Animated Solution for Mathematics - Matrices and Determinants: For a real number $a$, if the system $\left[\begin{array}{ccc}1& a& a^2\\ a& 1& a\\ a^2& a& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$ of linear equations, has infinitely many solutions, then $1+a+a^2=$

Unlocking the Matrix

Welcome, future engineers. Today, we are not just solving a system of equations; we are peeling back the layers of a geometric mystery.

When you see a matrix equation like $AX=B$, do not just see rows and columns of numbers. See a collection of planes in three-dimensional space.

The question is: how do these planes interact? Do they meet at a single point, do they never meet, or do they dance together along a common line? Let us embark on this journey to find the truth behind the parameter $a$.

The Determinant Dance

Our journey begins with the coefficient matrix $A=\left[\begin{array}{ccc}1& a& a^2\\ a& 1& a\\ a^2& a& 1\end{array}\right]$. For a system to have infinitely many solutions, the system must be singular.

This means the determinant of the coefficient matrix, $|A|$, must be zero. If $|A| eq 0$, the system would have a unique solution, which is not what we are looking for.

So, we set the stage by calculating the determinant:

Expanding along the first row, we get:

Look closely at the terms. The last term, $a^2(a^2-a^2)$, vanishes into thin air. We are left with $1-a^2-a^2+a^4$.

This simplifies beautifully to $a^4-2a^2+1$. Does this look familiar? It is a perfect square! We can write it as:

Setting this to zero, we find our candidates: $a^2=1$, which gives us $a=1$ or $a=-1$. But hold your horses! We have only just begun.

The JEE Trap

Here is where the battle is won or lost. Many students stop here, assuming that because the determinant is zero, the system must have infinite solutions.

But remember, in the world of linear algebra, $|A|=0$ is a necessary condition, but it is not sufficient. It is a gateway that leads to two possible paths: the path of 'No Solution' and the path of 'Infinite Solutions'.

We must test our candidates to see which path they lead us down.

Testing the Candidates

Let us test $a=1$ first. Substituting $a=1$ into our system, the equations become:

Stop and visualize this. We have two planes with the same normal vector $(1,1,1)$, but they are separated by a distance. They are parallel and distinct.

They will never intersect. Thus, for $a=1$, the system is inconsistent. There is no solution. We must reject this value.

Now, let us test $a=-1$. Substituting $a=-1$ into the system, we get:

Look at the second equation: $-x+y-z=-1$. If you multiply this entire equation by $-1$, you get $x-y+z=1$.

It is identical to the first and third equations! All three equations represent the exact same plane. They are coincident.

This means every point on that plane is a solution. We have found our winner: $a=-1$ is the key that unlocks the infinite solutions.

The Final Resolution

We have navigated the trap and found the correct parameter. The problem asks us to evaluate the expression $1+a+a^2$ for our valid $a$.

Substituting $a=-1$ into this expression, we get:

And there it is. The elegance of the result is a testament to the consistency of mathematics.

You have successfully navigated the geometry of planes, the algebra of determinants, and the logic of consistency. Keep this rigor in your toolkit, and no JEE problem will ever be able to stand in your way.