Question:
If, \(A = \begin{bmatrix} 1 & 4 \\ 3 & 15 \end{bmatrix}\), then \(|A^{-1}|\) is equal to
यदि \(A = \begin{bmatrix} 1 & 4 \\ 3 & 15 \end{bmatrix}\) है, तो \(|A^{-1}|\) किसके बराबर है?
Options:
- \(\dfrac{4}{3}\)
- \(-\dfrac{1}{3}\)
- \(\dfrac{2}{3}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{4}{3}\)
- \(-\dfrac{1}{3}\)
- \(\dfrac{2}{3}\)
- \(\dfrac{1}{3}\)
Correct Answer: d)
Explanation:
\(|A| = (1)(15) – (4)(3) = 15 – 12 = 3\).
Using the property of determinants: \(|A^{-1}| = \dfrac{1}{|A|} = \dfrac{1}{3}\).
\(|A| = 1(15) – 4(3) = 15 – 12 = 3\)।
सारणिक के गुणधर्म से: \(|A^{-1}| = \dfrac{1}{|A|} = \dfrac{1}{3}\)।