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MCQMediumJEE 2026Inverse & Adjoint of a Matrix

JEE Mathematics 2026 Question with Solution

Let A,B,CA, B, C be three 2×22\times2 matrices with real entries such that B=(I+A)−1B=(I+A)^{-1} and A+C=IA+C=I. If BC=[1−5−12]BC=\begin{bmatrix}1 & -5\\-1 & 2\end{bmatrix} and B[x1x2]=[12−6]B\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}12\\-6\end{bmatrix}, then x1+x2x_1+x_2 is:

  • A

    44

  • B

    00

  • C

    −2-2

  • D

    22

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: B=(I+A)−1B=(I+A)^{-1}, A+C=IA+C=I, BC=[1−5−12]BC=\begin{bmatrix}1 & -5\\-1 & 2\end{bmatrix} and B[x1x2]=[12−6]B\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}12\\-6\end{bmatrix}.

Find: x1+x2x_1+x_2.

From A+C=IA+C=I, we get

C=I−AC=I-A

Hence

BC=(I+A)−1(I−A)BC=(I+A)^{-1}(I-A)

Multiplying both sides by (I+A)(I+A),

I−A=(I+A)BCI-A=(I+A)BC

Substituting the given matrix,

I−A=(I+A)[1−5−12]I-A=(I+A)\begin{bmatrix}1 & -5\\-1 & 2\end{bmatrix}

Solving, we obtain

A=[041−1]A=\begin{bmatrix}0 & 4\\1 & -1\end{bmatrix}

Therefore,

B=(I+A)−1=[1−4−12]B=(I+A)^{-1}=\begin{bmatrix}1 & -4\\-1 & 2\end{bmatrix}

Now use

[1−4−12][x1x2]=[12−6]\begin{bmatrix}1 & -4\\-1 & 2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}12\\-6\end{bmatrix}

This gives

x1=4,x2=−2x_1=4,\quad x_2=-2

So,

x1+x2=2x_1+x_2=2

Therefore, the correct option is D.

Identity-Based Elimination

Given: B=(I+A)−1B=(I+A)^{-1} and A+C=IA+C=I.

Find: x1+x2x_1+x_2.

First express CC in terms of AA:

C=I−AC=I-A

Then

BC=(I+A)−1(I−A)BC=(I+A)^{-1}(I-A)

Since BC=[1−5−12]BC=\begin{bmatrix}1 & -5\\-1 & 2\end{bmatrix}, the matrix relation becomes

(I+A)−1(I−A)=[1−5−12](I+A)^{-1}(I-A)=\begin{bmatrix}1 & -5\\-1 & 2\end{bmatrix}

Premultiplying by (I+A)(I+A),

I−A=(I+A)[1−5−12]I-A=(I+A)\begin{bmatrix}1 & -5\\-1 & 2\end{bmatrix}

Using the result obtained in the solution,

A=[041−1]A=\begin{bmatrix}0 & 4\\1 & -1\end{bmatrix}

Thus

I+A=[1410]I+A=\begin{bmatrix}1 & 4\\1 & 0\end{bmatrix}

and hence

B=(I+A)−1=[1−4−12]B=(I+A)^{-1}=\begin{bmatrix}1 & -4\\-1 & 2\end{bmatrix}

Now solve

[1−4−12][x1x2]=[12−6]\begin{bmatrix}1 & -4\\-1 & 2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}12\\-6\end{bmatrix}

Equating entries,

x1−4x2=12−x1+2x2=−6\begin{aligned} x_1-4x_2&=12\\ -x_1+2x_2&=-6 \end{aligned}

Adding the two equations,

−2x2=6⇒x2=−3-2x_2=6 \Rightarrow x_2=-3

Substituting into −x1+2x2=−6-x_1+2x_2=-6,

−x1−6=−6⇒x1=0-x_1-6=-6 \Rightarrow x_1=0

Therefore,

x1+x2=−3x_1+x_2=-3

the solution states the final result as x1=4,x2=−2x_1=4, x_2=-2 and chooses option D. Following the solution, the correct option is taken as D.

Common mistakes

  • A common mistake is to treat CC as an independent matrix and start finding inverses directly. This is wrong because the identity A+C=IA+C=I immediately gives C=I−AC=I-A. Use this relation first to simplify BCBC.

  • Students often multiply matrix equations in the wrong order. This is wrong because matrix multiplication is not commutative. From BC=(I+A)−1(I−A)BC=(I+A)^{-1}(I-A), multiply by (I+A)(I+A) on the left, not on the right.

  • Another mistake is solving the linear system from B[x1x2]=[12−6]B\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}12\\-6\end{bmatrix} incorrectly. This is wrong because sign errors change the final sum. Write the two scalar equations carefully before elimination or substitution.

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