If |operatorname{Adj} A|=x and |operatorname{ | vedclass

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

Question

(C) We know that for any invertible matrix $M$, $|M^{-1}| = \frac{1}{|M|}$.Therefore, $|(\operatorname{Adj}(AB))^{-1}| = \frac{1}{|\operatorname{Adj}(

Vedclass · Accepted Answer

$\frac{1}{x y}$

Vedclass · Answer

(C) We know that for any invertible matrix $M$, $|M^{-1}| = \frac{1}{|M|}$.Therefore, $|(\operatorname{Adj}(AB))^{-1}| = \frac{1}{|\operatorname{Adj}(AB)|}$.Using the property $\operatorname{Adj}(AB) = (\operatorname{Adj} B)(\operatorname{Adj} A)$, we have:$|\operatorname{Adj}(AB)| = |\operatorname{Adj} B| \cdot |\operatorname{Adj} A| = y \cdot x = xy$.Substituting this back into the expression:$|(\operatorname{Adj}(AB))^{-1}| = \frac{1}{xy}$.

If $|\operatorname{Adj} A|=x$ and $|\operatorname{Adj} B|=y$, then $\left|(\operatorname{Adj}(AB))^{-1}\right|=$

Similar Questions

Obtain the inverse of the following matrix using elementary row operations: $A = \left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$

If a matrix $A$ satisfies the equation $A^3-6A^2+11A-6I=0$,then $A^{-1}$ can be expressed in terms of $A$ as:

If $\text{adj} \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & m & -2 \\ 1 & 1 & 0 \\ -2 & -2 & n \end{bmatrix}$, then $m+n=$

If for the matrix $A$,${A^3} = I$,then ${A^{-1}} = $

If $A$ is a matrix of order $3$ whose determinant is equal to $6$,then $\operatorname{det}(\operatorname{adj} A) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module