If $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -1 \end{bmatrix}$,then $A^4 A^{-1} = $
- A$\begin{bmatrix} 8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
- B$\begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
- C$\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & -\frac{1}{2} & 0 \\ 0 & 0 & -1 \end{bmatrix}$
- D$\begin{bmatrix} -4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
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Let $A = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $B = \begin{bmatrix} 9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2 \end{bmatrix}$,then the value of $A^{\prime} BA$ is.
If $A = [x \quad y \quad z]$,$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$,$C = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $(AB) \cdot C$ is an $m \times n$ order matrix,then:
If $A = \begin{bmatrix} 1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \end{bmatrix}$,then $AB =$
Easy
View SolutionFor the matrix $A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix}$,which of the following is correct?
Difficult
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:
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