Compute the indicated products $\left[\begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$
- A$\left[\begin{array}{ccc}-3 & -4 & 1 \\ 8 & 13 & 9\end{array}\right]$
- B$\left[\begin{array}{ccc}3 & 4 & -1 \\ -8 & -13 & -9\end{array}\right]$
- C$\left[\begin{array}{ccc}-3 & 4 & 1 \\ 8 & -13 & 9\end{array}\right]$
- D$\left[\begin{array}{ccc}3 & -4 & 1 \\ -8 & 13 & -9\end{array}\right]$
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Let $n \geq 2$ be an integer. $A = \begin{bmatrix} \cos (2 \pi / n) & \sin (2 \pi / n) & 0 \\ -\sin (2 \pi / n) & \cos (2 \pi / n) & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $I$ is the identity matrix of order $3$. Then,
DifficultWBJEE 2014
View SolutionLet $A = [a_{ij}]$ be a real matrix of order $3 \times 3$,such that $a_{i1} + a_{i2} + a_{i3} = 1$,for $i = 1, 2, 3$. Then,the sum of all the entries of the matrix $A^3$ is equal to:
Find the values of $a, b, c,$ and $d$ from the following equation:
$\begin{bmatrix} 2a+b & a-2b \\ 5c-d & 4c+3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix}$
$\begin{bmatrix} 2a+b & a-2b \\ 5c-d & 4c+3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix}$
Easy
View SolutionIf $\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}$,then $\frac{x^2+y^2+z^2}{\gamma} =$
The number of all possible matrices of order $3 \times 3$ with each entry $0$ or $1$ is:
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