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Multiplication of Matrices

A =[[ 1 0 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,

The value of $| \cup |$ equals

0

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1

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2

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- 1

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Solution

The correct option is B $1$
$A^{n} - A^{n - 2} = A^{2} - I$
$\Rightarrow A^{50} = A^{48} + A^{2} - I$

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

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

Now,
$A^{50} \cup_{1} = ⎡ ⎢ ⎣ \begin{matrix} 12525 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 25 & 1 & 0 25 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 12525 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow ⎡ ⎢ ⎣ \begin{matrix} x 25 x + y 25 x + z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 12525 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow x = 1, y = 0, z = 0$
$\Rightarrow \cup_{1} = ⎡ ⎢ ⎣ \begin{matrix} 100 \end{matrix} ⎤ ⎥ ⎦$

Similarly,
$\cup_{2} = ⎡ ⎢ ⎣ \begin{matrix} 010 \end{matrix} ⎤ ⎥ ⎦ and \cup_{3} = ⎡ ⎢ ⎣ \begin{matrix} 001 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow \cup = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow | \cup | = 1$

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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.

X

\cup_{3 \times 3}

\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} ⎤ ⎥ ⎦

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.

X

\cup_{3 \times 3}

\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} ⎤ ⎥ ⎦

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.

X

\cup_{3 \times 3}

\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} ⎤ ⎥ ⎦

| \cup |

A

2 \times 2

A

3

A^{3}

- 18,

A

If A is a square matrix such that $A^{2} = A$ , then write the value of $7 A - (1 + A)^{3}$ , where I is an identify matrix.

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