Let A be a 3 × 3 real matrix such that A[1, 0, 1]^T = 2[1, 0, 1]^T, A[-1, 0, 1]^T = 4[-1, 0, 1]^T, A

Q: Wait, why do we care about eigenvalues here? Isn't this just a system of linear equations?

That is a brilliant observation! While you could try to solve for the individual entries of $A$ by setting up a system of 9 equations, that would be a nightmare. Eigenvalues are the 'DNA' of a matrix. By identifying them, we bypass the tedious algebra and look directly at the matrix's core behavior.

Q: How do I know for sure that the three vectors given are linearly independent?

Great catch! If they weren't independent, we wouldn't have a full set of eigenvalues. However, you can check their determinant or simply notice they represent distinct directions in 3D space. Since they are eigenvectors corresponding to eigenvalues, they span the space effectively here.

Q: Why does $\det(A - 3I) \neq 0$ guarantee a unique solution?

Think of the determinant as the 'scaling factor' of the transformation. If the determinant is non-zero, the transformation is invertible. If you can 'undo' the transformation (by multiplying by the inverse), there is only one possible starting point for any destination vector $B$.

Q: Is it possible for $(A - 3I)$ to have a zero eigenvalue?

Yes, if $3$ were an eigenvalue of $A$. If $3$ were an eigenvalue, then $A - 3I$ would have a zero eigenvalue, making its determinant zero. In that case, the system would either have no solution or infinitely many. But here, our eigenvalues are $2, 4, 2$, so $3$ is not among them!

Q: What if the matrix $A$ wasn't diagonalizable?

Even if $A$ is not diagonalizable, the relationship between the determinant and the product of eigenvalues (the characteristic polynomial) still holds true! The determinant is always the product of the eigenvalues, counting algebraic multiplicity.

Question

Let $A$ be a $3 	imes 3$ real matrix such that $A \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} = 2 \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}, A \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix} = 4 \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}, A \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} = 2 \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}$. Then, the system $(A - 3I) \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix}$ has

Options:

(A) unique solution

(B) exactly two solutions

(C) no solution

(D) infinitely many solutions

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine the matrix $A$ not as a grid of numbers, but as a transformation—a physical force that stretches and rotates space. When we are given equations like $A \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} = 2 \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}$, we are being given the 'secret axes' of this transformation.

These are the eigenvectors, the special directions where the matrix acts simply by scaling, not by twisting.

### Decoding the DNA of the Matrix

Look at the given equations. They represent the fundamental eigenvalue problem: $AX = \lambda X$. By comparing our given data to this identity, we extract the eigenvalues of $A$.

For the vector $\begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}$, the scaling factor is $\lambda_1 = 2$. For $\begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}$, it is $\lambda_2 = 4$. For $\begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}$, it is $\lambda_3 = 2$.

We have successfully mapped the 'DNA' of our matrix $A$. We know exactly how it behaves along these three distinct directions.

### The Shifted Reality

The problem asks us to consider the system $(A - 3I)X = B$. Resist the urge to calculate the matrix $A$ explicitly. Instead, use the **Shifted Matrix Property**.

If $A$ has an eigenvalue $\lambda$, then the matrix $(A - kI)$ has an eigenvalue of $(\lambda - k)$. By subtracting $3I$, we are essentially 're-centering' our transformation. Our new eigenvalues for the matrix $M = (A - 3I)$ become:

$$\lambda'_1 = 2 - 3 = -1$$

$$\lambda'_2 = 4 - 3 = 1$$

$$\lambda'_3 = 2 - 3 = -1$$

We have transformed our problem into a new space where the eigenvalues are $\{-1, 1, -1\}$.

### The Elegance of the Determinant

The determinant is the ultimate gatekeeper of linear systems. The determinant of any square matrix is simply the product of its eigenvalues. It tells us whether the transformation collapses space or preserves its volume.

Calculating the determinant of our shifted matrix $M$ is now a trivial task of multiplication:

$$\det(A - 3I) = (-1) 	imes (1) 	imes (-1) = 1$$

Because the determinant is $1$, which is non-zero, the matrix $(A - 3I)$ is invertible. This implies that for any vector $B$, there exists one, and only one, **unique solution** $X$ that satisfies the equation.

### Conclusion

We did not need to perform a single row reduction or solve a massive system of linear equations. By understanding the geometric soul of the matrix—its eigenvalues—we bypassed the brute force and arrived at the truth with elegance.

The system $(A - 3I)X = B$ has a **unique solution** because the transformation $(A - 3I)$ is non-singular. Keep this perspective in your toolkit, and you will find that even the most daunting JEE problems become stories of beautiful, logical unfolding.

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a $3 \times 3$ real matrix such that $A 101 = 2101, A - 1 01 = 4 - 1 01, A 010 = 2010$ . Then, the system $(A - 3 I) x y z = 123$ has

Select Answer:

Visualized Solution

Given Matrix Equations

Eigenvalue Definition

Eigenvalues of $A$

Shifted Matrix Property

Eigenvalues of $(A - 3 I)$ Setup

Computing New Eigenvalues

Determinant and Eigenvalues

Calculating Determinant

System of Linear Equations

Final Answer

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

Analyzing the Setup

Decoding the DNA of the Matrix

The Shifted Reality

The Elegance of the Determinant

Conclusion

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