If $B$ is a $3 \times 3$ matrix such that $B^2 = 0$,then $\det[(I + B)^{50} - 50B]$ is equal to

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $50$

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Similar Questions

Let $A$ be the set of all $3 \times 3$ determinants with entries $0$ or $1$ only and $B$ be the subset of $A$ consisting of all determinants with value $1$. If $C$ is the subset of $A$ consisting of all determinants with value $-1$,then:

If $A+B=\left[\begin{array}{cr}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]$ where $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix,then the matrix $\left(A^{-1} B+A B^{-1}\right)$ at $\theta=\frac{\pi}{6}$ is given by

If $M$ and $N$ are square matrices of order $3$,then which one of the following statements is not true?

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If $A, B$ are two non-singular matrices of order $3$ and $|B|=k$,where $k$ is a positive integer,then match the items of List-$I$ with the items of List-$II$.
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$. $\frac{1}{|A|}(A^{-1})$
$V$. $\frac{1}{|A|^2}$

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