Let A=pmatrix1&21&pmatrix and B=pmatrix3&3&2pmatrix.

Question

Let $A=\begin{pmatrix}1&2\1&\alpha\end{pmatrix}$ and $B=\begin{pmatrix}3&3\\beta&2\end{pmatrix}$. If $A^2-4A+I=O$ and $B^2-5B-6I=O$, then among the two statements:

(S1): $\big[(B-A)(B+A)\big]^T=\begin{pmatrix}13&15\7&10\end{pmatrix}$, and

(S2): $\det(\operatorname{adj}(A+B))=-5$

Accepted Answer

From $A^2-4A+I=0$, the trace condition gives $\alpha+1=4\Rightarrow\alpha=3$.

From $B^2-5B-6I=0$, the determinant condition gives $\det(B)=6-3\beta=-6\Rightarrow\beta=4$.

So $A=\begin{pmatrix}1&2\1&3\end{pmatrix}$, $B=\begin{pmatrix}3&3\4&2\end{pmatrix}$.

$B-A=\begin{pmatrix}2&1\3&-1\end{pmatrix}$, $B+A=\begin{pmatrix}4&5\5&5\end{pmatrix}$, so $(B-A)(B+A)=\begin{pmatrix}13&15\7&10\end{pmatrix}$ --- this matrix is *not* equal to its own transpose in general (checking the actual product shows the transpose does not match the stated matrix), so (S1) is false.

$\operatorname{adj}(A)=\begin{pmatrix}3&-2\-1&1\end{pmatrix}$, $\operatorname{adj}(B)=\begin{pmatrix}2&-3\-4&3\end{pmatrix}$, and $\operatorname{adj}(A+B)=\begin{pmatrix}5&-5\-5&4\end{pmatrix}$, whose determinant is $20-25=-5$, confirming (S2) is true.

So only (S2) is correct --- Answer B.

Solutions · 1