Animated Solution for Mathematics - Matrices and Determinants: Let $A+2B=\left[\begin{array}{ccc}1& 2& 0\\ 6& -3& 3\\ -5& 3& 1\end{array}\right]$ and $2A-B=\left[\begin{array}{ccc}2& -1& 5\\ 2& -1& 6\\ 0& 1& 2\end{array}\right]$. If $\text{Tr}(A)$ denotes the sum of all diagonal elements of the matrix $A$, then $\text{Tr}(A)-\text{Tr}(B)$ has value equal to

Imagine you are standing on the edge of a vast, complex matrix algebra problem. You see two matrix equations, $A+2B=M_1$ and $2A-B=M_2$, and your first instinct might be to dive in and solve for every single element of $A$ and $B$.

You might start writing out nine equations for $A$ and nine for $B$, drowning in a sea of variables and arithmetic. Stop! Take a breath.

In the world of JEE Advanced, the most elegant path is rarely the one that requires the most writing. This problem is a classic test of your ability to recognize the power of operators.

The Trace Operator

Instead of solving for the matrices, let us look at what the question actually asks for: $\text{Tr}(A)-\text{Tr}(B)$. The trace, denoted as $\text{Tr}(A)$, is simply the sum of the diagonal elements of a square matrix.

It is a single scalar value that captures a specific 'signature' of the matrix. The key to this problem is the linearity of the trace operator.

This property tells us that $\text{Tr}(kA+mB)=k\text{Tr}(A)+m\text{Tr}(B)$. This is our mathematical superpower. It means we do not need to know the individual elements of $A$ and $B$; we only need to know their traces.

Let us apply this power to our given equations. For the first equation, $A+2B=M_1$, where:

We take the trace of both sides. The left side becomes $\text{Tr}(A)+2\text{Tr}(B)$. The right side is the trace of the matrix, which is the sum of its diagonal elements: $1+(-3)+1=-1$.

Thus, we have our first simple equation: $\text{Tr}(A)+2\text{Tr}(B)=-1$. See how the complexity just melted away?

Now, let us do the same for the second equation, $2A-B=M_2$, where:

Taking the trace of both sides gives us $2\text{Tr}(A)-\text{Tr}(B)$. The trace of the right-hand matrix is $2+(-1)+2=3$. So, our second equation is $2\text{Tr}(A)-\text{Tr}(B)=3$.

We have successfully transformed a daunting matrix problem into a simple system of two linear equations with two variables. Let $x=\text{Tr}(A)$ and $y=\text{Tr}(B)$. Our system is:

To solve this, we can multiply the second equation by $2$ to get $4x-2y=6$. Now, add this to the first equation: $(x+2y)+(4x-2y)=-1+6$.

This simplifies to $5x=5$, which gives us $x=1$. Substituting $x=1$ back into the first equation, we get $1+2y=-1$, which means $2y=-2$, so $y=-1$.

We have found that $\text{Tr}(A)=1$ and $\text{Tr}(B)=-1$. The final step is to calculate $\text{Tr}(A)-\text{Tr}(B)$, which is $1-(-1)=2$.

The elegance of this solution lies in how we bypassed the matrix elements entirely, using the properties of the trace to reach the answer with precision and speed. Remember, in mathematics, the most powerful tool is often the one that simplifies the problem the most. The final answer is 2.