Matrices – Determinants – JEE Main 22 Jan 2026 Shift 1

Solution:

We need to find $|A^{2025}-3A^{2024}+A^{2023}|$.
Factor out the lowest power of $A$, which is $A^{2023}$:
$$ A^{2025}-3A^{2024}+A^{2023} = A^{2023}(A^2 – 3A + I) $$

Using the property of determinants $|XY| = |X||Y|$:
$$ |A^{2023}(A^2 – 3A + I)| = |A|^{2023} |A^2 – 3A + I| $$

First, calculate $|A|$:
$$ |A| = \begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix} = (2)(5) – (3)(3) = 10 – 9 = 1 $$
So, $|A|^{2023} = 1^{2023} = 1$.

Now, calculate the matrix $B = A^2 – 3A + I$.
$$ A^2 = \begin{pmatrix} 2 & 3 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 3 & 5 \end{pmatrix} = \begin{pmatrix} 4+9 & 6+15 \\ 6+15 & 9+25 \end{pmatrix} = \begin{pmatrix} 13 & 21 \\ 21 & 34 \end{pmatrix} $$
$$ 3A = 3 \begin{pmatrix} 2 & 3 \\ 3 & 5 \end{pmatrix} = \begin{pmatrix} 6 & 9 \\ 9 & 15 \end{pmatrix} $$
$$ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$

$$ B = \begin{pmatrix} 13 & 21 \\ 21 & 34 \end{pmatrix} – \begin{pmatrix} 6 & 9 \\ 9 & 15 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$
$$ B = \begin{pmatrix} 13-6+1 & 21-9+0 \\ 21-9+0 & 34-15+1 \end{pmatrix} = \begin{pmatrix} 8 & 12 \\ 12 & 20 \end{pmatrix} $$

Finally, calculate the determinant of $B$:
$$ |B| = \begin{vmatrix} 8 & 12 \\ 12 & 20 \end{vmatrix} = (8)(20) – (12)(12) $$
$$ |B| = 160 – 144 = 16 $$

The required determinant is $1 \times 16 = 16$.

Ans. (4)