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Let A=[1−312]A = \begin{bmatrix} 1 & -3 \\ 1 & 2 \end{bmatrix}, B=[46−22]B = \begin{bmatrix} 4 & 6 \\ -2 & 2 \end{bmatrix} and CC be a matrix such that C=ABC = AB. Then ∣(BC)−1∣|(BC)^{-1}| is equal to

Solution

✅ Correct Option: 4

∣A∣=1(2)−(−3)(1)=5|A| = 1(2)-(-3)(1) = 5 and ∣B∣=4(2)−6(−2)=20|B| = 4(2)-6(-2) = 20.

Since C=ABC = AB, ∣C∣=∣A∣∣B∣=100|C| = |A||B| = 100.

So ∣BC∣=∣B∣∣C∣=20×100=2000|BC| = |B||C| = 20 \times 100 = 2000, and ∣(BC)−1∣=12000=0.0005|(BC)^{-1}| = \dfrac{1}{2000} = 0.0005.

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