If A- 1 0 0- 1 0 1- 0 1 0 - then A50 equals

The correct options are A [ 1 0 0; 25 1 0; 25 0 1 ] C A^2+24 ( A^2 I ) Given A= [ 1 amp; 0 amp; 0; 1 amp; 0 amp; 1; 0 amp ...

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Byju's Answer

Standard XII

Mathematics

Equality of Matrices

If A=[ 1 0 ...

Question

If $A = ⎛ ⎜ ⎝ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎞ ⎟ ⎠$ then $A^{50}$ equals

⎛ ⎜ ⎝ \begin{matrix} 1 & 0 & 0 25 & 1 & 0 25 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠

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A^{2} + 24 (A^{2} - I)

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⎛ ⎜ ⎝ \begin{matrix} 1 & 25 & 25 1 & 0 & 0 0 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠

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None of these

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Solution

The correct options are
A $⎛ ⎜ ⎝ \begin{matrix} 1 & 0 & 0 25 & 1 & 0 25 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠$
C $A^{2} + 24 (A^{2} - I)$
Given $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦$
$A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
$A^{4} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 1 & 1 & 0 1 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 2 & 1 & 0 2 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow A^{4} = A^{2} + A^{2} - I = 2 A^{2} - I$
$\Rightarrow A^{6} = A^{4} + A^{2} - I = 3 A^{2} - 2 I$
similarly
$A^{50} = 25 A^{2} - 24 I = A^{2} + 24 (A^{2} - I)$
$∴ A^{50} = ⎡ ⎢ ⎣ \begin{matrix} 25 & 0 & 0 25 & 25 & 0 25 & 0 & 25 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 24 & 0 & 0 0 & 24 & 0 0 & 0 & 24 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 25 & 1 & 0 25 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow A^{50} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 25 & 1 & 0 25 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
Hence, options A and B.

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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,

The value of

| \cup |

equals

Q. If

\frac{(a^{2} + 1)^{2}}{2 a - i} = x + i y, then x^{2} + y^{2}

is equal to
(a)

\frac{(a^{2} + 1)^{4}}{4 a^{2} + 1}

(b)

\frac{(a + 1)^{2}}{4 a^{2} + 1}

(c)

\frac{(a^{2} - 1)^{2}}{(4 a^{2} - 1)^{2}}

(d) none of these

Q. If

a_{1}, a_{2}, . . . . . . . . . . . . ., a_{50}

are in

G . P .

, then

\frac{a_{1} - a_{3} + a_{5} - . . . . . + a_{49}}{a_{2} - a_{4} + a_{6} - . . . . . + a_{50}}

Q. If

A^{3} = 0

, then

I + A + A^{2}

equals

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If A=⎛⎜⎝100101010⎞⎟⎠ then A50 equals

If $A = ⎛ ⎜ ⎝ \begin{matrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 \end{matrix} ⎞ ⎟ ⎠$ then $A^{50}$ equals