The point (a, b) is the foot of the perpendicular drawn from the point (3, 1) to the line x + 3y + 4 = 0. If (p, q) is the image of (a, b) with respect to the line 3x - 4y + 11 = 0, then $\frac{p}{a} + \frac{q}{b} = $
A, B, C, D are square matrices such that A + B is symmetric, A - B is skew-symmetric, and D is the transpose of C.
If
\[ A = \begin{bmatrix} -1 & 2 & 3 \\ 4 & 3 & -2 \\ 3 & -4 & 5 \end{bmatrix} \]
and
\[ C = \begin{bmatrix} 0 & 1 & -2 \\ 2 & -1 & 0 \\ 0 & 2 & 1 \end{bmatrix} \]
then the matrix \( B + D \) is:
The rank of matrix \(\begin{bmatrix} k & -1 & 0 \\[0.3em] 0 & k & -1 \\[0.3em] -1 & 0 & k \end{bmatrix}\) is 2, for \( k = \)
If \(A = \begin{bmatrix} 4 & 2 \\[0.3em] -3 & 3 \end{bmatrix}\), then \(A^{-1} =\)
If \(P(A) = \frac{7}{11}\), \( P(B) = \frac{6}{11} \), and \( P(A \cup B) = \frac{8}{11} \), then \( P(A|B) = \) ?}