JEE Main 2026 — Jan 21, Shift 2 — Mathematics solution

JEE Main2026Jan 21, Shift 2Mathematics

For the matrices $A = [3 - 4 1 - 1]$ and $B = [- 2949 - 1318]$ , if $(A^{15} + B) [x y] = [00]$ , then among the following which one is true?

A
x = 5, y = 7
B
x = 18, y = 11
C
x = 11, y = 2
D
x = 16, y = 3

Solution

Notice $A - I = [2 - 4 1 - 2]$ is nilpotent of index 2 (it has rank 1 and trace 0; $(A - I)^{2} = 0$ ). Writing $A = I + N$ with $N^{2} = 0$ , the binomial theorem gives $A^{n} = I + n N$ , so $A^{n} = [2 n + 1 - 4 n n - 2 n + 1]$ .

For $n = 15$ : $A^{15} = [31 - 60 15 - 29]$ , so $A^{15} + B = [2 - 11 2 - 11]$ .

The homogeneous system has both rows giving $2 x - 11 y = 0$ , i.e. $x = \frac{11 y}{2}$ . With $y = 2$ we get $x = 11$ , matching option (3).

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →