Question

Let \(A\) be a \(3\times3\) matrix such that
\(\mathrm{adj}\,A=\begin{bmatrix}2&-1&1\\-1&0&2\\1&-2&-1\end{bmatrix}\)
and \(B=\mathrm{adj}(\mathrm{adj}\,A)\).
If \(|A|=\lambda\) and \(|(B^{-1})^T|=\mu\), then the ordered pair \((|\lambda|,\mu)\) is equal to :

• A. \((3,81)\)
• B. \(\left(9,\frac19\right)\)
• C. \(\left(3,\frac{1}{81}\right)\)
• D. \(\left(9,\frac{1}{81}\right)\)

Hint:

For an \(n\times n\) matrix, \(|\mathrm{adj}\,M|=|M|^{n-1}\). Find \(|\mathrm{adj}\,A|\) directly, then deduce \(|A|\) and \(|B|\).

Answer 1

The Correct Option is C

\(|\mathrm{adj}\,A|=\begin{vmatrix}2&-1&1\\-1&0&2\\1&-2&-1\end{vmatrix}=2(0+4)+1(1-2)+1(2-0)=8-1+2=9\).
For a \(3\times3\) matrix \(|\mathrm{adj}\,A|=|A|^2\), so \(|A|^2=9\Rightarrow|\lambda|=3\).
\(B=\mathrm{adj}(\mathrm{adj}\,A)\Rightarrow|B|=|\mathrm{adj}\,A|^2=81\).
\(|(B^{-1})^T|=|B^{-1}|=\frac{1}{|B|}=\frac1{81}\).
So \((|\lambda|,\mu)=\left(3,\frac1{81}\right)\).
Answer: (3) \(\left(3,\frac{1}{81}\right)\)