JEE Main 2026 — Mathematics Algebra

The determinant of matrix $A$A is:

$$|A| = -1(0 - 0) - 1(1 - 0) - 1(0 - 0) = -1$$∣A∣=−1(0−0)−1(1−0)−1(0−0)=−1

Using the properties of the adjoint of a $3 \times 3$3×3 matrix:

$$adj(adj(A)) = |A|^{3-2}A = -A$$adj(adj(A))=∣A∣3−2A=−A

$$adj(A)(adj(adj(A))) = adj(A)(-A) = -|A|I = -(-1)I = I$$adj(A)(adj(adj(A)))=adj(A)(−A)=−∣A∣I=−(−1)I=I

The given equation simplifies to:

$$A^2 - \alpha A + \beta I = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$$A2−αA+βI=​2−20​−200​2−1−1​​

Calculating $A^2$A2:

$$A^2 = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$A2=​−110​100​−111​​​−110​100​−111​​=​2−10​−110​101​​

Substituting $A^2$A2, $A$A, and $I$I into the simplified equation:

$$\begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$$​2−10​−110​101​​−α​−110​100​−111​​+β​100​010​001​​=​2−20​−200​2−1−1​​

Comparing the corresponding elements on both sides:

From the $(1,2)$(1,2) element:

$$-1 - \alpha = -2 \Rightarrow \alpha = 1$$−1−α=−2⇒α=1

From the $(2,2)$(2,2) element:

$$1 + \beta = 0 \Rightarrow \beta = -1$$1+β=0⇒β=−1

Therefore, the value of $(\alpha - \beta)^2$(α−β)2 is:

$$(1 - (-1))^2 = 2^2 = 4$$(1−(−1))2=22=4

Answer: $4$4