The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system A [x; y; z]

Q: Wait, why can't we have exactly two solutions? Can't two planes intersect at a line and the third plane cut that line at two points?

That is a fantastic geometric intuition, but remember that a plane is a flat, infinite surface. If a plane intersects a line, it either contains the entire line or intersects it at exactly one point. It is physically impossible for a flat plane to 'curve' or 'bend' to hit a straight line at two distinct points. Therefore, you can never have exactly two solutions.

Q: Does this logic change if the matrix entries are only 0 or 1? Does that restriction allow for some weird exception?

Not at all! The restriction that entries are $0$ or $1$ limits the *number* of possible matrices we can choose from, but it does not change the fundamental laws of linear algebra. Once you pick a specific matrix $A$, the system $AX = B$ must still obey the rules of vector spaces. The nature of the solution set is determined by the rank of the matrix, not the values of its entries.

Q: Why did the question ask for 'exactly two' if it's impossible? Is it a trick?

It is not a trick; it is a conceptual test. JEE Advanced loves to move beyond rote calculation. They want to see if you understand the *structure* of linear systems. If you try to brute-force this by testing all $2^9$ matrices, you will waste precious time. Recognizing the impossibility is the mark of a true mathematician.

Question

The number of $3 	imes 3$ matrices $A$ whose entries are either 0 or 1 and for which the system $A \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$ has exactly two distinct solutions, is

Options:

(A) (a) 0

(B) (b) $2^9 - 1$

(C) (c) 168

(D) (d) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians. Today, we are not just solving a matrix problem; we are peeling back the curtain on the fundamental nature of linear algebra. You have been presented with a $3 	imes 3$ matrix $A$, filled with zeros and ones, and asked to find how many such matrices yield exactly two distinct solutions to the system $AX = B$.

At first glance, this looks like a counting problem. You might be tempted to start writing out matrices, calculating determinants, or perhaps even attempting to brute-force the possibilities. But stop. Take a breath.

In the world of JEE Advanced, the most powerful tool in your arsenal is not your speed of calculation, but your depth of conceptual understanding. Let us embark on a journey to understand why the answer to this problem is hidden in plain sight.

### The Geometric Reality: Planes in Space

Imagine you are standing in a three-dimensional room. You have three linear equations, each representing a plane. The system $AX = B$ is essentially asking: "Where do these three planes meet?"

When you have two non-parallel planes, they intersect along a single, continuous straight line. Now, bring in the third plane. What can happen?

1. The third plane could cut through that line at a single point (a unique solution).
2. The third plane could be parallel to that line, meaning there is no intersection at all (no solution).
3. The third plane could contain the entire line (infinitely many solutions).

Notice something? There is no scenario where the planes intersect at exactly two points. A plane is a flat, rigid surface. It cannot bend to touch a line at one point, leave, and then come back to touch it again. Geometry forbids it.

### The Algebraic Proof: The Power of Linear Combinations

Suppose, for the sake of argument, that the system $AX = B$ actually has two distinct solutions, $X_1$ and $X_2$. By definition, this means:

$$A X_1 = B$$

$$A X_2 = B$$

Now, let us construct a new vector, $X_3$, which is a linear combination of our two solutions. We define it as:

$$X_3 = k X_1 + (1 - k) X_2$$

Here, $k$ is any real number. This vector $X_3$ represents any point on the line connecting $X_1$ and $X_2$. Now, let us see what happens when we apply the matrix $A$ to this new vector $X_3$. Using the distributive property of matrix multiplication, we get:

$$A X_3 = A [k X_1 + (1 - k) X_2]$$

$$A X_3 = k (A X_1) + (1 - k) (A X_2)$$

Since we know that $A X_1 = B$ and $A X_2 = B$, we can substitute these values directly into our equation:

$$A X_3 = k B + (1 - k) B$$

Factor out the vector $B$:

$$A X_3 = (k + 1 - k) B$$

$$A X_3 = 1 \cdot B = B$$

### The Conclusion: The Infinite Truth

Look at that result! We have just proven that $X_3$ is a solution to the system for any real number $k$. Since there are infinitely many real numbers, there are infinitely many distinct solutions.

This is the "Aha!" moment. We started by assuming there were exactly two solutions, but our algebra forced us to conclude that if there are two, there must be infinitely many. This is a contradiction.

Therefore, our initial assumption—that a system can have exactly two solutions—is mathematically impossible. It does not matter if the entries of the matrix are $0$ or $1$, or if they are complex numbers, or transcendental constants. The structure of a linear system $AX = B$ dictates that the number of solutions can only be $0$, $1$, or $\infty$. It can never be $2$.

### Why This Matters for Your JEE Journey

This problem is a classic trap. It tests whether you will panic and start calculating, or whether you will pause and reflect on the theorems you have learned.

In the heat of the exam, remember this: linear algebra is not just about crunching numbers; it is about understanding the space in which those numbers live. Because it is impossible for any such matrix to yield exactly two solutions, the number of such matrices is **$0$**.

You have successfully navigated the trap. Keep this geometric and algebraic intuition with you, and you will find that even the most intimidating problems become clear. You are not just solving for an answer; you are mastering the language of the universe.

Animated Solution for Mathematics - Matrices and Determinants: The number of $3 \times 3$ matrices $A$ whose entries are either 0 or 1 and for which the system $A x y z = 100$ has exactly two distinct solutions, is

Select Answer:

Visualized Solution

The Matrix Equation $A X = B$

Geometric Interpretation of $A X = B$

How Planes Intersect

Fundamental Theorem of Linear Systems

Algebraic Proof: The Setup

Linear Combination of Solutions

Algebraic Verification

Generating Infinitely Many Solutions

Conclusion: Count is Zero

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Geometric Reality

The Algebraic Proof

The Conclusion

Why This Matters for Your JEE Journey

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