\ \end{matrix} \right]$
Determinant of the matrix M is calculated as follows:
Matrix M is shown below;
$\left[ \begin{matrix} {{\sin }^{4}}\theta & -1-{{\sin }^{2}}\theta \\\ 1+{{\cos }^{2}}\theta & {{\cos }^{4}}\theta \\\ \end{matrix} \right]$
Taking determinant of the matrix we get,
$\begin{aligned} & {{\sin }^{4}}\theta {{\cos }^{4}}\theta -\left( 1+{{\cos }^{2}}\theta \right)\left( -1 \right)\left( 1+{{\sin }^{2}}\theta \right) \\\ & ={{\sin }^{4}}\theta {{\cos }^{4}}\theta +\left( 1+{{\cos }^{2}}\theta \right)\left( 1+{{\sin }^{2}}\theta \right) \\\ & ={{\sin }^{4}}\theta {{\cos }^{4}}\theta +1+{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\\ \end{aligned}$
We know that there is a trigonometric identity that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ so using this relation in the above we get,
$\begin{aligned} & 1+1+{{\sin }^{4}}\theta {{\cos }^{4}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\\ & =2+{{\sin }^{4}}\theta {{\cos }^{4}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\\ \end{aligned}$
Hence, we got the determinant value of matrix M as:
$\left| M \right|=2+{{\sin }^{4}}\theta {{\cos }^{4}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta$
Now, the inverse of matrix M is equal to:
$\dfrac{\left[ \begin{matrix} {{\cos }^{4}}\theta & 1+{{\sin }^{2}}\theta \\\ -1-{{\cos }^{2}}\theta & {{\sin }^{4}}\theta \\\ \end{matrix} \right]}{\left| M \right|}$
Multiplying $\left| M \right|$ inside the matrix we get,
$\left[ \begin{matrix} \dfrac{{{\cos }^{4}}\theta }{\left| M \right|} & \dfrac{1+{{\sin }^{2}}\theta }{\left| M \right|} \\\ \dfrac{-1-{{\cos }^{2}}\theta }{\left| M \right|} & \dfrac{{{\sin }^{4}}\theta }{\left| M \right|} \\\ \end{matrix} \right]$
Substituting this inverse of M in the given matrix equation we get,
$\left[ \begin{matrix} {{\sin }^{4}}\theta & -1-{{\sin }^{2}}\theta \\\ 1+{{\cos }^{2}}\theta & {{\cos }^{4}}\theta \\\ \end{matrix} \right]=\alpha \left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]+\beta \left[ \begin{matrix} \dfrac{{{\cos }^{4}}\theta }{\left| M \right|} & \dfrac{1+{{\sin }^{2}}\theta }{\left| M \right|} \\\ \dfrac{-1-{{\cos }^{2}}\theta }{\left| M \right|} & \dfrac{{{\sin }^{4}}\theta }{\left| M \right|} \\\ \end{matrix} \right]$
Solving the right hand side of the above equation we get,
$\left[ \begin{matrix} \alpha & 0 \\\ 0 & \alpha \\\ \end{matrix} \right]+\left[ \begin{matrix} \dfrac{\beta {{\cos }^{4}}\theta }{\left| M \right|} & \dfrac{\beta \left( 1+{{\sin }^{2}}\theta \right)}{\left| M \right|} \\\ \dfrac{\beta \left( -1-{{\cos }^{2}}\theta \right)}{\left| M \right|} & \dfrac{\beta {{\sin }^{4}}\theta }{\left| M \right|} \\\ \end{matrix} \right]$
Adding the above two matrices by adding the elements of first matrix with the corresponding row and column of the second matrix we get,
$\left[ \begin{matrix} \alpha +\dfrac{\beta {{\cos }^{4}}\theta }{\left| M \right|} & \dfrac{\beta \left( 1+{{\sin }^{2}}\theta \right)}{\left| M \right|} \\\ \dfrac{\beta \left( -1-{{\cos }^{2}}\theta \right)}{\left| M \right|} & \alpha +\dfrac{\beta {{\sin }^{4}}\theta }{\left| M \right|} \\\ \end{matrix} \right]$
Now, equating the above matrix to the left hand side of the above equation we get,
$\left[ \begin{matrix} {{\sin }^{4}}\theta & -1-{{\sin }^{2}}\theta \\\ 1+{{\cos }^{2}}\theta & {{\cos }^{4}}\theta \\\ \end{matrix} \right]=\left[ \begin{matrix} \alpha +\dfrac{\beta {{\cos }^{4}}\theta }{\left| M \right|} & \dfrac{\beta \left( 1+{{\sin }^{2}}\theta \right)}{\left| M \right|} \\\ \dfrac{\beta \left( -1-{{\cos }^{2}}\theta \right)}{\left| M \right|} & \alpha +\dfrac{\beta {{\sin }^{4}}\theta }{\left| M \right|} \\\ \end{matrix} \right]$
Above two matrices are equal when each of the elements of matrix on the left hand side is equal to the corresponding elements of the matrix on the right hand side of the above equation we get,
$\begin{aligned} & -\left( 1+{{\sin }^{2}}\theta \right)=\dfrac{\beta \left( 1+{{\sin }^{2}}\theta \right)}{\left| M \right|} \\\ & \Rightarrow -1=\dfrac{\beta }{\left| M \right|} \\\ & \Rightarrow \beta =-1\left| M \right| \\\ \end{aligned}$
Substituting the value of $\left| M \right|$ in the above equation we get,
$\beta =-\left( 2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\sin }^{4}}\theta {{\cos }^{4}}\theta \right)$
Hence, we got the value of $\beta$ as $-\left( 2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\sin }^{4}}\theta {{\cos }^{4}}\theta \right)$
${{\sin }^{4}}\theta =\alpha +\dfrac{\beta {{\cos }^{4}}\theta }{\left| M \right|}$
Substituting the value of $\beta$ that we have solved above in the above equation we get,

& {{\sin }^{4}}\theta =\alpha +\dfrac{-\left| M \right|{{\cos }^{4}}\theta }{\left| M \right|} \\\ & \Rightarrow {{\sin }^{4}}\theta =\alpha -{{\cos }^{4}}\theta \\\ & \Rightarrow {{\sin }^{4}}\theta +{{\cos }^{4}}\theta =\alpha \\\ \end{aligned}$$ Making the L.H.S of the above equation as the perfect square we get, $\begin{aligned} & {{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta =\alpha \\\ & \Rightarrow 1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta =\alpha \\\ \end{aligned}$ Hence, we got the value of $\alpha \And \beta $ as follows: $\begin{aligned} & \alpha =1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\\ & \beta =-\left( 2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\sin }^{4}}\theta {{\cos }^{4}}\theta \right) \\\ \end{aligned}$ Now, we have to find the minimum value of $\alpha \And \beta $ which is denoted as ${{\alpha }^{*}}\And {{\beta }^{*}}$ as follows: ${{\alpha }^{*}}=$ Minimum of $\alpha $ ${{\beta }^{*}}=$ Minimum of $\beta $ Finding the minimum value of $\alpha $ we get, $\begin{aligned} & \alpha =1-2\dfrac{{{\sin }^{2}}2\theta }{4} \\\ & \Rightarrow \alpha =1-\dfrac{{{\sin }^{2}}2\theta }{2} \\\ \end{aligned}$ Now, the minimum value of the above expression occurs when ${{\sin }^{2}}2\theta $ is maximum and we know that maximum value of $\sin 2\theta $ is 1 so the value of its square is also 1 substituting this maximum value of ${{\sin }^{2}}2\theta $ in the above equation we get, $\begin{aligned} & \alpha =1-\dfrac{1}{2} \\\ & \Rightarrow \alpha =\dfrac{1}{2} \\\ \end{aligned}$ Hence, we got the value of ${{\alpha }^{*}}=\dfrac{1}{2}$ Finding the minimum value of $\beta $ we get, $\beta =-\left( 2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\sin }^{4}}\theta {{\cos }^{4}}\theta \right)$ Substituting ${{\sin }^{2}}\theta {{\cos }^{2}}\theta =t$ in the above equation we get, $\beta =-\left( 2+t+{{t}^{2}} \right)$ The above equation will give minimum value when $2+t+{{t}^{2}}$ is having maximum value. Now, we are going to find the maximum value of $2+t+{{t}^{2}}$ as follows: $y=2+t+{{t}^{2}}$ As we have taken ${{\sin }^{2}}\theta {{\cos }^{2}}\theta =t$ so rearranging the L.H.S of this equation we get, $\dfrac{{{\sin }^{2}}2\theta }{4}$ Now, the maximum value of this expression is $\dfrac{1}{4}$ which is found when $\sin 2\theta =1$ and so with the similar reason the minimum value of the above expression is 0. And we need the maximum value of $2+t+{{t}^{2}}$ which is possible when $t=\dfrac{1}{4}$. Substituting this value of t in $\beta =-\left( 2+t+{{t}^{2}} \right)$ then we get, $\begin{aligned} & \beta =-\left( 2+\dfrac{1}{4}+{{\left( \dfrac{1}{4} \right)}^{2}} \right) \\\ & \Rightarrow \beta =-\left( 2+\dfrac{1}{4}+\dfrac{1}{16} \right) \\\ \end{aligned}$ $\Rightarrow \beta =-\left( \dfrac{32+4+1}{16} \right)$ $\Rightarrow \beta =-\left( \dfrac{37}{16} \right)$ Hence, we got the minimum value of $\beta =-\dfrac{37}{16}$ or ${{\beta }^{*}}=-\left( \dfrac{37}{16} \right)$. Now, adding ${{\alpha }^{*}}\And {{\beta }^{*}}$ we get, $\begin{aligned} & {{\alpha }^{*}}+{{\beta }^{*}}=\dfrac{1}{2}-\dfrac{37}{16} \\\ & \Rightarrow {{\alpha }^{*}}+{{\beta }^{*}}=\dfrac{8-37}{16}=\dfrac{-29}{16} \\\ \end{aligned}$ **So, the correct answer is “Option B”.** **Note:** The most possible mistake that could happen in this problem is the calculation mistake. While finding the cofactor of the matrix, there is a high chance of making calculation mistakes so be careful in your calculations. While finding the minimum value of $\beta $ which we have shown above as: $\beta =-\left( 2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\sin }^{4}}\theta {{\cos }^{4}}\theta \right)$ Do not jump to put the value of $\theta =\dfrac{\pi }{2}$ directly into the above equation. This is also a very common mistake that students make. In such type of problems, try to reduce this expression as much as you can like you can write $\sin \theta \cos \theta =\dfrac{\sin 2\theta }{2}$ and then put the values of $\theta $ which can give the minimum value.