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NVAMediumJEE 2025Algebra of Matrices

JEE Mathematics 2025 Question with Solution

Let S={m∈Z:Am2+Am=3I−A−6},whereA=[2−110]S = \left\{ m \in \mathbb{Z} : A^{m^2} + A^m = 3I - A^{-6} \right\}, \quad \text{where} \quad A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} Then n(S)n(S) is equal to:

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given:

A=[2−110]A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}

and

S={m∈Z:Am2+Am=3I−A−6}S = \left\{ m \in \mathbb{Z} : A^{m^2} + A^m = 3I - A^{-6} \right\}

Find: n(S)n(S)

From the extracted working,

A2=[3−22−1],A3=[4−33−2],A4=[5−44−3]A^2 = \begin{bmatrix} 3 & -2 \\ 2 & -1 \end{bmatrix}, \quad A^3 = \begin{bmatrix} 4 & -3 \\ 3 & -2 \end{bmatrix}, \quad A^4 = \begin{bmatrix} 5 & -4 \\ 4 & -3 \end{bmatrix}

and

A6=[7−66−5]A^6 = \begin{bmatrix} 7 & -6 \\ 6 & -5 \end{bmatrix}

the solution gives the general form

Am=[m+1−mm−m+1]A^m = \begin{bmatrix} m+1 & -m \\ m & -m+1 \end{bmatrix}

So,

Am2=[m2+1−m2m2−(m2−1)]A^{m^2} = \begin{bmatrix} m^2 + 1 & -m^2 \\ m^2 & -(m^2-1) \end{bmatrix}

Also, the extracted solution substitutes

3I−A−6=3[1001]−[−56−67]=[8−66−4]3I - A^{-6} = 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -5 & 6 \\ -6 & 7 \end{bmatrix} = \begin{bmatrix} 8 & -6 \\ 6 & -4 \end{bmatrix}

Now,

Am2+Am=[m2+1−m2m2−(m2−1)]+[m+1−mm−m+1]A^{m^2} + A^m = \begin{bmatrix} m^2+1 & -m^2 \\ m^2 & -(m^2-1) \end{bmatrix} + \begin{bmatrix} m+1 & -m \\ m & -m+1 \end{bmatrix}

Equating corresponding entries as shown in the solution,

m2+1+m+1=8m^2 + 1 + m + 1 = 8

Therefore,

m2+m−6=0m^2 + m - 6 = 0

So,

(m+3)(m−2)=0(m+3)(m-2) = 0

which gives

m=−3,2m = -3, 2

Hence the set SS has two elements.

Therefore, n(S)=2n(S) = 2.

Using the matrix pattern

Given:

A=[2−110]A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}

Find: the number of integers mm satisfying the matrix equation.

The provided solution observes a pattern in powers of AA:

A2=[3−22−1],A3=[4−33−2],A4=[5−44−3]A^2 = \begin{bmatrix} 3 & -2 \\ 2 & -1 \end{bmatrix}, \quad A^3 = \begin{bmatrix} 4 & -3 \\ 3 & -2 \end{bmatrix}, \quad A^4 = \begin{bmatrix} 5 & -4 \\ 4 & -3 \end{bmatrix}

This leads to

Ak=[k+1−kk−k+1]A^k = \begin{bmatrix} k+1 & -k \\ k & -k+1 \end{bmatrix}

for the exponents used in the working.

Replacing kk by mm and m2m^2,

Am=[m+1−mm−m+1],Am2=[m2+1−m2m2−m2+1]A^m = \begin{bmatrix} m+1 & -m \\ m & -m+1 \end{bmatrix}, \qquad A^{m^2} = \begin{bmatrix} m^2+1 & -m^2 \\ m^2 & -m^2+1 \end{bmatrix}

Hence,

Am2+Am=[m2+m+2−(m2+m)m2+m−m2−m+2]A^{m^2} + A^m = \begin{bmatrix} m^2+m+2 & -(m^2+m) \\ m^2+m & -m^2-m+2 \end{bmatrix}

From the solution,

3I−A−6=[8−66−4]3I - A^{-6} = \begin{bmatrix} 8 & -6 \\ 6 & -4 \end{bmatrix}

So we compare entries:

m2+m+2=8m^2 + m + 2 = 8

which gives

m2+m−6=0m^2 + m - 6 = 0

Thus,

m=−3 or m=2m = -3 \text{ or } m = 2

Therefore, the number of valid integers is 22.

Common mistakes

  • Using the matrix written in the question as Am2+An=31−A6Am^2 + A^n = 31 - A^6 by reading the superscripts incorrectly. This is wrong because the solution working clearly uses powers Am2A^{m^2} and AmA^m with the right side 3I−A−63I - A^{-6}. Read matrix exponents carefully before equating entries.

  • Equating only one scalar-looking expression and ignoring that this is a matrix equation. This is wrong because two matrices are equal only when their corresponding entries are equal. First write both sides as full matrices, then compare matching entries.

  • Missing the pattern for AmA^m and computing many powers separately. This is inefficient and can lead to errors. Instead, observe the entry pattern from successive powers and then substitute mm and m2m^2 directly.

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