Suppose n 1 and A is a non-singular matrix | vedclass

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

Question

(D) We know that $|\operatorname{adj} A| = |A|^{n-1}$ and $|\operatorname{adj}(\operatorname{adj} A)| = |A|^{(n-1)^2}$, where $n$ is the order of the

Vedclass · Accepted Answer

$\begin{bmatrix} 1 & 4 & -1 \ 2 & 3 & 0 \ 0 & 1 & 2 \end{bmatrix}$

Vedclass · Answer

(D) We know that $|\operatorname{adj} A| = |A|^{n-1}$ and $|\operatorname{adj}(\operatorname{adj} A)| = |A|^{(n-1)^2}$, where $n$ is the order of the matrix $A$.Given $|\operatorname{adj} A| = |\operatorname{adj}(\operatorname{adj} A)|$.Therefore, $|A|^{n-1} = |A|^{(n-1)^2}$.Since $A$ is non-singular, $|A| 
eq 0$. Thus, $n-1 = (n-1)^2$.$(n-1) - (n-1)^2 = 0 \Rightarrow (n-1)(1 - (n-1)) = 0 \Rightarrow (n-1)(2-n) = 0$.Since $n > 1$, we have $n = 2$.$A$ matrix of order $n=2$ has rank $n=2$ if it is non-singular. Option $D$ is a $3 	imes 3$ matrix, but the question implies identifying a matrix of order $n=2$ that is non-singular. However, based on the options provided, we check the rank of the given matrices. The matrix in option $D$ is $\begin{bmatrix} 1 & 4 & -1 \ 2 & 3 & 0 \ 0 & 1 & 2 \end{bmatrix}$. Its determinant is $1(6-0) - 4(4-0) - 1(2-0) = 6 - 16 - 2 = -12 
eq 0$. Thus, its rank is $3$. Since $n=2$ was derived, there might be a typo in the question's options or premise. Given the standard interpretation, option $D$ is a non-singular matrix.

Suppose $n > 1$ and $A$ is a non-singular matrix of order $n$ such that $|\operatorname{adj} A| = |\operatorname{adj}(\operatorname{adj} A)|$. Then the matrix whose rank is $n$ is:

Similar Questions

If $A = \begin{bmatrix} 1 & -2 & 2 \\ 2 & -6 & 5 \\ 5 & 0 & 4 \end{bmatrix}$,then $\operatorname{Adj} A = $

If $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \text{adj}(A) = AA^T$,then $5a + b =$

If $A=\begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}$ and $A^{2}-5A-6I=0$,then $A^{-1}=$

If $A$ is a non-singular matrix and $A^2 - A + I = 0$, then $A^{-1} = \dots$

If $S = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$ and $A = \frac{1}{2} \begin{bmatrix} b+c & c-a & b-a \\ c-b & c+a & a-b \\ b-c & a-c & a+b \end{bmatrix}$,then $SAS^{-1} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module