Let \(X=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right], Y=\alpha I+\beta X+\gamma X^{2}\) and \(\mathrm{Z}=\alpha^{2} \mathrm{I}-\alpha \beta \mathrm{X}+\left(\beta^{2}-\alpha \gamma\right) \mathrm{X}^{2}, \alpha, \beta, \gamma \in \mathbb{R}\). If \(\mathrm{Y}^{-1}= \left[\begin{array}{ccc}\frac{1}{5} & \frac{-2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{-2}{5} \\ 0 & 0 & \frac{1}{5}\end{array}\right]\), then \((\alpha-\beta+\gamma)^{2}\) is equal to ______.

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Matrices

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} = \frac{1}{5} 00 \frac{- 2}{5} \frac{1}{5} 0 \frac{1}{5} \frac{- 2}{5} \frac{1}{5}

, then

(α - β + γ)^{2}

is equal to ______.

Answer: 100

First compute

X^{2} = 000000100

.

Then

Y = α 00 β α 0 γ β α

.

From

Y \cdot Y^{- 1} = I

, multiply the matrices and equate:

From the (1,1) entry:

\frac{α}{5} = 1

, so

α = 5

.

From the (1,2) entry:

\frac{- 2 α + β}{5} = 0

, so

β = 10

.

From the (1,3) entry:

\frac{α - 2 β + γ}{5} = 0

, so

γ = 15

.

Therefore

(α - β + γ)^{2} = (5 - 10 + 15)^{2} = 1 0^{2} = 100

.

Answer: 100

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer