Let A and B be two 3 times 3 non-singular mat | vedclass

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

Question

(A) Given, $\operatorname{det}(A^T B A) = 27$. Since $\operatorname{det}(A^T) = \operatorname{det}(A)$, we have $|A|^2 |B| = 27$  $(i)$. Also, $\opera

Vedclass · Accepted Answer

$\frac{3}{32}$

Vedclass · Answer

(A) Given, $\operatorname{det}(A^T B A) = 27$. Since $\operatorname{det}(A^T) = \operatorname{det}(A)$, we have $|A|^2 |B| = 27$  $(i)$. Also, $\operatorname{det}(A B^{-1}) = 8$, which implies $\frac{|A|}{|B|} = 8$, so $|B| = \frac{|A|}{8}$ $(ii)$. Substituting $(ii)$ into $(i)$: $|A|^2 \left(\frac{|A|}{8}\right) = 27 \Rightarrow |A|^3 = 216 \Rightarrow |A| = 6$. Then, $|B| = \frac{6}{8} = \frac{3}{4}$. We need to find $\operatorname{det}(B^T A^{-1} B) = |B^T| |A^{-1}| |B| = |B| \cdot \frac{1}{|A|} \cdot |B| = \frac{|B|^2}{|A|}$. Substituting the values: $\frac{(\frac{3}{4})^2}{6} = \frac{9/16}{6} = \frac{9}{96} = \frac{3}{32}$.

Let $A$ and $B$ be two $3 \times 3$ non-singular matrices such that $\operatorname{det}(A^T B A) = 27$ and $\operatorname{det}(A B^{-1}) = 8$. Then $\operatorname{det}(B^T A^{-1} B) = $

Similar Questions

Which of the following is(are) $NOT$ the square of a $3 \times 3$ matrix with real entries?

Let $A^{-1} = \begin{bmatrix} 1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8 \end{bmatrix}$. Then,$|2A| - |2A^{-1}|$ is equal to

Let $I$ be an identity matrix of order $2 \times 2$ and $P = \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix}$. Then the value of $n \in N$ for which $P^n = 5I - 8P$ is equal to ..... .

If $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$ and $\alpha A^2 + \beta A = 2I$ for some $\alpha, \beta \in \mathbb{R}$,then $\alpha + \beta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module