If $A-B=\begin{bmatrix} 2 & 5 \\ 9 & 0 \end{bmatrix}$ and $A+B=\begin{bmatrix} 6 & 3 \\ -1 & 0 \end{bmatrix}$,then matrix $A =$ . . . . . .

  • A
    $\begin{bmatrix} 4 & 4 \\ 4 & 0 \end{bmatrix}$
  • B
    $\begin{bmatrix} 4 & 0 \\ 4 & 4 \end{bmatrix}$
  • C
    $\begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}$
  • D
    $\begin{bmatrix} 0 & 4 \\ 4 & 4 \end{bmatrix}$

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In the matrix $A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$,write:
$(i)$ The order of the matrix
$(ii)$ The number of elements
$(iii)$ The elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$

Let $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$,$B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$,and $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$. Find $A - B$.

Find the values of $x, y, z$ if the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies the equation $A^{\prime} A = I$.

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If $A = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}$,then the value of $A^{40}$ is

In an upper triangular matrix of order $n \times n$,the minimum number of zeros is:

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