Q46 / 75 · JEE Main 2026 (06 April Shift 1) · Mathematics
Let $A = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ satisfy $A^2 + \alpha(adj(adj(A))) + \beta(adj(A)(adj(adj(A)))) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$ for some $\alpha, \beta \in \mathbb{R}$. Then $(\alpha - \beta)^2$ is equal to _______
Correct answer: 4
Solution
The determinant of matrix $A$ is:
$|A| = -1(0 - 0) - 1(1 - 0) - 1(0 - 0) = -1$
Using the properties of the adjoint of a $3 \times 3$ matrix:
$adj(adj(A)) = |A|^{3-2}A = -A$
$adj(A)(adj(adj(A))) = adj(A)(-A) = -|A|I = -(-1)I = I$
The given equation simplifies to:
$A^2 - \alpha A + \beta I = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$
Calculating $A^2$:
$A^2 = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Substituting $A^2$, $A$, and $I$ into the simplified equation:
$\begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$
Comparing the corresponding elements on both sides:
From the $(1,2)$ element:
$-1 - \alpha = -2 \Rightarrow \alpha = 1$
From the $(2,2)$ element:
$1 + \beta = 0 \Rightarrow \beta = -1$
Therefore, the value of $(\alpha - \beta)^2$ is:
$(1 - (-1))^2 = 2^2 = 4$
Answer: $4$
$|A| = -1(0 - 0) - 1(1 - 0) - 1(0 - 0) = -1$
Using the properties of the adjoint of a $3 \times 3$ matrix:
$adj(adj(A)) = |A|^{3-2}A = -A$
$adj(A)(adj(adj(A))) = adj(A)(-A) = -|A|I = -(-1)I = I$
The given equation simplifies to:
$A^2 - \alpha A + \beta I = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$
Calculating $A^2$:
$A^2 = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Substituting $A^2$, $A$, and $I$ into the simplified equation:
$\begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$
Comparing the corresponding elements on both sides:
From the $(1,2)$ element:
$-1 - \alpha = -2 \Rightarrow \alpha = 1$
From the $(2,2)$ element:
$1 + \beta = 0 \Rightarrow \beta = -1$
Therefore, the value of $(\alpha - \beta)^2$ is:
$(1 - (-1))^2 = 2^2 = 4$
Answer: $4$
Hence, the correct answer is 4.