Let α be a root of equation x² + x + 1 = 0 and the matrix A = (1/√3) * [[1, 1, 1], [1, α, α²], [1, α

Q: Wait, why do we assume $\alpha = \omega$? Could it be $\omega^2$?

That is a brilliant observation! If you chose $\alpha = \omega^2$, the matrix $A$ would simply swap the positions of $\omega$ and $\omega^2$. However, because the properties $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$ hold true for both roots, the final result for $A^{31}$ remains invariant. The symmetry of the roots ensures the outcome is the same regardless of your choice.

Q: Is there a shortcut to multiplying these matrices without doing all 9 elements?

While you could use properties of circulant matrices, calculating the first few rows is actually the safest way to avoid errors in an exam setting. Trust the process! The 'shortcut' is realizing that the sum of the roots property will collapse most of your terms to zero, making the arithmetic much faster than you fear.

Q: Why does $A^4 = I$ matter so much?

It defines the 'period' of the matrix. In JEE Advanced, whenever you see high powers like $A^{31}$, you are almost certainly looking for a cycle. Once you find that $A^4 = I$, you have unlocked the ability to reduce any power $A^n$ to a much smaller power, effectively turning a massive calculation into a trivial one.

Question

Let $\alpha$ be a root of equation $x^2 + x + 1 = 0$ and the matrix $A = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \ 1 & \alpha & \alpha^2 \ 1 & \alpha^2 & \alpha \end{pmatrix}$, then the matrix $A^{31}$ is equal to :

Options:

(A) $A$

(B) $A^2$

(C) $A^3$

(D) $A^4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The matrix is given by:

$$A = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \ 1 & \alpha & \alpha^2 \ 1 & \alpha^2 & \alpha \end{pmatrix}$$

We are given that $\alpha$ is a root of $x^2 + x + 1 = 0$. This is the classic cyclotomic equation, and its roots are the complex cube roots of unity, $\omega$ and $\omega^2$.

Let us set $\alpha = \omega$. Recall the two fundamental properties: $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. These identities will simplify our matrix operations significantly.

### The Dance of Multiplication: Calculating $A^2$

To find $A^{31}$, we must first identify a pattern by calculating $A^2 = A \cdot A$. The scalar $\frac{1}{\sqrt{3}}$ squared gives us $\frac{1}{3}$ outside the matrix.

Performing the matrix multiplication:

$$A^2 = \frac{1}{3} \begin{pmatrix} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{pmatrix}$$

For the first row, second column, we obtain $1(1) + 1(\omega) + 1(\omega^2) = 1 + \omega + \omega^2 = 0$. As we continue this process, the off-diagonal elements vanish, and the diagonal elements align.

We arrive at the following result:

$$A^2 = \begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{pmatrix}$$

### The Revelation: Finding the Cycle

We have determined $A^2$. Now, let us find $A^4$ by calculating $A^2 \cdot A^2$:

$$A^4 = \begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = I$$

This is the breakthrough. We have discovered that the matrix $A$ is periodic with a period of $4$, meaning $A^4 = I$.

### The Final Leap: Conquering $A^{31}$

We need to evaluate $A^{31}$. We can express the exponent as $31 = 28 + 3$.

Since $28$ is a multiple of $4$, we know that $A^{28} = (A^4)^7 = I^7 = I$. Therefore:

$$A^{31} = A^{28} \cdot A^3 = I \cdot A^3 = A^3$$

Finally, we calculate $A^3 = A^2 \cdot A$:

$$A^3 = \begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{pmatrix} \cdot \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{pmatrix} = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \ 1 & \omega^2 & \omega \ 1 & \omega & \omega^2 \end{pmatrix}$$

The final result is **$\frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 1 & 1 \ 1 & \omega^2 & \omega \ 1 & \omega & \omega^2 \end{pmatrix}$**.

Animated Solution for Mathematics - Matrices and Determinants: Let $α$ be a root of equation $x^{2} + x + 1 = 0$ and the matrix $A = \frac{1}{3} 111 1 α α^{2} 1 α^{2} α$ , then the matrix $A^{31}$ is equal to :

Select Answer:

Visualized Solution

Identify $α$ as a Cube Root of Unity

Recall Properties of $ω$

Substitute $α = ω$ in Matrix $A$

Setup for $A^{2}$ Calculation

Compute Row 1 of $A^{2}$

Compute Row 2 of $A^{2}$

Compute Row 3 of $A^{2}$

Simplify $A^{2}$ Result

Calculate $A^{4}$

Evaluate $A^{31}$ using Periodicity

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Dance of Multiplication

The Revelation

The Final Leap

Similar Questions

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Select Answer:

Visualized Solution

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α as a Cube Root of Unity

Recall Properties of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=ω in Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2 Calculation

Compute Row 1 of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Compute Row 2 of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Compute Row 3 of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2 Result

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A4

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A31 using Periodicity