<p>Let <em>A</em> be a square matrix of order 3 such that \(\text{adj. }(\text{adj. }(\text{adj. }A)) = \begin{bmatrix}16 & 0 & -24\\ 0 & 4 & 0\\ 0 & 12 & 4\end{bmatrix}\). Find \(\text{adj. }A\).</p>
Step-by-Step Solution
Key Concept: Use the property that adj(adj(A)) = (det A)^(n-2) · A for an n×n matrix, applied iteratively three times to establish a relationship between adj(adj(adj(A))) and A, then solve for adj(A).
<p><strong>Step 1:</strong> Use the fundamental property: For an n×n matrix, adj(adj(A)) = (det A)^(n-2) · A. For n=3: adj(adj(A)) = (det A) · A</p><p><strong>Step 2:</strong> Apply this property successively. Let det A = d.</p><p>adj(adj(A)) = d · A, so</p><p>adj(adj(adj(A))) = adj(d · A) = d² · adj(A)</p><p><strong>Step 3:</strong> We're given adj(adj(adj(A))) = $\begin{bmatrix}16 & 0 & -24\ 0 & 4 & 0\ 0 & 12 & 4\end{bmatrix}$</p><p>Therefore: d² · adj(A) = $\begin{bmatrix}16 & 0 & -24\ 0 & 4 & 0\ 0 & 12 & 4\end{bmatrix}$</p><p><strong>Step 4:</strong> Observe that the given matrix factors as: $\begin{bmatrix}16 & 0 & -24\ 0 & 4 & 0\ 0 & 12 & 4\end{bmatrix}$ = 4 · $\begin{bmatrix}4 & 0 & -6\ 0 & 1 & 0\ 0 & 3 & 1\end{bmatrix}$</p><p>So d² = 4, giving d = ±2</p><p><strong>Step 5:</strong> Thus adj(A) = $\begin{bmatrix}4 & 0 & -6\ 0 & 1 & 0\ 0 & 3 & 1\end{bmatrix}$</p><p>∴ Answer: $\begin{bmatrix}4 & 0 & -6\ 0 & 1 & 0\ 0 & 3 & 1\end{bmatrix}$</p>
Correct Answer: \(\begin{bmatrix}4 & 0 & -6\\ 0 & 1 & 0\\ 0 & 3 & 1\end{bmatrix}\)