If A is an invertible matrix of order n,then | vedclass

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

Question

(C) We know that for any square matrix $A$ of order $n$,the property of the adjoint matrix is given by $A(\operatorname{adj} A) = |A|I_n$,where $I_n$

Vedclass · Accepted Answer

$|A|^{n-1}$

Vedclass · Answer

(C) We know that for any square matrix $A$ of order $n$,the property of the adjoint matrix is given by $A(\operatorname{adj} A) = |A|I_n$,where $I_n$ is the identity matrix of order $n$. Taking the determinant on both sides,we get $|A(\operatorname{adj} A)| = ||A|I_n|$. Using the property $|AB| = |A||B|$ and $|kA| = k^n|A|$ for a matrix of order $n$,we have $|A| \cdot |\operatorname{adj} A| = |A|^n \cdot |I_n|$. Since $|I_n| = 1$,we have $|A| \cdot |\operatorname{adj} A| = |A|^n$. Since $A$ is invertible,$|A| 
eq 0$,so we can divide by $|A|$ to get $|\operatorname{adj} A| = |A|^{n-1}$.

If $A$ is an invertible matrix of order $n$,then the determinant of $\operatorname{adj} A$ is equal to :

Similar Questions

Which one of the following graphs represents the Freundlich adsorption isotherm?

At $25^{\circ} C$,the molar conductances at infinite dilution for the strong electrolytes $NaOH$,$NaCl$ and $BaCl_2$ are $248 \times 10^{-4}$,$126 \times 10^{-4}$ and $280 \times 10^{-4} \ S \ m^2 \ mol^{-1}$ respectively. The value of $\lambda_m^{\circ} Ba(OH)_2$ in $S \ m^2 \ mol^{-1}$ is:

$A$ vertical pole subtends an angle $\tan ^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the angles subtended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$,then $(\tan \alpha, \tan \beta)$ is equal to

Define environmental chemistry.

Four convergent lenses have focal lengths $100 \, cm, 10 \, cm, 4 \, cm$ and $0.3 \, cm$. For a telescope with maximum possible magnification,we choose the lenses of focal lengths:

Vedclass Test Series

Exam Paper Generator

Online Exam Module