JEE Main 2026 — Jan 28, Shift 1 — Mathematics solution

JEE Main2026Jan 28, Shift 1Mathematics

Let $A$ , $B$ and $C$ be three $2 \times 2$ matrices with real entries such that $B = (I + A)^{- 1}$ and $A + C = I$ . If $B C = [1 - 5 - 12]$ and $C B [x_{1} x_{2}] = [12 - 6]$ , then $x_{1} + x_{2}$ is

A
2
B
0
C
$- 2$
D
4

Solution

From $B = (I + A)^{- 1}$ : $B (I + A) = I = (I + A) B$ , so $B + B A = B + A B$ , giving $B A = A B$ .

With $C = I - A$ : $B C = B (I - A) = B - B A$ and $C B = (I - A) B = B - A B$ . Since $A B = B A$ , we get $B C = C B$ .

$∴ C B = B C = [1 - 5 - 12]$ .

Solve $[1 - 5 - 12] [x_{1} x_{2}] = [12 - 6]$ . Determinant $= 2 - 5 = - 3$ .

$[x_{1} x_{2}] = - \frac{1}{3} [2511] [12 - 6] = - \frac{1}{3} [24 - 30 12 - 6] = - \frac{1}{3} [- 6 6] = [2 - 2]$ .

$x_{1} + x_{2} = 0$ . Option (2).

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