If - A - begin bmatrix i - -il-i - ilend bmatrix and - B - bmatrix 18-1 -1 - 1lend bmatrix - then - A - 8 - equals A- 4 BB- 128 BC- -128 BD- - 64 B

The correct option is B 128 BWe have A = iB ⇒ A^2 = (iB)^2 = i^2B^2 = B^2 = [ 2 amp; 2; 2 amp; 2 ]= 2B ⇒ A^4 = ( 2B)^2 = 4B^2 = 4(2B) = 8B ⇒ ...

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Byju's Answer

Standard X

Mathematics

Multiplication of Matrices

If ∼ A = begi...

Question

$I f A = [\begin{matrix} i & - i - i & i \end{matrix}] a n d B = [\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix}], t h e n A^{8} e q u a l s$

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128 B

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- 128 B

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- 64 B

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Solution

The correct option is B 128 B
We have A = iB
$\Rightarrow A^{2} = (i B)^{2} = i^{2} B^{2} = - B^{2} = - [\begin{matrix} 2 & - 2 - 2 & 2 \end{matrix}] = - 2 B \Rightarrow A^{4} = (- 2 B)^{2} = 4 B^{2} = 4 (2 B) = 8 B \Rightarrow (A^{4})^{2} = (8 B)^{2} \Rightarrow A^{8} = 64 B^{2} = 128 B$

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Similar questions

Q. If

A = [\begin{matrix} i & - i - i & i \end{matrix}]

and

B = [\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix}]

, then

A^{8}

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If $A = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & 3 5 & 7 & 9 - 2 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ and $B = ⎡ ⎢ ⎣ \begin{matrix} - 4 & 1 & - 5 1 & 2 & 0 1 & 3 & 1 \end{matrix} ⎤ ⎥ ⎦$ , then verify that
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A = [\begin{matrix} 0 & 1 1 & 0 \end{matrix}]

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If $A = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & 3 5 & 7 & 9 - 2 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ and $B = ⎡ ⎢ ⎣ \begin{matrix} - 4 & 1 & - 5 1 & 2 & 0 1 & 3 & 1 \end{matrix} ⎤ ⎥ ⎦$ , then verify that
(i)(A+B)'=A'+B'

(ii)(A-B)'=A'-B'

A = [\begin{matrix} i & - i - i & i \end{matrix}], B [\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix}] \Rightarrow A^{8} =

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If A=[i−i−ii] and B=[1−1−11],then A8 equals

The correct option is B 128 BWe have A = iB ⇒A2=(iB)2=i2B2=−B2=−[2−2−22]=−2B⇒A4=(−2B)2=4B2=4(2B)=8B⇒(A4)2=(8B)2⇒A8=64B2=128B

$I f A = [\begin{matrix} i & - i - i & i \end{matrix}] a n d B = [\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix}], t h e n A^{8} e q u a l s$

The correct option is B 128 B
We have A = iB
$\Rightarrow A^{2} = (i B)^{2} = i^{2} B^{2} = - B^{2} = - [\begin{matrix} 2 & - 2 - 2 & 2 \end{matrix}] = - 2 B \Rightarrow A^{4} = (- 2 B)^{2} = 4 B^{2} = 4 (2 B) = 8 B \Rightarrow (A^{4})^{2} = (8 B)^{2} \Rightarrow A^{8} = 64 B^{2} = 128 B$