Using determinant properties to evaluate adjugate of a Vandermonde matrix

Key takeaway: The matrix A is a Vandermonde-type matrix, and the shortcut det(adj A) = det(A)^2 makes the computation manageable.

Step 1: Recognize the determinant structure

For the matrix

$A=\begin{bmatrix}1&2&3\\1&4&9\\1&8&27\end{bmatrix},$

the determinant can be found by factoring the first column or by row reduction. Subtract the first row from the second and third rows:

$\begin{bmatrix}1&2&3\\0&2&6\\0&6&24\end{bmatrix}$

Now expand along the first column:

$|A| = 1\begin{vmatrix}2&6\\6&24\end{vmatrix} = 2\cdot 24 - 6\cdot 6 = 48 - 36 = 12$

Step 2: Apply the adjugate determinant formula

Since A is 3 x 3,

$|\operatorname{adj}A| = |A|^2$

So

$|\operatorname{adj}A| = 12^2 = 144$

Step 3: Check the pattern against the answer choices

The matrix has the classic determinant pattern of a Vandermonde-type system in the powers of 2, 4, 8 and 3, 9, 27. Even if you do not identify that pattern immediately, row operations lead to the same result quickly.

The correct answer is

$\boxed{144}$

Step 4: Why the shortcut matters

Computing the full adjugate matrix would be far longer than necessary. Here the problem asks only for its determinant, so the identity det(adj A)=det(A)^{n-1} turns a complicated cofactor expansion into a single square.

That is the cleanest route for any square matrix once det(A) is known.

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Pitfalls the pros know 👇 A typical mistake with this matrix is to jump straight to cofactors for the adjugate instead of first finding det(A). That wastes time and often introduces sign errors. Another trap is to think the matrix is singular because the second and third rows look related through powers, but they are not multiples of one another. The determinant is not zero here. Also, when you subtract the first row from the other rows, do not forget that only the new rows change; the first row stays the same for the expansion step. Once det(A)=12 is secured, squaring it gives the answer immediately.

What if the problem changes? If the last row were changed from one, eight, twenty-seven to one, sixteen, sixty-four, the matrix would become [[1,2,3],[1,4,9],[1,16,64]]. The same row-reduction approach would still apply, but the determinant of A would change, so det(adj A) would change as well. If the matrix were instead scaled by a factor of 3, the determinant of the adjugate would scale by 3 to the power of 4 because for a three by three matrix, adj of cA scales by c squared and then its determinant scales accordingly. These variants show that the final number depends heavily on the underlying determinant, not on the visible pattern alone.

Tags: Vandermonde matrix determinant, adjugate matrix formula, row reduction determinant