Question

Let \(A=\begin{pmatrix}\alpha & 1 & 2\\ 1 & 2 & 3\\ 3 & 1 & 1\end{pmatrix}\), \(\alpha\in\mathbb{R}\) and \(B=[b_{ij}]_{3\times3}\) such that \(AB=rI\), \(r\neq0\) in \(\mathbb{R}\). If \(b_{32}=-\frac{3}{2}r\) and \(|\operatorname{adj}B|=r^{5}\), then \(|\operatorname{adj}(AB)|\) equals :

• A. \(4^{12}\)
• B. \(2^{12}\)
• C. \(2^{6}\)
• D. \(4^{8}\)

Hint:

Since \(AB=rI\), \(B=rA^{-1}\); use \(b_{32}\) to find \(\alpha\), then \(|\operatorname{adj}B|=|B|^2\) to find \(r\), and finally \(|\operatorname{adj}(AB)|=|AB|^2=r^6\).

Answer 1

The Correct Option is B

\(AB=rI\) gives \(B=rA^{-1}=\dfrac{r}{|A|}\operatorname{adj}A\), so \(b_{32}=\dfrac{r\,C_{23}}{|A|}\) where \(C_{23}\) is the cofactor of the (2,3) entry of \(A\).
\(C_{23}=-\begin{vmatrix}\alpha & 1\\ 3 & 1\end{vmatrix}=-(\alpha-3)=3-\alpha\).
\(|A|=\alpha(2-3)-1(1-9)+2(1-6)=-\alpha+8-10=-(\alpha+2)\).
Given \(b_{32}=-\frac32 r\): \(\dfrac{3-\alpha}{-(\alpha+2)}=-\dfrac32\), so \(\dfrac{3-\alpha}{\alpha+2}=\dfrac32\), i.e. \(6-2\alpha=3\alpha+6\), giving \(\alpha=0\). Then \(|A|=-2\).
From \(AB=rI\): \(|A||B|=r^3\), so \(|B|=-\dfrac{r^3}{2}\).
For a \(3\times3\) matrix \(|\operatorname{adj}B|=|B|^2=\dfrac{r^6}{4}\). Setting this equal to \(r^5\) gives \(r=4\).
Finally \(|\operatorname{adj}(AB)|=|AB|^2=(r^3)^2=r^6=4^6=2^{12}\).
Answer: (2) \(2^{12}\)