Matrices & Determinants
Adjoint Matrix — Finding Vector-Matrix-Vector Product
nta_pyq_2023_apr
Grade 12

Question:

Let $B=\begin{pmatrix}1&3&\alpha\\1&2&3\\\alpha&\alpha&4\end{pmatrix}$, $\alpha>2$, be the adjoint of matrix $A$ with $|A|=2$. Then $[\alpha\ -2\alpha\ \alpha]B\begin{pmatrix}\alpha\\-2\alpha\\\alpha\end{pmatrix}$ is equal to
0
16
$-16$
32

Step-by-Step Solution

Key Concept: $|B|=|\text{adj}(A)|=|A|^{n-1}=2^2=4$. Compute $|B|$ in terms of $\alpha$: $|B|=\alpha^2-6\alpha+8$. Set $=4$.
$\alpha=4$. Expression $=-16$.
Correct Answer: 3

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