Let A be a 3 \times 3 matrix and \operatornamedet(A)=2 . If n=\operatornamedet(\underbrace\operatornameadj(\operatornameadj(\ldots . .(\operatornameadj A))_2024- times )) , then the remainder when n is divided by 9 is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q59 with Solution

View written solutionFree

Correct answer: 7

Key fact about adjugate and determinant

For an $m \times m$ matrix $M$ ,

\det(\operatorname{adj} M)=\big(\det M\big)^{m-1}.

Here $m=3$ , so for any $3\times 3$ matrix,

\det(\operatorname{adj} M)=\big(\det M\big)^2.

Also, for a $3\times 3$ matrix,

\operatorname{adj}(\operatorname{adj} M)=\big(\det M\big)^{3-2}M=(\det M)M,

but using only determinants is enough here.

Track the determinant after repeated adjugates

Let

M_0=A,\qquad M_{k+1}=\operatorname{adj}(M_k).

Define

d_k=\det(M_k).

Given

d_0=\det(A)=2.

Since each matrix is $3\times 3$ ,

d_{k+1}=(d_k)^2.

So:

$d_1=2^2=2^{2}$
$d_2=(2^2)^2=2^{2^2}$
$d_3=2^{2^3}$

Hence, by induction,

d_k=2^{2^k}.

Therefore after $2024$ adjugates,

n=d_{2024}=2^{2^{2024}}.

Find $n \pmod 9$

We need

2^{2^{2024}} \pmod 9.

Powers of $2$ modulo $9$ repeat with period $6$ :

2^1\equiv 2,\quad 2^2\equiv 4,\quad 2^3\equiv 8,\quad 2^4\equiv 7,\quad 2^5\equiv 5,\quad 2^6\equiv 1 \pmod 9.

So we need the exponent $2^{2024}$ modulo $6$ .

Now,

2^1\equiv 2 \pmod 6,

and for any $k\ge 1$ ,

2^k \equiv 2 \text{ or }4 \pmod 6,

more specifically for $k\ge 2$ ,

2^k\equiv 4 \pmod 6.

Since $2024\ge 2$ ,

2^{2024}\equiv 4 \pmod 6.

Therefore,

2^{2^{2024}}\equiv 2^4\equiv 16\equiv 7 \pmod 9.

Final answer

The remainder when $n$ is divided by $9$ is

\boxed{7}.

Comparison with stored answer

Stored correct answer: $7$

Our derived answer is also $7$ , so they agree.

More from Matrices and Determinants

View written solutionFree

Correct answer: 7

Key fact about adjugate and determinant

For an $m \times m$ matrix $M$ ,

\det(\operatorname{adj} M)=\big(\det M\big)^{m-1}.

Here $m=3$ , so for any $3\times 3$ matrix,

\det(\operatorname{adj} M)=\big(\det M\big)^2.

Also, for a $3\times 3$ matrix,

\operatorname{adj}(\operatorname{adj} M)=\big(\det M\big)^{3-2}M=(\det M)M,

but using only determinants is enough here.

Track the determinant after repeated adjugates

Let

M_0=A,\qquad M_{k+1}=\operatorname{adj}(M_k).

Define

d_k=\det(M_k).

Given

d_0=\det(A)=2.

Since each matrix is $3\times 3$ ,

d_{k+1}=(d_k)^2.

So:

$d_1=2^2=2^{2}$
$d_2=(2^2)^2=2^{2^2}$
$d_3=2^{2^3}$

Hence, by induction,

d_k=2^{2^k}.

Therefore after $2024$ adjugates,

n=d_{2024}=2^{2^{2024}}.

Find $n \pmod 9$

We need

2^{2^{2024}} \pmod 9.

Powers of $2$ modulo $9$ repeat with period $6$ :

2^1\equiv 2,\quad 2^2\equiv 4,\quad 2^3\equiv 8,\quad 2^4\equiv 7,\quad 2^5\equiv 5,\quad 2^6\equiv 1 \pmod 9.

So we need the exponent $2^{2024}$ modulo $6$ .

Now,

2^1\equiv 2 \pmod 6,

and for any $k\ge 1$ ,

2^k \equiv 2 \text{ or }4 \pmod 6,

more specifically for $k\ge 2$ ,

2^k\equiv 4 \pmod 6.

Since $2024\ge 2$ ,

2^{2024}\equiv 4 \pmod 6.

Therefore,

2^{2^{2024}}\equiv 2^4\equiv 16\equiv 7 \pmod 9.

Final answer

The remainder when $n$ is divided by $9$ is

\boxed{7}.

Comparison with stored answer

Stored correct answer: $7$

Our derived answer is also $7$ , so they agree.

Matrices and Determinants question

Matrices and Determinants question

More from Matrices and Determinants

Matrices and Determinants question

Matrices and Determinants question

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