Animated Solution for Mathematics - Complex Numbers: Let $\alpha$ be a solution of $x^2+x+1=0$, and for some $a$ and $b$ in $\mathbb{R}$, $[4ab]\left[\begin{array}{ccc}1& 16& 13\\ -1& -1& 2\\ -2& -14& -8\end{array}\right]=[000]$. If $\frac{4}{{\alpha }^4}+\frac{m}{{\alpha }^a}+\frac{n}{{\alpha }^b}=3$, then $m+n$ is equal to

Welcome, student. Today, we are not just solving a problem; we are conducting a symphony. We have a matrix equation dancing with complex numbers, and our job is to find the harmony between them.

This problem is a classic JEE Advanced challenge—it tests your ability to switch gears between linear algebra and the elegant properties of the cube roots of unity. Let us dive in.

The Gateway of Roots

We begin with the equation $x^2+x+1=0$. If you have spent enough time in the trenches of JEE preparation, this equation should feel like an old friend.

It is the cyclotomic polynomial for the cube roots of unity. The roots are not real; they are the complex numbers $\omega$ and ${\omega }^2$, where $\omega =e^{i\frac{2\pi }{3}}$.

Let us set $\alpha =\omega$. The properties we rely on are the bedrock of complex number theory: 1. ${\omega }^3=1$ 2. $1+\omega +{\omega }^2=0$

From the second property, we can derive the crucial identity ${\omega }^2=-1-\omega$. Keep this in your toolkit; we will need it to simplify our final expression later.

Decoding the Matrix

Now, let us turn to the matrix equation:

Do not let the matrix notation intimidate you. It is merely a compact way of presenting a system of linear equations. We perform row-by-column multiplication.

For the first column, we have:

For the second column, we have:

We now have a simple system of two linear equations. Subtracting the first from the second, we get $(a+14b)-(a+2b)=64-4$, which simplifies to $12b=60$, giving us $b=5$.

Substituting this back into $a+2b=4$, we find $a+10=4$, so $a=-6$. The matrix has surrendered its secrets: $a=-6$ and $b=5$.

The Synthesis

Now, we bring the two worlds together. We substitute $\alpha =\omega$, $a=-6$, and $b=5$ into the target expression:

This looks messy, but let us apply our properties: - ${\omega }^4={\omega }^3\cdot \omega =1\cdot \omega =\omega$ - ${\omega }^{-6}=({\omega }^3{)}^{-2}=1^{-2}=1$ - ${\omega }^5={\omega }^3\cdot {\omega }^2=1\cdot {\omega }^2={\omega }^2$

Our equation transforms into: $\frac{4}{\omega }+m+\frac{n}{{\omega }^2}=3$.

To clear the denominators, remember that $\frac{1}{\omega }={\omega }^2$ and $\frac{1}{{\omega }^2}=\omega$. Thus, the equation becomes:

The Final Cancellation

We are almost there. We have a mix of ${\omega }^2$ and $\omega$. To compare coefficients, we must eliminate ${\omega }^2$ using our identity ${\omega }^2=-1-\omega$:

Grouping the real and $\omega$ terms:

By comparing the coefficients, we see that the real part $m-4$ must equal $3$, so $m=7$. The coefficient of $\omega$, which is $n-4$, must equal $0$, so $n=4$.

The final step is simply to calculate $m+n=7+4=11$.

Look at what you have achieved. You navigated through matrix algebra, invoked the properties of complex roots, and performed a clean algebraic reduction. The final answer is 11.