Question

If A is a square matrix of order 3 and |A|=5, then |adj(adjA)| is:

• A. 125
• B. 625
• C. 75
• D. 375

Hint:

Do not compute |adj(A)| as a separate number first. Combine both adjoint steps into one exponent using |adj(adj A)| = |A| raised to the power (n-1) squared, then substitute n = 3 and the given determinant directly.

Answer 1

The Correct Option is B

To find the value of |adj(adjA)| for a 3x3 matrix A where the determinant |A| is given as 5, we use the properties of determinants and adjugates (adjoint matrices).

For a square matrix A of order n, the property of the adjugate is that |adj(A)| = |A|^n-1. Given A is a 3x3 matrix (n=3), |adj(A)| = |A|^3-1 = |A|².

Therefore, |adj(A)| = 5² = 25.

Now, applying the same property to find |adj(adjA)|:

Since adj(adjA) involves another application of the adjugate, we consider an (n-1)th application, yielding: |adj(adjA)| = |adj(A)|^3-1 = |adj(A)|² for a 3x3 matrix.

Thus, |adj(adjA)| = 25² = 625.

Therefore, the value of |adj(adjA)| is 625.

Answer 2

Concept:

• Instead of squaring an intermediate numeric determinant, the two applications of the adjoint-determinant rule can be combined into a single exponent formula first.
• For an $n \times n$ matrix $A$, applying the rule $|adj(M)| = |M|^{n-1}$ twice in a row gives $|adj(adj A)| = |A|^{(n-1)^2}$, so the whole answer is reached in one power calculation.

Step 1: Write down the base rule for a single adjoint.
For any $n \times n$ matrix $M$, $|adj(M)| = |M|^{n-1}$. Here $n = 3$, so this rule gives an exponent of $2$ each time it is used.

Step 2: Apply the rule a second time, symbolically, before putting in numbers.
Let $M = adj(A)$. Then $|adj(adj A)| = |adj(A)|^{n-1} = \left(|A|^{n-1}\right)^{n-1} = |A|^{(n-1)^2}$. For $n = 3$, this becomes $|adj(adj A)| = |A|^4$.

Step 3: Substitute the given value of $|A|$ directly into the combined formula.
$|adj(adj A)| = 5^4$.

Step 4: Compute the power.
$5^4 = 5 \times 5 \times 5 \times 5 = 625$.

Final Answer: $625$.