If B is the inverse of a third order matrix A | vedclass

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

Question

(C) Given that $A$ is a matrix of order $n = 3$ and $B = A^{-1}$.We know that $\det B = \det(A^{-1}) = \frac{1}{\det A} = k$,so $\det A = \frac{1}{k}$

Vedclass · Accepted Answer

$k B$

Vedclass · Answer

(C) Given that $A$ is a matrix of order $n = 3$ and $B = A^{-1}$.We know that $\det B = \det(A^{-1}) = \frac{1}{\det A} = k$,so $\det A = \frac{1}{k}$.The property for the adjoint of an adjoint matrix is $\operatorname{adj}(\operatorname{adj} A) = (\det A)^{n-2} A$.Since $n = 3$,we have $\operatorname{adj}(\operatorname{adj} A) = (\det A)^{3-2} A = (\det A) A$.Substituting $\det A = \frac{1}{k}$,we get $\operatorname{adj}(\operatorname{adj} A) = \frac{1}{k} A$.Now,we need to find the inverse: $(\operatorname{adj}(\operatorname{adj} A))^{-1} = (\frac{1}{k} A)^{-1}$.Using the property $(c A)^{-1} = \frac{1}{c} A^{-1}$,we get $(\frac{1}{k} A)^{-1} = k A^{-1}$.Since $A^{-1} = B$,the expression becomes $k B$.

If $B$ is the inverse of a third order matrix $A$ and $\det B = k$,then $(\operatorname{adj}(\operatorname{adj} A))^{-1} =$

Similar Questions

Let $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Then find the value of $(A^{-1}B)^{-1} + (AB^{-1})^{-1}$.

If $K \in R_0$,then $\det(adj(KI_n))$ is equal to:

If $3$ and $-2$ are the eigenvalues of a non-singular matrix $A$ and $|A| = 4$,then the eigenvalues of $adj(A)$ are:

If $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$, then which of the following is true?

For matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$,$(A^{-1})^2 = $ . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module