Matrices – Characteristics Equations – JEE Main 21 Jan 2026 Shift 1

Solution:

For a $2 \times 2$ matrix $M$, the characteristic equation is $M^2 – \text{Tr}(M)M + \det(M)I = 0$.
Comparing with the given equation $A^2 – 4A + 2I = 0$:
$$\text{Trace}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$$
$$\det(A) = 2 \Rightarrow 2\alpha – 2 = 2 \Rightarrow 2(2) – 2 = 2 \quad (\text{Consistent})$$
So, $A = \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix}$.

Comparing with the given equation $B^2 – 3B + I = 0$:
$$\text{Trace}(B) = 3 \Rightarrow 1 + \beta = 3 \Rightarrow \beta = 2$$
$$\det(B) = 1 \Rightarrow \beta – 1 = 1 \Rightarrow 2 – 1 = 1 \quad (\text{Consistent})$$
So, $B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$.

Now we calculate $A^3$ and $B^3$ using the characteristic equations:
$$A^2 = 4A – 2I$$
Multiplying by $A$:
$$A^3 = 4A^2 – 2A = 4(4A – 2I) – 2A = 16A – 8I – 2A = 14A – 8I$$
$$A^3 = 14\begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} – \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 28 & 28 \\ 14 & 28 \end{bmatrix} – \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix}$$

Similarly for $B$:
$$B^2 = 3B – I$$
Multiplying by $B$:
$$B^3 = 3B^2 – B = 3(3B – I) – B = 9B – 3I – B = 8B – 3I$$
$$B^3 = 8\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} – \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 8 & 16 \end{bmatrix} – \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$$

Now find $A^3 – B^3$:
$$A^3 – B^3 = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix} – \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} = \begin{bmatrix} 15 & 20 \\ 6 & 7 \end{bmatrix}$$

Let $M = A^3 – B^3$. We need $(\det(\text{adj}(M)))^2$.
For a matrix of order $n=2$, $\det(\text{adj}(M)) = (\det(M))^{n-1} = \det(M)$.
$$\det(M) = (15)(7) – (20)(6) = 105 – 120 = -15$$

So, $(\det(\text{adj}(M)))^2 = (-15)^2 = 225$.

Ans. 225