Q38 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics

Let $A, B$ and $C$ be three $2 \times 2$ matrices with real entries such that $B=(I+A)^{-1}$ and $\mathrm{A}+\mathrm{C}=\mathrm{I}$. If $\mathrm{BC}=\left[\begin{array}{cc}1 & -5 \\ -1 & 2\end{array}\right]$ and $\mathrm{CB}\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]=\left[\begin{array}{c}12 \\ -6\end{array}\right]$, then $x_{1}+x_{2}$ is

  • A
    2
  • B
    -2
  • Correct answer
    C
    0
  • D
    4

Solution

From $B = (I + A)^{-1}$ we get $B(I + A) = I$, so $BA = I - B$.
From $A + C = I$: $C = I - A$.
Calculating $CB = C(I+A)^{-1} = (I-A)(I+A)^{-1}$.
Since $B(I+A) = I$, we have $(I+A) = B^{-1}$, so $C = (I-A)B^{-1}$.
Multiplying by $B$: $CB = (I-A)$.
However, using the given $BC = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}$ and $B(I+A) = I$ implies $B - BA = B - (I-B) = 2B - I = BC$, i.e. $BC = 2B - I$.
Therefore $B = \frac{1}{2}(BC + I) = \frac{1}{2}\begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -5/2 \\ -1/2 & 3/2 \end{bmatrix}$.
Also $CB = 2B - I = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}$.
From $CB\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \end{bmatrix}$: $(CB)^{-1} = -\frac{1}{3}\begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix}$.
So $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = -\frac{1}{3}\begin{bmatrix} -6 \\ 6 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}$.
Therefore $x_1 + x_2 = 0$

Hence, the correct answer is option: 'C'.

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