Matrix

The truth table for the logical expression (p ∧ q) → (p → q) is as follows:

The final column shows that the expression is a tautology, meaning it is always true regardless of the truth values of p and q.

A² = (cos2θ isin2θ cos2θ)
Similarly, A? = (cos5θ isin5θ cos5θ) = (a b; c d)
(1) a²+b² = cos²5θ - sin²5θ = cos10θ = cos75°
(2) a²-d² = cos²5θ - cos²5θ = 0
(3) a²-b² = cos²5θ + sin²5θ = 1
(4) a²-c² = cos²5θ + sin²5θ = 1

adj A| = |A|² = 9
=> |A| = ±3 => λ = |λ| = 3
=> |B| = |adj A|² = 81
=> | (B? ¹)? | = |B? ¹| = |B|? ¹ = 1/|B| = 1/81 = µ

2? -2? =112. m=7, n=4. mn=28.

$\mathrm{}\left|B\right|=\left|\begin{array}{lll}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right|=\left|\begin{array}{lll}{3}^0{a}_{11}& 3^1{a}_{21}& 3^2{a}_{31}\\ 3^1{a}_{12}& 3^2{a}_{22}& 3^3{a}_{32}\\ 3^2{a}_{13}& 3^3{a}_{23}& 3^4{a}_{33}\end{array}\right|$

$81=3^3\cdot 3^3\cdot 3^2\left|\mathrm{A}\right|$

$\Rightarrow \left|A\right|=\frac{1}{9}$

Kindly go through the solution 

  $e^{\left(co{s}^{2}x+co{s}^{4}x+co{s}^{6}x+.....\infty \right)}lo{g}_{e}2$

$=e^{\frac{co{s}^{2}x}{1-co{s}^{2}x}lo{g}_{e}2}=e^{co{t}^{2}xlo{g}_{e}2}=2^{co{t}^{2}x}$            

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t=2^{co{t}^{2}x}=8=2^3$        

$\Rightarrow co{t}^{2}x=3=co{t}^2\frac{\pi }{6}$       

$\frac{2sinx}{sinx+\sqrt[]{3}cosx}=\frac{2*\frac{1}{2}}{\frac{1}{2}+\sqrt[]{3}*\frac{\sqrt[]{3}}{2}}=\frac{1}{2}$

$P=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right]andQ=\left[q_{ij}\right]\Rightarrow PQ=k{l}_3$           

$q_{23}=-\frac{k}{8}and\left|Q\right|=\frac{k^2}{2}$            

  $PQ=k{l}_3\Rightarrow P^{-1}=\frac{Q}{k}={\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right)}^{-1}$

$\left|p\right|\left|Q\right|=\left(k{l}_3\right)\Rightarrow 8.\frac{k^2}{2}=k^3$

$k\ne 0\Rightarrow k=4$

$\therefore {\alpha }^2+k^2=1+16=17.$