If $A = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix}$ and $B$ is a matrix such that $AB = BA$. If $AB$ is not an identity matrix,then the matrix that can be taken as $B$ is:

  • A
    $\begin{bmatrix} -9 & -3 & 6 \\ -6 & 8 & -4 \\ 12 & -4 & -2 \end{bmatrix}$
  • B
    $\begin{bmatrix} 9 & -3 & 6 \\ -6 & 8 & -4 \\ -12 & -4 & 2 \end{bmatrix}$
  • C
    $\begin{bmatrix} 9 & -3 & -6 \\ -6 & 8 & -4 \\ -12 & 4 & -2 \end{bmatrix}$
  • D
    $\begin{bmatrix} 9 & -3 & -6 \\ -6 & -8 & 4 \\ -12 & 4 & -2 \end{bmatrix}$

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If $\begin{bmatrix} x+3 & z+4 & 2y-7 \\ -6 & a-1 & 0 \\ b-3 & -21 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 6 & 3y-2 \\ -6 & -3 & 2c+2 \\ 2b+4 & -21 & 0 \end{bmatrix}$,then find the values of $a, b, c, x, y$ and $z$.

If $A = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}$,then find $AB$ and $BA$. Show that $AB \neq BA$.

If $A = \begin{vmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{vmatrix}$,then $1 + A^2 =$ . . . . . . .

If $A$ is a $3 \times 4$ matrix and $B$ is a matrix such that $A^{\prime}B$ and $BA^{\prime}$ are both defined,then $B$ is of the type:

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