If \(P\), \(Q\) and \(R\) are three singular matrices given by \(P = \begin{bmatrix} 2 & 3a \\ 4 & 3 \end{bmatrix}\), \(Q = \begin{bmatrix} b & 5 \\ 2a & 6 \end{bmatrix}\) and \(R = \begin{bmatrix} a^2 + b^2 - c & 1 - c \\ c + 1 & c \end{bmatrix}\), then the value of \((2a + 6b + 17c)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 24
A singular matrix has determinant 0. From P:
\(|P| = 6 - 12a = 0\)
\( \displaystyle a = \frac{1}{2} \)
From Q:
\(|Q| = 6b - 10a = 0\)
\( \displaystyle 6b = 5,\quad b = \frac{5}{6} \)
From R:
\(|R| = (a^2 + b^2 - c)c - (1 - c)(1 + c) = 0\)
\((a^2 + b^2)c - c^2 - 1 + c^2 = 0\)
\((a^2 + b^2)c = 1\)
Now:
\( \displaystyle a^2 + b^2 = \frac{1}{4} + \frac{25}{36} = \frac{9 + 25}{36} = \frac{17}{18} \)
\( \displaystyle c = \frac{18}{17} \)
Hence:
\( \displaystyle 2a + 6b + 17c = 1 + 5 + 17 \times \frac{18}{17} \)
\(= 1 + 5 + 18 = 24\)