Tag: business maths

If $\omega$ is the complex cube root of unity, then inverse of $\begin{bmatrix} \omega  & 0 & 0 \ 0 & { \omega  }^{ 2 } & 0 \ 0 & 0 & { \omega  }^{ 2 } \end{bmatrix}$ is

The inverse of the matrix $\begin{bmatrix}1 & 0 & 1\ 0 & 2 & 3\ 1 & 2& 1\end{bmatrix}$ is

If A =$\left[ \begin{matrix} i \ 0 \end{matrix}\begin{matrix} 0 \ -1 \end{matrix} \right] $, than check whether: ${{\text{A}}^2} =  - {\text{I,(}}{{\text{i}}^2} =  - 1)$

If $M = \left[ \begin{array}{l}0\,\,\,\,2\5\,\,\,\,\,0\end{array} \right]\,\,\,and\,\,N = \left[ \begin{array}{l}0\,\,\,\,5\2\,\,\,\,\,0\end{array} \right]$,then ${M^{2011}}$ is-

If $A = \left[ \begin{array}{l}\cos \theta \,\,\,\,\sin \theta \ - \sin \theta \,\,\,\cos \theta \end{array} \right]$ where $\theta  = \frac{{2\pi }}{{19}}$ then ${A^{2017}} = $

If A and B are matrices of the same order, then $\displaystyle :\left ( A+B \right )^{2}= A^{2}+2AB+B^{2}$ is possible, iff

If $A$ and $B$ are any two matices, then  

If $A^{2}-A+I=0$, then inverse of $A$ is

The matrices $\begin{bmatrix} \cos { \theta  }  & -\sin { \theta  }  \ \sin { \theta  }  & \cos { \theta  }  \end{bmatrix}$ and $\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}$ commute under multiplication

If $A$ and $B$ are two square matrices of order $3 \times  3$ which satisfy $AB = A$ and $BA = B$, then Which of the following is true?