Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{cc}2& 3\\ a& 0\end{array}\right],a\in \mathbb{R}$ be written as $P+Q$ where $P$ is a symmetric matrix and $Q$ is a skew symmetric matrix. If $\det (Q)=9$, then the modulus of the sum of all possible values of determinant of $P$ is equal to :

A Journey into Decomposition

Welcome, future engineer. Today, we are not just solving a matrix problem; we are peeling back the layers of linear algebra to understand the very anatomy of a matrix.

You have been given a matrix $A=\left[\begin{array}{cc}2& 3\\ a& 0\end{array}\right]$ and told that it can be decomposed into two distinct entities: $P$ and $Q$. This is not merely an algebraic trick; it is a fundamental truth of linear algebra.

Every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix. Think of this like breaking a complex force vector into its horizontal and vertical components. It simplifies the chaos into order.

The Anatomy of Symmetry

Let us define our players. We are told $P$ is symmetric, meaning $P^T=P$, and $Q$ is skew-symmetric, meaning $Q^T=-Q$. How do we extract these from $A$?

The beauty of linear algebra gives us the formulas directly:

Before we rush into the calculation, pause and visualize what is happening. When we add $A$ to its transpose $A^T$, the off-diagonal elements align in a way that creates symmetry. When we subtract $A^T$ from $A$, the diagonal elements cancel out to zero, and the off-diagonal elements become negatives of each other.

This is the signature of skew-symmetry. Let us write down $A^T=\left[\begin{array}{cc}2& a\\ 3& 0\end{array}\right]$ and prepare for the construction.

The Skew-Symmetric Constraint

We are given a specific constraint: $\det (Q)=9$. This is our anchor. Let us construct $Q$ explicitly.

Using our formula, we have:

Look at that matrix. The diagonal is zero, and the off-diagonal elements are negatives of each other. It is perfectly skew-symmetric.

Now, let us calculate the determinant. For a $2\times 2$ matrix, the determinant is $ad-bc$. Here, that is $(0)(0)-(\frac{3-a}{2})(\frac{a-3}{2})$.

We set this equal to $9$. This leads us to $(a-3{)}^2=36$.

Here is where the trap lies. When you take the square root, you must account for both the positive and negative roots. Thus, $a-3=6$ or $a-3=-6$.

This gives us two possible values for our parameter: $a=9$ and $a=-3$. Do not discard either one! Both are valid realities of this matrix.

The Symmetric Reality

Now that we have our values for $a$, we turn our attention to $P$. Using the formula $P=\frac{1}{2}(A+A^T)$, we construct:

Again, notice the symmetry. The off-diagonal elements are identical. Now, let us find the determinant of $P$:

This is our general expression for the determinant of the symmetric part. We have two cases to evaluate.

The Final Synthesis

For $a=9$, we have:

For $a=-3$, we have:

The problem asks for the modulus of the sum of all possible values of the determinant of $P$. The sum is $-36+0=-36$.

The modulus of $-36$ is simply $36$.

We have navigated the decomposition, respected the constraints, and arrived at the solution. Remember, in JEE Advanced, the math is rarely just about calculation; it is about understanding the structure. You have done exactly that. Keep this clarity, and you will conquer any matrix that comes your way.