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Multiplication of Matrices

Let A = [ 1 0...

Question

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

A

0

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B

1

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C

2

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D

3

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Solution

The correct option is D $3$
$A^{n} - A^{n - 2} = A^{2} - I$
$\Rightarrow A^{50} = A^{48} + A^{2} - I$

Further,
$A^{48} = A^{46} + A^{2} - I$
$A^{46} = A^{44} + A^{2} - I$
$⋮ ⋮ ⋮$
$A^{4} = A^{2} + A^{2} - I$
_________________
$A^{50} = 25 A^{2} - 24 I$

Here,
$A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦$
$= ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow A^{50} = ⎡ ⎢ ⎣ \begin{matrix} 25 & 0 & 0 25 & 25 & 0 25 & 0 & 25 \end{matrix} ⎤ ⎥ ⎦ - 24 ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
$= ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 25 & 1 & 0 25 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow tr (A^{50}) = 1 + 1 + 1 = 3.$

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Similar questions

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

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

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}

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 |

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

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

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}

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}

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

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

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}

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 |

A^{50}

A = ⎛ ⎜ ⎝ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎞ ⎟ ⎠

A^{50}

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