Animated Solution for Mathematics - Matrices and Determinants: Let $A,B$ and $C$ be three $2\times 2$ matrices with real entries such that $B=(I+A{)}^{-1}$ and $A+C=I$. If $BC=\left[\begin{array}{cc}1& -5\\ -1& 2\end{array}\right]$ and $CB\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}12\\ -6\end{array}\right]$, then $x_1+x_2$ is

Imagine you are standing before a complex system of equations. Your instinct might be to dive in, calculate every element, and solve for $A$, $B$, and $C$ individually.

But in the world of JEE Advanced, the most elegant path is rarely the one that requires the most arithmetic. Today, we are going to unravel a matrix problem that rewards insight over brute force.

We are given $B=(I+A{)}^{-1}$ and $A+C=I$. The first thing we must do is understand the relationship between $A$ and $B$.

By the definition of an inverse, we know that $B(I+A)=(I+A)B=I$. If we expand this, we get $B+BA=B+AB=I$.

By subtracting $B$ from both sides, we arrive at the beautiful realization that $BA=AB$. This means $A$ and $B$ commute, which is the key that unlocks the entire problem.

Next, we look at $C$. We are told $A+C=I$, which means $C=I-A$.

Now, we want to evaluate $CB$. Substituting our expression for $C$, we get $CB=(I-A)B=B-AB$.

Similarly, for $BC$, we have $BC=B(I-A)=B-BA$. Because we already established that $AB=BA$, it follows that $B-AB=B-BA$.

Therefore, $CB=BC$. This is a massive simplification; we do not need to find $B$ or $C$ individually. We simply need to know that $CB$ is equal to the matrix $BC$, which is already given to us as:

Now, the problem becomes a simple system of linear equations. We have:

This translates to two linear equations:

1) $x_1-5x_2=12$

2) $-x_1+2x_2=-6$

If we add these two equations together, the $x_1$ terms cancel out perfectly, leaving us with $-3x_2=6$, which means $x_2=-2$.

Substituting $x_2=-2$ into the second equation, we get $-x_1+2(-2)=-6$, which simplifies to $-x_1-4=-6$, or $-x_1=-2$, so $x_1=2$.

The question asks for the sum $x_1+x_2$. With $x_1=2$ and $x_2=-2$, the sum is:

By looking for the underlying structure of the matrices rather than getting lost in the algebra, we have arrived at the answer of 0 with elegance and precision. Keep this mindset for your exams—always look for the property that simplifies the problem before you start calculating!