Let A = [[i, -i], [-i, i]], i = √-1. Then, the system of linear equations A⁸ [x, y]ᵀ = [8, 64]ᵀ has

Q: Why did we calculate $A^8$ instead of using the determinant to check for a unique solution?

That is a great question! While the determinant is a powerful tool for checking if a system has a unique solution (if $\det(A) \neq 0$), in this case, the determinant of $A$ is $i^2 - (-i)(-i) = -1 - 1 = -2$. Wait, the determinant is not zero! However, the system is $A^8 X = B$. Even if $A$ is invertible, we still need to solve the system. Calculating $A^8$ is the most direct way to see the actual equations and determine if they are consistent or inconsistent.

Q: Is there a faster way to find $A^8$ than repeated squaring?

You could use diagonalization, but for a $2 \times 2$ matrix, repeated squaring is often faster and less prone to algebraic errors. The matrix $A$ has a very simple structure, and the powers $A^2, A^4, A^8$ follow a clear pattern. Sometimes, the simplest path is the most reliable one!

Q: What if the constants in the two equations were the same? Would there be infinitely many solutions?

Exactly! If the two equations were identical (e.g., $x - y = \frac{1}{16}$ and $x - y = \frac{1}{16}$), the lines would be coincident, meaning they lie on top of each other. In that case, every point on the line would be a solution, leading to infinitely many solutions. But here, the constants are different, which makes the lines parallel and distinct, leading to no solution.

Question

Let $A = \begin{bmatrix} i & -i \ -i & i \end{bmatrix}, i = \sqrt{-1}$. Then, the system of linear equations $A^8 \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 8 \ 64 \end{bmatrix}$ has :

Options:

(A) A unique solution

(B) Infinitely many solutions

(C) No solution

(D) Exactly two solutions

Sigma Percentile · Accepted Answer

### The Mystery of the Imaginary Matrix

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to demystify a matrix problem that looks intimidating at first glance because of the imaginary unit $i$.

But fear not! As we peel back the layers, you will see that this problem is not about complex calculations, but about recognizing patterns and the geometric soul of linear equations.

### Phase 1: The Power of Patterns

We are given the matrix $A = \begin{bmatrix} i & -i \ -i & i \end{bmatrix}$. Our mission is to solve the system $A^8 \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 8 \ 64 \end{bmatrix}$.

In JEE Advanced, we look for elegance, not brute force. Let us use the strategy of repeated squaring to avoid multiplying the matrix eight times.

First, let us find $A^2$:
$$A^2 = \begin{bmatrix} i & -i \ -i & i \end{bmatrix} \begin{bmatrix} i & -i \ -i & i \end{bmatrix} = \begin{bmatrix} i^2 + (-i)(-i) & -i^2 + (-i)(i) \ -i^2 + (i)(-i) & (-i)(-i) + i^2 \end{bmatrix} = \begin{bmatrix} -2 & 2 \ 2 & -2 \end{bmatrix}$$

To make our lives easier, let us factor out a $2$:
$$A^2 = 2 \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix}$$

Now, let us find $A^4$ by squaring $A^2$. Remember to square the scalar $2$ as well:
$$A^4 = \left( 2 \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} ight)^2 = 4 \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} = 4 \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} = 8 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix}$$

Finally, we reach $A^8$ by squaring $A^4$:
$$A^8 = \left( 8 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} ight)^2 = 64 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} = 64 \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} = 128 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix}$$

### Phase 2: The System Reveal

Now that we have $A^8 = 128 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix}$, let us plug this into our system:
$$128 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 8 \ 64 \end{bmatrix}$$

This matrix equation expands into two linear equations:

1. $128(x - y) = 8 \implies x - y = \frac{8}{128} = \frac{1}{16}$

2. $128(-x + y) = 64 \implies -x + y = \frac{64}{128} = \frac{1}{2} \implies x - y = -\frac{1}{2}$

### Phase 3: The Geometric Insight

Look closely at these two equations: $x - y = \frac{1}{16}$ and $x - y = -\frac{1}{2}$. The left sides are identical, but the right sides are different!

Geometrically, these represent two parallel lines with the same slope but different y-intercepts. Parallel lines, as we know, never intersect.

Since the solution to a system of linear equations is the intersection point of the lines, and these lines never meet, the system is inconsistent. Therefore, there is **no solution**. You have just conquered a complex-looking matrix problem with nothing but patience and a keen eye for patterns.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [i - i - i i], i = - 1$ . Then, the system of linear equations $A^{8} [x y] = [864]$ has :

Select Answer:

Visualized Solution

Matrix $A$ and the Power $A^{8}$

Calculating $A^{2}$

Calculating $A^{4}$

Calculating $A^{8}$

Setting up the System

Equation 1: $x - y = \frac{1}{16}$

Equation 2: $x - y = - \frac{1}{2}$

Consistency Check

Conclusion: No Solution

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

The Mystery of the Imaginary Matrix

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=[i−i​−ii​],i=−1​. Then, the system of linear equations A8[xy​]=[864​] has :

Select Answer:

Visualized Solution

Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and the Power .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A8

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Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A8