Animated Solution for Mathematics - Matrices and Determinants: Find the solution set of the system $x+2y+z=1;2x-3y-w=2;x\ge 0;y\ge 0;z\ge 0;w\ge 0$.

Solving Underdetermined Systems

Welcome, fellow traveler of the JEE landscape. You look at this system: $x+2y+z=1$ and $2x-3y-w=2$. Your heart sinks. Four variables, two equations.

The ghost of linear algebra past whispers, "This is underdetermined; there must be infinite solutions." But wait! Before you rush to write down a parametric solution, look closer.

The constraints $x\ge 0,y\ge 0,z\ge 0,w\ge 0$ are not just decorations; they are the keys to the kingdom. In this article, we will learn how to use these constraints to "squeeze" the solution out of thin air.

Usually, when we see fewer equations than variables, we expect a line, a plane, or a higher-dimensional manifold of solutions. However, in competitive exams like the JEE, constraints are often the hidden boundary conditions that turn an infinite problem into a unique one.

We are looking for a point in the first orthant of 4D space. Let's see how we can isolate our variables.

Isolating $x$

Let's focus on the first equation: $x+2y+z=1$. We want to isolate $x$ because it appears in both equations. Rearranging this, we get:

Now, apply the constraints. We know that $y\ge 0$ and $z\ge 0$. Therefore, the term $(2y+z)$ must be greater than or equal to zero.

If you subtract a non-negative number from $1$, the result must be less than or equal to $1$. Thus, we have our first crucial inequality:

The Lower Bound

Now, let's shift our attention to the second equation: $2x-3y-w=2$. Let's rearrange this to isolate the $x$ term:

Again, look at the constraints. Since $y\ge 0$ and $w\ge 0$, the expression $(3y+w)$ is non-negative. This means $2x$ is equal to $2$ plus some non-negative value.

Therefore, $2x$ must be greater than or equal to $2$. Dividing by $2$, we get:

Look at what we have achieved! We have established that $x\le 1$ and $x\ge 1$. On the real number line, there is only one point that satisfies both conditions simultaneously: $x=1$.

We have successfully "squeezed" the variable $x$ into a single, unique value. This is the power of constraints in action.

Now that we have $x=1$, the rest of the system collapses like a row of dominoes. Substitute $x=1$ back into our first equation:

As we discussed in our FAQ, the sum of two non-negative numbers can only be zero if both are zero. Thus, $y=0$ and $z=0$.

Finally, substitute $x=1$ and $y=0$ into the second equation:

We have arrived at the unique solution $(x,y,z,w)=(1,0,0,0)$. It is a beautiful result.

What initially looked like a daunting, underdetermined system was actually a perfectly constrained puzzle. Always remember: when you see constraints in a system of equations, don't ignore them.

They are often the most important part of the problem, waiting to be used to simplify your path to the answer. Keep practicing, keep visualizing, and keep falling in love with the elegance of mathematics!