Let B=pmatrix1&3&1&2&3&&4pmatrix, 2, be the adjoint of a matrix A with |A|=2.

Question

Let $B=\begin{pmatrix}1&3&\alpha\1&2&3\\alpha&\alpha&4\end{pmatrix}$, $\alpha>2$, be the adjoint of a matrix $A$ with $|A|=2$. Then $\begin{pmatrix}\alpha&-2\alpha&\alpha\end{pmatrix}B\begin{pmatrix}\alpha\-2\alpha\\alpha\end{pmatrix}$ is equal to

Accepted Answer

For a $3\times3$ matrix, $|\operatorname{adj}A|=|A|^2=4$. Expanding $|B|=4$:
$$1(8-3\alpha)-3(4-3\alpha)+\alpha(\alpha-2\alpha)=4 \Rightarrow -\alpha^2+6\alpha-8=0 \Rightarrow \alpha=2,4.$$

Since $\alpha>2$: $\alpha=4$.

With $\alpha=4$: $\begin{pmatrix}4&-8&4\end{pmatrix}\begin{pmatrix}1&3&4\1&2&3\4&4&4\end{pmatrix}\begin{pmatrix}4\-8\4\end{pmatrix} = \begin{pmatrix}4&-8&4\end{pmatrix}\begin{pmatrix}4\-8\4\end{pmatrix}\cdot\text{(after inner multiplication)}=-16$ (per direct matrix multiplication in the source).

Solutions · 1