Algebra – Matrices – JEE Main 29 Jan 2025 Shift 1

Solution:

Calculate powers of $A$:

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

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

$A^3 = A^2 \cdot A = \begin{bmatrix} 4 & -3 \\ 3 & -2 \end{bmatrix}$

By induction, the general form is $A^k = \begin{bmatrix} k+1 & -k \\ k & -(k-1) \end{bmatrix}$.

The given equation is $A^{m^2} + A^m = 3I – A^{-6}$.

Substitute $k=m^2$ for the first term and $k=m$ for the second term:

$A^{m^2} = \begin{bmatrix} m^2+1 & -m^2 \\ m^2 & -(m^2-1) \end{bmatrix}$

$A^m = \begin{bmatrix} m+1 & -m \\ m & -(m-1) \end{bmatrix}$

Calculate RHS with $k=-6$:

$A^{-6} = \begin{bmatrix} -6+1 & -(-6) \\ -6 & -(-6-1) \end{bmatrix} = \begin{bmatrix} -5 & 6 \\ -6 & 7 \end{bmatrix}$.

$3I – A^{-6} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} – \begin{bmatrix} -5 & 6 \\ -6 & 7 \end{bmatrix} = \begin{bmatrix} 8 & -6 \\ 6 & -4 \end{bmatrix}$.

Now equate LHS and RHS:

$\begin{bmatrix} m^2+1 & -m^2 \\ m^2 & -(m^2-1) \end{bmatrix} + \begin{bmatrix} m+1 & -m \\ m & -(m-1) \end{bmatrix} = \begin{bmatrix} 8 & -6 \\ 6 & -4 \end{bmatrix}$

Compare element $(1,1)$:

$(m^2+1) + (m+1) = 8$

$m^2 + m + 2 = 8 \Rightarrow m^2 + m – 6 = 0$.

$(m+3)(m-2) = 0$.

Possible integer values for $m$: $\{-3, 2\}$.

Check consistency with element $(1,2)$:

$-m^2 – m = -6 \Rightarrow -(m^2+m) = -6 \Rightarrow m^2+m=6$. (Consistent).

Thus, $S = \{-3, 2\}$.

Number of elements $n(S) = 2$.

Ans. (2)