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 $BC=\left[\begin{matrix}1 & −5 \ −1 & 2\end{matrix}\right]$ and $CB\left[\begin{matrix}{x}_{1} \ {x}_{2}\end{matrix}\right]=\left[\begin{matrix}12 \ −6\end{matrix}\right]$, then ${x}_{1}+{x}_{2}$ is [JEE Main 2026, 28 Jan (Shift 1)]

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Correct answer: d) 0. $B={(I+A)}^{−1},A+C=I$ $\Rightarrow B\left(I+A\right)=\left(I+A\right)B=I$ $\Rightarrow B+BA=B+AB$ $\Rightarrow B+B\left(I−C\right)=B+\left(I−C\right)B$ $\Rightarrow 2B−BC=2B−CB$ $\Rightarrow BC=CB$ $∴CB\left[\begin{matrix}{x}_{1} \ {x}_{2}\end{matrix}\right]=\left[\begin{matrix}1 & −5 \ −1 & 2\end{matrix}\right]\left[\begin{matrix}{x}_{1} \ {x}_{2}\end{matrix}\right]=\left[\begin{matrix}12 \ −6\end{matrix}\right]$ $\Rightarrow \left[\begin{matrix}{x}_{1} \ {x}_{2}\end{matrix}\right]={\left[\begin{matrix}1 & −5 \ −1 & 2\end{matrix}\right]}^{−1}\left[\begin{matrix}12 \ −6\end{matrix}\right]=−\frac{1}{3}\left[\begin{matrix}2 & 5 \ 1 & 1\end{matrix}\right]\left[\begin{matrix}12 \ −6\end{matrix}\right]$ $\Rightarrow \left[\begin{matrix}{x}_{1} \ {x}_{2}\end{matrix}\right]=\left[\begin{matrix}2 \ −2\end{matrix}\right]$ $∴{x}_{1}+{x}_{2}=0$

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…

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