Q.If for a square matrix $A$, $A. (\text{adj}A) = \begin{bmatrix} 2025 & 0 & 0 \\ 0 & 2025 & 0 \\ 0 & 0 & 2025 \end{bmatrix}$, then the value of $|A| + |\text{adj}A|$ is equal to:
(A) $1$
(B) $2025+1$
(C) $(2025)^2 + 45$
(D) $2025+(2025)^2$

Concept: Adjoint Matrix Property — For any $n \times n$ matrix $A$,

$A \cdot (\text{adj } A) = |A| \, I_n$.

Here $n = 3$, and the given product is $2025 \, I_3$.

Thus $|A| = 2025$.

For a $3 \times 3$ matrix, $|\text{adj } A| = |A|^{n-1} = |A|^2 = (2025)^2$.

Therefore, …

We start with a fundamental property of adjoint matrices. For any square matrix $A$ of order $n$, the product of $A$ and its adjoint is always a scalar matrix — specifically, the determinant of $A$ times the identity matrix. That is:

This is not a coincidence; it comes from the fact that each entry of $A \cdot (\text{adj } A)$ is the expansion of a determinant along a row (or column), giving $|A|$ on the diagonal and zero elsewhere. This single relation unlocks the entire problem.

Now, look at what we are given:

This is clearly $2025$ times the $3 \times 3$ identity matrix. So we have:

Comparing this with the formula $A \cdot (\text{adj } A) = |A| \, I_3$, we immediately see that:

That is the first piece. Now we need $|\text{adj } A|$.

There is another standard result: for an $n \times n$ matrix $A$, the determinant of its adjoint is $|A|^{n-1}$. Let’s see why this is true.

Method: Chaining |A| and |adj A| from a Given Scalar Matrix A(adj A)

This method handles any question that gives you the matrix $A\cdot\operatorname{adj}(A)$ as a scalar multiple of the identity and asks for some combination of $|A|$ and $|\operatorname{adj}A|$.

Steps

Step 1: Read off |A| directly from the given product

Since $A\cdot\operatorname{adj}(A)=|A|\,I_n$ always, if you are told $A\cdot\operatorname{adj}(A)=k\,I_n$ for some number $k$, then immediately $|A|=k$ — no computation needed, just matching the identity to the given matrix.

Step 2: Use the order relation to get |adj A| …

Common Mistakes

Mistake 1: Using exponent $n$ instead of $n-1$ for $|\text{adj }A|$

Why it's wrong: for a $3\times3$ matrix, $|\text{adj }A|=|A|^{n-1}=|A|^2$, not $|A|^3$ — using the wrong power gives $|\text{adj }A|=2025^3$ and a completely wrong final sum. Correct approach: always subtract one from the order before applying the exponent; here $n=3$ gives $|\text{adj }A|=2025^2$.

Mistake 2: Stopping after finding only one of $|A|$ or $|\text{adj }A|$

Why it's wrong: the question asks for the sum $|A|+|\text{adj }A|$, but a student who correctly finds $|A|=2025$ can forget the question isn't just asking for $|A|$ and pick an option matching only that value. Correct approach: compute both quantities separately ($|A|=2025$, $|\text{adj }A|=2025^2$) and add them before matching an option.

Mistake 3: Confusing $A(\text{adj }A)$ with $|\text{adj }A|$ …