If $A=\begin{bmatrix}1 & 0\\ -1 & 7\end{bmatrix}$, $I=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$ and $A^2=8A+kI$, then $k$ is equal to:

The area (in sq. units) of the triangle whose vertices are $(0, 0)$, $(a, 0)$, $(0, b)$, is equal to

Let $A=\begin{bmatrix}1 & \sin\theta & 1\\-\sin\theta & 1 & \sin\theta\\-1 & -\sin\theta & 1\end{bmatrix}$, where $\theta\in[0,2\pi]$ then which of the following statements are correct?

A. $|A|=2+2\sin^2\theta$

B. Maximum value of $|A|$ is 1.

C. Maximum value of $|A|$ is –1.

D. $|A|\in[2,4]$

Choose the correct answer from the options given below:

If $[\mathrm{A}]_{3 \times 2}[\mathrm{~B}]_{\mathrm{x} \times \mathrm{y}}=[\mathrm{C}]_{3 \times 1}$, then :

If $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ x & y & -1 \end{bmatrix}$, then $A^2$ is

If $\begin{bmatrix} 2a + b & a - 2b \\ 5c - d & 4c + 3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix}$, then the value of $a + 2b - 3c + 4d$ is equal to

If $a, b$ and $c$ are non-zero real number such that $\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = 0$, then value of $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$ is