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JEE Mathematics 2026 Question with Solution

For the matrices A=[3−41−1]A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} and B=[−2949−1318]B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix}, if (A15+B)[xy]=[00](A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, then among the following which one is true?

  • A

    x=16,y=3x = 16, y = 3

  • B

    x=18,y=11x = 18, y = 11

  • C

    x=5,y=7x = 5, y = 7

  • D

    x=11,y=2x = 11, y = 2

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A=[3−41−1]A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}, B=[−2949−1318]B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} and

(A15+B)[xy]=[00](A^{15}+B)\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}

Find: which ordered pair (x,y)(x,y) satisfies the equation.

From the solution, the stated correct option is D.

Compute the characteristic polynomial of AA:

det⁡(A−λI)=∣3−λ−41−1−λ∣\det(A-\lambda I)=\begin{vmatrix}3-\lambda & -4 \\ 1 & -1-\lambda\end{vmatrix} =(3−λ)(−1−λ)+4=λ2−2λ+1=(λ−1)2=(3-\lambda)(-1-\lambda)+4=\lambda^2-2\lambda+1=(\lambda-1)^2

So AA has repeated eigenvalue 11.

Now write

A−I=[2−41−2]=NA-I=\begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}=N

Then

N2=[2−41−2][2−41−2]=[0000]N^2=\begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}\begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}

Hence A=I+NA = I+N with N2=0N^2=0.

Using the binomial expansion for a nilpotent matrix,

A15=(I+N)15=I+15NA^{15}=(I+N)^{15}=I+15N

Therefore

A15=I+15(A−I)=15A−14IA^{15}=I+15(A-I)=15A-14I

Substituting AA,

A15=15[3−41−1]−14[1001]=[31−6015−29]A^{15}=15\begin{bmatrix}3 & -4 \\ 1 & -1\end{bmatrix}-14\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=\begin{bmatrix}31 & -60 \\ 15 & -29\end{bmatrix}

Now add BB:

A15+B=[31−6015−29]+[−2949−1318]=[2−112−11]A^{15}+B=\begin{bmatrix}31 & -60 \\ 15 & -29\end{bmatrix}+\begin{bmatrix}-29 & 49 \\ -13 & 18\end{bmatrix}=\begin{bmatrix}2 & -11 \\ 2 & -11\end{bmatrix}

So the equation becomes

[2−112−11][xy]=[00]\begin{bmatrix}2 & -11 \\ 2 & -11\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}

which gives

2x−11y=02x-11y=0

Hence

2x=11y2x=11y

We now test the given options.

Checking option D:

2(11)−11(2)=22−22=02(11)-11(2)=22-22=0

So x=11,y=2x=11, y=2 satisfies the equation. Therefore, the correct option is D.

The solution also declares D as the correct option.

Use the form $$A = I + N$$

Given: A=[3−41−1]A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}. Find: a quick way to compute A15A^{15}.

Observe that

A−I=[2−41−2]A-I=\begin{bmatrix}2 & -4 \\ 1 & -2\end{bmatrix}

and

(A−I)2=0(A-I)^2=0

So if A=I+NA = I+N with N2=0N^2=0, then

(I+N)n=I+nN(I+N)^n = I+nN

This works because every term containing N2N^2 or higher powers vanishes.

Hence

A15=I+15(A−I)=15A−14IA^{15}=I+15(A-I)=15A-14I

Then

A15+B=[2−112−11]A^{15}+B=\begin{bmatrix}2 & -11 \\ 2 & -11\end{bmatrix}

so the condition is only

2x−11y=02x-11y=0

Among the options, only x=11,y=2x=11, y=2 satisfies this relation. Therefore, the correct option is D.

Common mistakes

  • Students may try to multiply AA repeatedly up to A15A^{15}. That is inefficient and hides the key idea. Instead, observe that A−IA-I is nilpotent, so A15A^{15} can be found using A=I+NA = I+N with N2=0N^2=0.

  • A common error is computing the characteristic polynomial incorrectly. If det⁡(A−λI)\det(A-\lambda I) is expanded wrongly, the repeated eigenvalue 11 is missed. Expand the determinant carefully to get (λ−1)2(\lambda-1)^2.

  • Some students stop after finding A15A^{15} and forget to add BB. The equation involves (A15+B)(A^{15}+B), so the null-vector condition must be formed only after adding the two matrices.

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