The correct option is A 3 Given nbsp; A= nbsp;[ 1 amp; 0 amp; 0; 1 amp; 0 amp; 1; 0 amp; 1 amp; 0 ] satisfies A^n = A^n 2 + ...

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Equality of Matrices

Trace of A5...

Question

Trace of $A^{50}$ equals

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Solution

The correct option is A $3$
Given $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦$ satisfies
$A^{n} = A^{n - 2} + A^{2} - I$ for $n \geq 3$ ....(1)
Now, put $n = 50$ in (1) , we get
$A^{50} = A^{48} + A^{2} - I$
$= (A^{46} + A^{2} - I) + A^{2} - I$ (by (1))
$\Rightarrow A^{50} = A^{46} + 2 A^{2} - 2 I$
$= (A^{44} + A^{2} - I) + 2 A^{2} - 2 I$ (by (1))
$\Rightarrow A^{50} = A^{44} + 3 A^{2} - 3 I$
So, we can conclude
$A^{50} = A^{4} + 23 A^{2} - 23 I$
$\Rightarrow A^{50} = 25 A^{2} - 24 I$ ...(2)
Now, $A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦$
$A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 1 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
So, using this in (2),
$A^{50} = ⎡ ⎢ ⎣ \begin{matrix} 25 & 0 & 0 25 & 25 & 25 25 & 0 & 25 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 24 & 0 & 0 0 & 24 & 0 0 & 0 & 24 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow A^{50} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 25 & 1 & 0 25 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
Trace of $A^{50} = 1 + 1 + 1 = 3$

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Similar questions

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦

satisfies

A^{n} = A^{n - 2} + A^{2} - I

for

n \geq 3.

And trace of a square matrix

X

is equal to the sum of elements in its principal diagonal. Further consider a matrix

\cup_{3 \times 3}

with its column as

\cup_{1}, \cup_{2}, \cup_{3}

such that

A^{50} \cup_{1} = ⎡ ⎢ ⎣ \begin{matrix} 12525 \end{matrix} ⎤ ⎥ ⎦, A^{50} \cup_{2} = ⎡ ⎢ ⎣ \begin{matrix} 010 \end{matrix} ⎤ ⎥ ⎦, A^{50} \cup_{3} = ⎡ ⎢ ⎣ \begin{matrix} 001 \end{matrix} ⎤ ⎥ ⎦

Then,

Trace of

A^{50}

equals

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦

satisfies

A^{n} = A^{n - 2} + A^{2} - I

for

n \geq 3.

And trace of a square matrix

X

is equal to the sum of elements in its principal diagonal. Further consider a matrix

\cup_{3 \times 3}

with its column as

\cup_{1}, \cup_{2}, \cup_{3}

such that

A^{50} \cup_{1} = ⎡ ⎢ ⎣ \begin{matrix} 12525 \end{matrix} ⎤ ⎥ ⎦, A^{50} \cup_{2} = ⎡ ⎢ ⎣ \begin{matrix} 010 \end{matrix} ⎤ ⎥ ⎦, A^{50} \cup_{3} = ⎡ ⎢ ⎣ \begin{matrix} 001 \end{matrix} ⎤ ⎥ ⎦

Then,

Trace of

A^{50}

equals

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦

satisfies

A^{n} = A^{n - 2} + A^{2} - I

for

n \geq 3.

And trace of a square matrix

X

is equal to the sum of elements in its principal diagonal. Further consider a matrix

\cup_{3 \times 3}

with its column as

\cup_{1}, \cup_{2}, \cup_{3}

such that

A^{50} \cup_{1} = ⎡ ⎢ ⎣ \begin{matrix} 12525 \end{matrix} ⎤ ⎥ ⎦, A^{50} \cup_{2} = ⎡ ⎢ ⎣ \begin{matrix} 010 \end{matrix} ⎤ ⎥ ⎦, A^{50} \cup_{3} = ⎡ ⎢ ⎣ \begin{matrix} 001 \end{matrix} ⎤ ⎥ ⎦

Then,

The value of

| \cup |

equals

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦

satisfies

A^{n} = A^{n - 2} + A^{2} - I

for

n \geq 3.

And trace of a square matrix

X

is equal to the sum of elements in its principal diagonal. Further consider a matrix

\cup_{3 \times 3}

with its column as

\cup_{1}, \cup_{2}, \cup_{3}

such that

A^{50} \cup_{1} = ⎡ ⎢ ⎣ \begin{matrix} 12525 \end{matrix} ⎤ ⎥ ⎦, A^{50} \cup_{2} = ⎡ ⎢ ⎣ \begin{matrix} 010 \end{matrix} ⎤ ⎥ ⎦, A^{50} \cup_{3} = ⎡ ⎢ ⎣ \begin{matrix} 001 \end{matrix} ⎤ ⎥ ⎦

Then,

The value of

| \cup |

equals

Q. If

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦

satisfies

A^{n} = A^{n - 2} + A^{2} - I

for

n \geq 3

then :

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Trace of A50 equals

Trace of $A^{50}$ equals