Let M=begin{bmatrix} sin^4 theta & -1-sin^2 t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $M = \alpha I + \beta M^{-1}$. Multiplying by $M$,we get $M^2 = \alpha M + \beta I$,or $M^2 - \alpha M - \beta I = O$. By the Cayley-Hamilto

Vedclass · Accepted Answer

$-\frac{29}{16}$

Vedclass · Answer

(B) Given $M = \alpha I + \beta M^{-1}$. Multiplying by $M$,we get $M^2 = \alpha M + \beta I$,or $M^2 - \alpha M - \beta I = O$. By the Cayley-Hamilton theorem,$M^2 - 	ext{tr}(M)M + \det(M)I = O$. Comparing coefficients,$\alpha = 	ext{tr}(M) = \sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2\sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2}\sin^2(2	heta)$. Since $\sin^2(2	heta) \in [0, 1]$,the minimum value $\alpha^* = 1 - \frac{1}{2} = \frac{1}{2}$. Also,$-\beta = \det(M) = \sin^4 	heta \cos^4 	heta + (1+\cos^2 	heta)(1+\sin^2 	heta) = \frac{\sin^4(2	heta)}{16} + 1 + \sin^2 	heta + \cos^2 	heta + \sin^2 	heta \cos^2 	heta = \frac{\sin^4(2	heta)}{16} + 2 + \frac{\sin^2(2	heta)}{4}$. Let $t = \sin^2(2	heta) \in [0, 1]$. Then $-\beta(t) = \frac{t^2}{16} + \frac{t}{4} + 2$. To find the minimum of $\beta$,we find the maximum of $-\beta$. Since $f(t) = \frac{t^2}{16} + \frac{t}{4} + 2$ is increasing on $[0, 1]$,its maximum is at $t=1$,which is $\frac{1}{16} + \frac{4}{16} + \frac{32}{16} = \frac{37}{16}$. Thus,$\beta^* = -\frac{37}{16}$. Therefore,$\alpha^* + \beta^* = \frac{1}{2} - \frac{37}{16} = \frac{8-37}{16} = -\frac{29}{16}$.

List-$I$	List-$II$
$(A)$ If $A$ is a non-singular matrix of order $3$ and $\|A\|=a$,then $\|\text{adj}(A)\|=$	$(I)$ null matrix
$(B)$ $A$ is a non-singular matrix of order $3$ and $B$ is any matrix of order $3$ such that $AB=O$,then $B$ is	$(II)$ $a^2$
$(C)$ $\begin{vmatrix} 1 & x & x^2 \\ \cos(a-b)y & \cos ay & \cos(a+b)y \\ \sin(a-b)y & \sin ay & \sin(a+b)y \end{vmatrix}$ does not depend on	$(III)$ $b$
$(D)$ $A$ is a square matrix of order $3$ and $B=A-A^T$,then $B$ is	$(IV)$ $a$
	$(V)$ $0$

Similar Questions

Let $A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{bmatrix}$. The sum of the prime factors of $|P^{-1}AP - 2I|$ is equal to

If matrix $A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$,then:

Let $A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$,where $x, y$ and $z$ are real numbers such that $x + y + z > 0$ and $xyz = 2$. If $A^2 = I_3$,then the value of $x^3 + y^3 + z^3$ is ............

Vedclass Test Series

Exam Paper Generator

Online Exam Module