If A, B and C are three singular matrices given by \(A=\begin{bmatrix}1 & 4\\ 3 & 2a\end{bmatrix}\), \(B=\begin{bmatrix}3b & 5\\ a & 2\end{bmatrix}\) and \(C=\begin{bmatrix}a+b+c & c+1\\ a+c & c\end{bmatrix}\), then the value of abc is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 45
A singular matrix has determinant 0.
From A:
\(|A| = 2a - 12 = 0\)
\(a = 6\)
From B:
\(|B| = 6b - 5a = 0\)
\(6b = 30,\quad b = 5\)
From C, with a + b = 11 and a = 6:
\(|C| = c(a + b + c) - (c + 1)(a + c) = 0\)
\(c(11 + c) - (c + 1)(6 + c) = 0\)
\(11c + c^2 - \left(c^2 + 7c + 6\right) = 0\)
\(4c - 6 = 0\)
\( \displaystyle c = \frac{3}{2} \)
Hence
\( \displaystyle abc = 6 \times 5 \times \frac{3}{2} = 45 \)