Matrices – Matrix Operations – JEE Main 28 Jan 2026 Shift 1

Solution:

Given $B = (I+A)^{-1}$, we have $B(I+A) = I$.
$$B + BA = I$$
Also given $A + C = I$, so $A = I – C$.
Substitute $A$ into the first equation:
$$B + B(I – C) = I$$
$$B + B – BC = I$$
$$2B – BC = I \quad \dots(1)$$

Alternatively, $(I+A)B = I \Rightarrow B + AB = I$.
Substitute $A = I – C$:
$$B + (I – C)B = I$$
$$2B – CB = I \quad \dots(2)$$

From (1) and (2):
$$2B – BC = 2B – CB \Rightarrow BC = CB$$

Given $BC = \begin{bmatrix}1 & -5 \\ -1 & 2\end{bmatrix}$.
The equation $CB\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}=\begin{bmatrix}12\\ -6\end{bmatrix}$ becomes $BC\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}=\begin{bmatrix}12\\ -6\end{bmatrix}$.

$$\begin{bmatrix}1 & -5 \\ -1 & 2\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}12 \\ -6\end{bmatrix}$$

This gives the system of equations:
$$x_1 – 5x_2 = 12 \quad \dots(i)$$
$$-x_1 + 2x_2 = -6 \quad \dots(ii)$$

Adding $(i)$ and $(ii)$:
$$-3x_2 = 6 \Rightarrow x_2 = -2$$
Substitute $x_2 = -2$ in $(i)$:mat
$$x_1 – 5(-2) = 12$$
$$x_1 + 10 = 12 \Rightarrow x_1 = 2$$

We need to find $x_1 + x_2$:
$$x_1 + x_2 = 2 + (-2) = 0$$

Ans. (2)