Let X = [[0, 1, 0], [0, 0, 1], [0, 0, 0]], Y = αI + βX + γX^2 and Z = α^2I - αβX + (β^2 - αγ)X^2, α,

Q: Wait, why do we completely ignore matrix $Z$? Is it not part of the equation?

This is a classic 'distractor' technique used in competitive exams like the JEE. The examiners want to see if you can identify the necessary information versus the noise. If you can filter out the irrelevant, you have already won half the battle. $Z$ is mathematically defined, but it is not required to solve for the variables $\alpha, \beta,$ and $\gamma$ given the inverse of $Y$.

Q: Why can't I just use the formula for the inverse of a $3 \times 3$ matrix? It seems more systematic.

You *could*, but in the JEE, time is your most precious currency. Calculating the determinant and the adjoint of a matrix with variables is prone to arithmetic errors and is incredibly time-consuming. The property $Y Y^{-1} = I$ is the 'surgical' approach—it cuts straight to the answer by leveraging the structure of the matrices provided.

Q: How do I know $X^3$ is zero without multiplying it out?

Look at the pattern of $X$. It shifts elements to the right. $X^2$ shifts them further. $X^3$ shifts them off the matrix entirely! This is the beauty of nilpotent matrices—they simplify complex polynomials into finite, manageable terms. Whenever you see a matrix with zeros on and below the diagonal, always check if it is nilpotent.

Question

Let $X = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}, Y = \alpha I + \beta X + \gamma X^2$ and $Z = \alpha^2 I - \alpha\beta X + (\beta^2 - \alpha\gamma)X^2, \alpha, \beta, \gamma \in \mathbb{R}$. If $Y^{-1} = \begin{bmatrix} 1/5 & -2/5 & 1/5 \ 0 & 1/5 & -2/5 \ 0 & 0 & 1/5 \end{bmatrix}$, then $(\alpha - \beta + \gamma)^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Nilpotent Matrix

We are given the matrix $X = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$. Observe that $X$ is a strictly upper triangular matrix.

When we compute the powers of $X$, we find:
$$X^2 = \begin{bmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$$
$$X^3 = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$$
This confirms that $X$ is a **nilpotent matrix** of index 3.

### Constructing the Matrix Y

Given the expression $Y = \alpha I + \beta X + \gamma X^2$, we substitute the matrices to form $Y$:
$$Y = \alpha \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix} + \gamma \begin{bmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \alpha & \beta & \gamma \ 0 & \alpha & \beta \ 0 & 0 & \alpha \end{bmatrix}$$

### Solving for Parameters

We avoid the adjoint method and use the fundamental property $Y Y^{-1} = I$. We are given:
$$Y^{-1} = \begin{bmatrix} \frac{1}{5} & -\frac{2}{5} & \frac{1}{5} \ 0 & \frac{1}{5} & -\frac{2}{5} \ 0 & 0 & \frac{1}{5} \end{bmatrix}$$

By performing matrix multiplication $\begin{bmatrix} \alpha & \beta & \gamma \ 0 & \alpha & \beta \ 0 & 0 & \alpha \end{bmatrix} \begin{bmatrix} \frac{1}{5} & -\frac{2}{5} & \frac{1}{5} \ 0 & \frac{1}{5} & -\frac{2}{5} \ 0 & 0 & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$, we equate the elements:

1. For the first column: $\alpha \left( \frac{1}{5} ight) = 1 \Rightarrow \alpha = 5$.
2. For the second column: $\alpha \left( -\frac{2}{5} ight) + \beta \left( \frac{1}{5} ight) = 0$. Substituting $\alpha = 5$, we get $-2 + \frac{\beta}{5} = 0 \Rightarrow \beta = 10$.
3. For the third column: $\alpha \left( \frac{1}{5} ight) + \beta \left( -\frac{2}{5} ight) + \gamma \left( \frac{1}{5} ight) = 0$. Substituting $\alpha = 5$ and $\beta = 10$, we get $1 - 4 + \frac{\gamma}{5} = 0 \Rightarrow \gamma = 15$.

### Final Calculation

We have determined the values to be $\alpha = 5$, $\beta = 10$, and $\gamma = 15$. The problem asks for the value of $(\alpha - \beta + \gamma)^2$.

Substituting the values:
$$(5 - 10 + 15)^2 = (10)^2 = 100$$

The final answer is **100**.

Animated Solution for Mathematics - Matrices and Determinants: Let $X = 000100010, Y = α I + β X + γ X^{2}$ and $Z = α^{2} I - α β X + (β^{2} - α γ) X^{2}, α, β, γ \in R$ . If $Y^{- 1} = 1/5 00 - 2/5 1/5 0 1/5 - 2/5 1/5$ , then $(α - β + γ)^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Analyzing the Given Matrices

Properties of Matrix $X$

Calculating $X^{2}$

Constructing Matrix $Y$

The Explicit Form of $Y$

The Inverse Property

Matrix Multiplication Setup

Solving for $α$

Solving for $β$

Solving for $γ$

Evaluating $α - β + γ$

Final Answer

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Nilpotent Matrix

Constructing the Matrix Y

Solving for Parameters

Final Calculation

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Visualized Solution

Analyzing the Given Matrices

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X2

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

The Explicit Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

The Inverse Property

Matrix Multiplication Setup

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

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

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

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α−β+γ