JEE Main 2024 — 27 January, Shift 1 — Question 27

Given the matrix equation $A B_i = C_i$, we can combine these into a single matrix equation $AB = C$, where:

$$A = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$$

Finding $\alpha = |B|$ Using the property of determinants $|AB| = |A||B|$:

$$|A| = \begin{vmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix} = 2(1-0) - 0(1-0) + 1(0-1) = 2 - 1 = 1$$ $$|C| = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{vmatrix} = 1 \times 3 \times 1 = 3$$

Since $|A||B| = |C|$, we have:

$$1 \cdot \alpha = 3 \implies \alpha = 3$$

Finding $\beta = \text{tr}(B)$ Since $AB = C$, then $B = A^{-1}C$. First, we find $A^{-1}$:

$$\text{adj}(A) = \begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{bmatrix}$$

Since $|A| = 1$, $A^{-1} = \text{adj}(A)$. Now compute the diagonal elements of $B$:

$$B = \begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$$

The diagonal elements are: $b_{11} = (1)(1) + (0)(0) + (-1)(0) = 1$ $b_{22} = (-1)(2) + (1)(3) + (1)(0) = 1$ $b_{33} = (-1)(3) + (0)(2) + (2)(1) = -1$

$$\beta = \text{tr}(B) = 1 + 1 - 1 = 1$$

Final Result

$$\alpha^3 + \beta^3 = 3^3 + 1^3 = 27 + 1 = 28$$