If A, B are two non-singular matrices of orde | vedclass

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

Question

(C) Given $A$ and $B$ are non-singular matrices of order $3$ and $|B|=k$.$A$. $|k^{-1} A^{-1}| = (k^{-1})^3 |A^{-1}| = \frac{1}{k^3 |A|} = \frac{1}{|B

Vedclass · Accepted Answer

$A-I, B-V, C-II, D-IV$

Vedclass · Answer

(C) Given $A$ and $B$ are non-singular matrices of order $3$ and $|B|=k$.$A$. $|k^{-1} A^{-1}| = (k^{-1})^3 |A^{-1}| = \frac{1}{k^3 |A|} = \frac{1}{|B|^3 |A|}$. Thus,$A-III$.$B$. $|	ext{Adj}(A^{-1})| = |A^{-1}|^{3-1} = |A^{-1}|^2 = \frac{1}{|A|^2}$. Thus,$B-V$.$C$. Given $BAB^{-1} = I$,then $A = B^{-1}IB = B^{-1}B = I$ is not necessarily true,but $BAB^{-1} = I \Rightarrow A = B^{-1}B = I$ is wrong. Actually,$BAB^{-1} = I \Rightarrow A = B^{-1}B = I$ is not implied. However,$BAB^{-1} = I \Rightarrow BA^kB^{-1} = (BAB^{-1})^k = I^k = I$. Wait,the options suggest $BA^kB^{-1} = B \frac{	ext{Adj}(B)}{|B|} = I$. Since $B 	ext{Adj}(B) = |B|I$,then $B \frac{	ext{Adj}(B)}{|B|} = I$. Thus,$C-II$.$D$. $	ext{Adj}(	ext{Adj}(A^{-1})) = |A^{-1}|^{3-2} (A^{-1}) = |A^{-1}| A^{-1} = \frac{1}{|A|} A^{-1}$. Thus,$D-IV$.Therefore,the correct match is $A-III, B-V, C-II, D-IV$.

List-$I$	List-$II$
$A$. $\|k^{-1} A^{-1}\|$	$I$. $BA^k + A^kB$
$B$. $\|\text{Adj}(A^{-1})\|$	$II$. $\frac{B\text{Adj}(B)}{\|B\|}$
$C$. $BAB^{-1} = I \Rightarrow BA^kB^{-1} =$	$III$. $\frac{1}{\|B\|^3\|A\|}$
$D$. $\text{Adj}(\text{Adj}(A^{-1})) =$	$IV$. $\frac{1}{\|A\|}(A^{-1})$
	$V$. $\frac{1}{\|A\|^2}$

Similar Questions

Let $X = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$. Let $Y$ be a $2 \times 2$ real matrix satisfying the condition $XY = YX$. Then the smallest possible value of $\det(Y)$ is

The value of $\left|\begin{array}{cc}\log _5 729 & \log _3 5 \\ \log _5 27 & \log _9 25\end{array}\right| \times \left|\begin{array}{cc}\log _3 5 & \log _{27} 5 \\ \log _5 9 & \log _5 9\end{array}\right|$ is

Let $A = \begin{bmatrix} \frac{1}{6} & \frac{-1}{3} & \frac{-1}{6} \\ \frac{-1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{-1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix}$. If $A^{2016l} + A^{2017m} + A^{2018n} = \frac{1}{\alpha} A$,for every $l, m, n \in N$,then the value of $\alpha$ is

If $\begin{bmatrix} x & 4 & -1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$,then $x=$

Let $A = \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix}$. If $B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} A \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}$,then the sum of all the elements of the matrix $\sum_{n=1}^{50} B^n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module