If a, b, c and d are real numbers such that a | vedclass

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

Question

(C) Given that $a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+i b & c+i d \ -c+i d & a-i b\end{array}ight]$.First,we calculate the determinant

Vedclass · Accepted Answer

$\left[\begin{array}{cc}a-i b & -c-i d \ c-i d & a+i b\end{array}ight]$

Vedclass · Answer

(C) Given that $a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+i b & c+i d \ -c+i d & a-i b\end{array}ight]$.First,we calculate the determinant of $A$,denoted as $|A|$.$|A| = (a+i b)(a-i b) - (c+i d)(-c+i d)$$|A| = (a^2 - (i b)^2) - ((i d)^2 - c^2)$$|A| = (a^2 + b^2) - (-d^2 - c^2)$$|A| = a^2 + b^2 + c^2 + d^2 = 1$.The inverse of a $2 	imes 2$ matrix $A = \left[\begin{array}{cc}x & y \ z & w\end{array}ight]$ is given by $A^{-1} = \frac{1}{|A|} \left[\begin{array}{cc}w & -y \ -z & x\end{array}ight]$.Substituting the values of $A$ and $|A|=1$,we get:$A^{-1} = \frac{1}{1} \left[\begin{array}{cc}a-i b & -(c+i d) \ -(-c+i d) & a+i b\end{array}ight]$$A^{-1} = \left[\begin{array}{cc}a-i b & -c-i d \ c-i d & a+i b\end{array}ight]$.

If $a, b, c$ and $d$ are real numbers such that $a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right]$,then $A^{-1}$ is equal to

Similar Questions

The number of nodal planes in a $p_x$ orbital is

$A$ wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of $15^{\circ}$ in time $t \ s$. The increase in angle through which it rotates in the next $2t \ s$ is (in $^{\circ}$)

$A$ ball of mass $M$ falls from a height $h$ on a floor for which the coefficient of restitution is $e$. The height attained by the ball after two rebounds is

Which pair of substances gives the same gaseous product when they react with water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module