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JEE Main 202628 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Matrices

Let A, B and C be three 2 × 2 matrices with real entries such that...

Let A,BA, B and CC be three 2×22 \times 2 matrices with real entries such that B=(I+A)−1andB=(I+A)^{-1} and\mathrm{A}+\mathrm{C}=\mathrm{I}.If. If\mathrm{BC}=\left[1−5−12\begin{array}{cc}1 & -5 \\ -1 & 2\end{array}\right] and \mathrm{CB}\left[x1x2\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]=\left[12−6\begin{array}{c}12 \\ -6\end{array}\right], then x1+x2x_{1}+x_{2} is

Options

  1. A2
  2. B-2
  3. C0
  4. D4

Correct answer

C. 0

Step-by-step solution

From B=(I+A)−1B = (I + A)^{-1} we get B(I+A)=IB(I + A) = I, so BA=I−BBA = I - B.
From A+C=IA + C = I: C=I−AC = I - A.
Calculating CB=C(I+A)−1=(I−A)(I+A)−1CB = C(I+A)^{-1} = (I-A)(I+A)^{-1}.
Since B(I+A)=IB(I+A) = I, we have (I+A)=B−1(I+A) = B^{-1}, so C=(I−A)B−1C = (I-A)B^{-1}.
Multiplying by BB: CB=(I−A)CB = (I-A).
However, using the given BC=[1−5−12]BC = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix} and B(I+A)=IB(I+A) = I implies B−BA=B−(I−B)=2B−I=BCB - BA = B - (I-B) = 2B - I = BC, i.e. BC=2B−IBC = 2B - I.
Therefore B=12(BC+I)=12[2−5−13]=[1−5/2−1/23/2]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=[1−5−12]CB = 2B - I = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}.
From CB[x1x2]=[12−6]CB\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \end{bmatrix}: (CB)−1=−13[2511](CB)^{-1} = -\frac{1}{3}\begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix}.
So [x1x2]=−13[−66]=[2−2]\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = -\frac{1}{3}\begin{bmatrix} -6 \\ 6 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}.
Therefore x1+x2=0x_1 + x_2 = 0
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