If A = begin{bmatrix} a+ib & c+id -c+id & a- | vedclass

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

Question

(A) Given the matrix $A = \begin{bmatrix} a+ib & c+id \ -c+id & a-ib \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = (a+ib)(a-ib) - (c

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the matrix $A = \begin{bmatrix} a+ib & c+id \ -c+id & a-ib \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = (a+ib)(a-ib) - (c+id)(-c+id) = (a^2+b^2) - (-(c^2+d^2)) = a^2+b^2+c^2+d^2$.We know that $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.The adjoint of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:$	ext{adj}(A) = \begin{bmatrix} a-ib & -c-id \ c-id & a+ib \end{bmatrix}$.Thus,$A^{-1} = \frac{1}{a^2+b^2+c^2+d^2} \begin{bmatrix} a-ib & -c-id \ c-id & a+ib \end{bmatrix}$.Comparing this with the given $A^{-1} = \begin{bmatrix} a+ib & -c-id \ -c+id & a-ib \end{bmatrix}$,we observe that for the equality to hold,the determinant $|A|$ must be $1$.Therefore,$a^2+b^2+c^2+d^2 = 1$.

If $A = \begin{bmatrix} a+ib & c+id \\ -c+id & a-ib \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} a+ib & -c-id \\ -c+id & a-ib \end{bmatrix}$,find $(a^2+b^2+c^2+d^2)$.

Similar Questions

If $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$,verify that $A^{3}-6 A^{2}+9 A-4 I=0$ and hence find $A^{-1}$.

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 9 \\ 1 & 8 & 27 \end{bmatrix}$,then the value of $|adj\, A|$ is

Let $A$ be a matrix of order $3 \times 3$ and $\det(A) = 2$. Then $\det(\det(A) \cdot \operatorname{adj}(5 \operatorname{adj}(A^3)))$ is equal to:

Matrix $A = \begin{bmatrix} 1 & 0 & -k \\ 2 & 1 & 3 \\ k & 0 & 1 \end{bmatrix}$ is invertible for

If $A$ is a non-singular matrix,then $A(\text{adj } A) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module