Determinant of the adjugate of a 3 by 3 matrix

Expert intro: The adjugate matrix connects directly to the determinant through a standard power rule for square matrices [1][2].

Detailed walkthrough

Key idea for adjugate determinants

For any square matrix $A$ of order $n$, the determinant of its adjugate is not random: it follows a precise rule. The identity is

$|\operatorname{adj}A| = |A|^{n-1}$

because the adjugate is built from cofactors, and its determinant scales in a predictable way with the size of the matrix.

Here, $A$ is a matrix of order 3, so $n = 3$. That means the exponent is $3 - 1 = 2$.

Step-by-step substitution

Since the problem gives $|A| = 8$, we substitute directly into the formula:

$|\operatorname{adj}A| = |A|^{3-1} = 8^2$

Now compute the power:

$8^2 = 64$

So the determinant of the adjugate matrix is $64$.

Why this formula works

The matrix identity behind this result is

$A\,\operatorname{adj}A = |A|I$

for every square matrix $A$. Taking determinants on both sides gives a product rule that leads to the formula above. When $A$ is nonsingular, this relation is especially useful because it lets you compute the adjugate determinant without expanding cofactors by hand.

It is also important to notice that the order of the matrix matters. A 2 by 2 matrix would give $|\operatorname{adj}A| = |A|^1$, while a 4 by 4 matrix would give $|\operatorname{adj}A| = |A|^3$. The exponent is always one less than the matrix order.

Common mistake to avoid

A frequent error is to think the answer should be the same as $|A|$ or the inverse of $|A|$. That is not correct. The adjugate is not the inverse itself; it is the matrix of cofactors transposed. Another mistake is forgetting that the formula uses the order of the matrix, not the value of the determinant.

In this question, the matrix is 3 by 3, so the correct exponent is 2, and the computed result is 64.

💡 Pitfall guide

A common trap with $|\operatorname{adj}A|$ is to mix it up with $|A|$ or with the determinant of the inverse. The first place students lose points is writing $|\operatorname{adj}A|=8$ just because the original determinant is 8. That skips the order-dependent exponent entirely. Another issue is applying the formula without checking that $A$ is square of order 3. The formula $|\operatorname{adj}A| = |A|^{n-1}$ only works for an $n \times n$ matrix, so the matrix size must be read carefully before substituting. In this problem, the order is 3, so the exponent is 2, not 3 or 1. A second mistake is doing unnecessary cofactor expansion. That is much longer and more error-prone than using the determinant identity directly. If you remember one pattern, remember that the adjugate determinant grows as the determinant to the power one less than the matrix size.

🔄 Real-world variant

If the same question changed to a 4 by 4 matrix with $|A|=8$, the new problem would be: "If $A$ is a matrix of order 4 and $|A|=8$, then $|\operatorname{adj}A|=?" The method stays identical, but the exponent changes to$4-1=3$. So the answer would become$8^3=512$. If instead the matrix stayed 3 by 3 but the determinant changed to 2, the question would become: "If$A$is a matrix of order 3 and$|A|=2$, then$|\operatorname{adj}A|=?" Then the result would be $2^2=4$. These variants show that the key variables are the matrix order and the determinant of the original matrix. Once those two values are known, the adjugate determinant follows immediately from the same rule.

🔍 Related terms

adjugate matrix determinant, cofactor matrix identity, determinant power rule