Let $A=\begin{bmatrix}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{bmatrix}$, $B=[B_{1}, B_{2}, B_{3}]$, where $B_{1}$, $B_{2}$, $B_{3}$ are column matrices, and $AB_{1}=\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$, $AB_{2}=\begin{bmatrix}2 \\ 3 \\ 0\end{bmatrix}$, $AB_{3}=\begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}$ If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^{3}+\beta^{3}$ is equal to ____.

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Matrices

Let

A = 211010101

B = [B_{1}, B_{2}, B_{3}]

, where

B_{1}

B_{2}

B_{3}

are column matrices, and

A B_{1} = 100

A B_{2} = 230

A B_{3} = 321

α = ∣ B ∣

and

β

is the sum of all the diagonal elements of

B

, then

α^{3} + β^{3}

is equal to ____.

Answer: 28

Solving

A B_{1} = (1, 0, 0)^{T}

2 x + z = 1, x + y = 0, x + z = 0 \Rightarrow B_{1} = (1, - 1, - 1)^{T}

.

Solving

A B_{2} = (2, 3, 0)^{T}

: gives

B_{2} = (2, 1, - 2)^{T}

.

Solving

A B_{3} = (3, 2, 1)^{T}

: gives

B_{3} = (2, 0, - 1)^{T}

B = 1 - 1 - 1 21 - 2 20 - 1

α = ∣ B ∣ = 1 (- 1) - 2 (1) + 2 (3) = - 1 - 2 + 6 = 3, β = 1 + 1 - 1 = 1

α^{3} + β^{3} = 27 + 1 = 28

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Integer answer