The matrix $\begin{bmatrix} 2 & -1 & 3 \\ \lambda & 0 & 7 \\ -1 & 1 & 4 \end{bmatrix}$ is not invertible for

If $\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}P\begin{bmatrix}2 & 1\\3 & 2\end{bmatrix}=\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}$ then matrix P is

Let $A = [a_{ij}]$ be a square matrix of order 3 with each entry either 0 or 1, then the number of all such possible matrices is:

If $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 3 & -1 \end{bmatrix}$, then $A^2$ is equal to

If A (a, a − b), B (a + b, − b) and C (b, a) are three collinear points, where a, b, c, $\in \mathbb{R}$, then :

If $A = \begin{bmatrix} 3 & 2a \\ 1 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ b & 5 \end{bmatrix}$ both are singular matrices, then $a + b$ is equal to

If $\begin{vmatrix} 1+x^2 & 1 & 1 \\ 1 & 1+y^2 & 1 \\ 1 & 1 & 1+z^2 \end{vmatrix} = k\left(1 + \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right)$ then $k$ is equal to