If $A$ and $B$ are two square matrices with $\det(A) = 5$ and $\det(B^T \cdot A^T) = -15$,then $\det(B)$ is equal to

  • A
    $3$
  • B
    -$3$
  • C
    $0$
  • D
    $1$

Explore More

Similar Questions

Using properties of determinants,prove that:
$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

Difficult
View Solution

If ${D_r} = \left| \begin{array}{ccc} {2^{r - 1}} & {2 \cdot 3^{r - 1}} & {4 \cdot 5^{r - 1}} \\ x & y & z \\ {2^n} - 1 & {3^n} - 1 & {5^n} - 1 \end{array} \right|$,then the value of $\sum\limits_{r = 1}^n {D_r} = $

Difficult
View Solution

If $A, B$ and $C$ are $n \times n$ matrices and $\det(A) = 2$,$\det(B) = 3$ and $\det(C) = 5$,then the value of $\det(A^2BC^{-1})$ is equal to

$2\,\,\left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ {a^2 - bc} & {b^2 - ac} & {c^2 - ab} \end{array}} \right| = $

Difficult
View Solution

The value of the determinant $\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|$ is equal to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo