Practicing Success
If A and B are invertible matrices of order 3, |A| = 2 and $\left|(A B)^{-1}\right|=-\frac{1}{6}$, then the value of |B| is : |
3 -3 2 $\frac{1}{6}$ |
-3 |
$|A|=2$ $\left|(A B)^{-1}\right|=-\frac{1}{6}$ so $\left|(A B)^{-1}\right|=-\frac{1}{6}$ $=\left|B^{-1} A^{-1}\right|=\frac{-1}{6}$ (as (AB)-1 = B-1 A-1) We know that $A A^{-1}=I$ so $\left|A A^{-1}\right|=|I|$ so $|A|=\frac{1}{|A^{-1}|}$ or $|A^{-1}|=\frac{1}{|A|}$ $\Rightarrow \left|B^{-1}\right||A-1|=-\frac{1}{6}$ $\Rightarrow \frac{1}{|A||B|}=-\frac{1}{6}$ So $\frac{1}{2|B|}=-\frac{1}{6}$ So |B| = -3 |