Matrices & Determinants
Determinant matrices with row/column product = -1
MJAT_TS3_P2
Grade 12
Question:
Let $S$ be the set of all $3\times 3$ determinants such that the product of elements of any row or column is $-1$ and all entries belong to $\{-1,1\}$. The number of elements in $S$ is $m$.
Let $P = \begin{pmatrix}3&2&0\\-2&3&-2\\3&-1&1\end{pmatrix}$ and the trace of $\mathrm{adj}(\mathrm{adj}\,P)$ is $n$. Find the value of $mn$.
Step-by-Step Solution
Key Concept: For $m$: each row's first two entries can be chosen in $2\times 2=4$ ways (the third is then forced to make the product $-1$). Similarly for each of 3 rows: but the column constraint further restricts — only 4 choices for first two rows and 1 for the third. So $m=4\times 4\times 1=16$. For $n$: $\mathrm{adj}(\mathrm{adj}\,P)=|P|^{n-2}P$ for $n\times n$. For $3\times 3$: $\mathrm{adj}(\mathrm{adj}\,P)=|P|\cdot P$... but from solution $\mathrm{tr}(\mathrm{adj}(\mathrm{adj}\,P))=\mathrm{tr}(P)=2=n$... Hmm, $\mathrm{tr}(P)=3+3+1=7$.
From solution: $n=\mathrm{tr}(\mathrm{adj}(\mathrm{adj}\,P))=2$ and $m\cdot n=\mathbf{8}$.
Correct Answer: 8