If d is the determinant of a square matrix A | vedclass

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

Question

(B) We know that for any square matrix $A$ of order $n$,the relationship between the matrix and its adjoint is given by $A(	ext{adj } A) = |A| I_n$,w

Vedclass · Accepted Answer

$d^{n-1}$

Vedclass · Answer

(B) We know that for any square matrix $A$ of order $n$,the relationship between the matrix and its adjoint is given by $A(	ext{adj } A) = |A| I_n$,where $I_n$ is the identity matrix of order $n$.Taking the determinant on both sides,we get $|A(	ext{adj } A)| = ||A| I_n|$.Using the property $|AB| = |A||B|$,we have $|A| |	ext{adj } A| = |A|^n |I_n|$.Since $|I_n| = 1$,this simplifies to $|A| |	ext{adj } A| = |A|^n$.Given that $|A| = d$,we have $d |	ext{adj } A| = d^n$.Therefore,$|	ext{adj } A| = d^{n-1}$.

If $d$ is the determinant of a square matrix $A$ of order $n$,then the determinant of its adjoint is

Similar Questions

If $A$ is an invertible matrix of order $2$,then $\det(A^{-1})$ is equal to

Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A ))|=81$. If $S =\{ n \in \mathbb{Z} :(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^2}{2}}=|A|^{(3n^2-5n-4)}\}$,then $\sum_{n \in S}|A^{(n^2+n)}|$ is equal to

If $A = \begin{bmatrix} -2 & 6 \\ -5 & 7 \end{bmatrix}$,then find $adj(A)$.

Find the inverse of the matrix $A = \left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$,if it exists.

If $A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]$ and $A^{-1}=x A+y I$,where $I$ is the unit matrix of order $2$,then the values of $x$ and $y$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module