Match List-I with List-II | List-I | List-II | | :--- | :--- | | **Type of matrix** | **Conditions** | | (A) Square matrix A | (I) $A = [a_{ij}]_{m \times m}$ where $\begin{cases} a_{ij} = 0 & , i \neq j \\ a_{ij} = k & , i = j \end{cases}$, where $k \neq 0$ is constant. | | (B) Scalar Matrix A | (II) $A = [a_{ij}]_{m \times m}$ | | (C) Diagonal matrix A | (III) $A = [a_{ij}]_{m \times m}$ where $\begin{cases} a_{ij} = 0 & , i \neq j \\ a_{ij} = 1 & , i = j \end{cases}$ | | (D) Identity matrix A | (IV) $A = [a_{ij}]_{m \times m}$ where $a_{ij} = 0$, $i \neq j$ | Choose the correct answer from the options given below:

For any invertible matrix $A$, which of the following is NOT true.

If A and B are two square matrices of order 'n', where $AB \neq BA$, then

If a, b and c are all different from zero and $\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = 0$, then the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is :

If A is a square matrix of order 4 and $|adj\,A| = 8$, then $|A| =$

If $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$, then the matrix AB is equal to

If $\left|\begin{matrix} x & 8 \\ 4 & x \end{matrix}\right| = \left|\begin{matrix} 6 & 2 \\ 18 & 6 \end{matrix}\right|$, then $x$ is/are equal to