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Adjoint of a Matrix

If A = [ 1 ...

Question

If A = $[\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}]$ and if $A^{6}$ = KA - 205I then

A

K = 11

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B

K= 22

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C

K = 33

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D

K = 44

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Solution

The correct option is D K = 44
Given,

$A = (\begin{matrix} 1 & 2 - 1 & 3 \end{matrix})$

$A^{6} = {(\begin{matrix} 1 & 2 - 1 & 3 \end{matrix})}^{6}$

$(\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}) (\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}) = (\begin{matrix} - 1 & 8 - 4 & 7 \end{matrix})$

$(\begin{matrix} - 1 & 8 - 4 & 7 \end{matrix}) (\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}) = (\begin{matrix} - 9 & 22 - 11 & 13 \end{matrix})$

$(\begin{matrix} - 9 & 22 - 11 & 13 \end{matrix}) (\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}) = (\begin{matrix} - 31 & 48 - 24 & 17 \end{matrix})$

$(\begin{matrix} - 31 & 48 - 24 & 17 \end{matrix}) (\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}) = (\begin{matrix} - 79 & 82 - 41 & 3 \end{matrix})$

$(\begin{matrix} - 79 & 82 - 41 & 3 \end{matrix}) (\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}) = (\begin{matrix} - 161 & 88 - 44 & - 73 \end{matrix})$

$∴ A^{6} = (\begin{matrix} - 161 & 88 - 44 & - 73 \end{matrix})$

Now,

$A^{6} = K A - 205 I$

$(\begin{matrix} - 161 & 88 - 44 & - 73 \end{matrix}) = K (\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}) - 205 (\begin{matrix} 1 & 0 0 & 1 \end{matrix})$

$\Rightarrow K (\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}) = (\begin{matrix} - 161 & 88 - 44 & - 73 \end{matrix}) + 205 (\begin{matrix} 1 & 0 0 & 1 \end{matrix})$

$K (\begin{matrix} 1 & 2 - 1 & 3 \end{matrix}) = (\begin{matrix} 44 & 88 - 44 & 132 \end{matrix})$

$K = (\begin{matrix} 44 & 88 - 44 & 132 \end{matrix}) {(\begin{matrix} 1 & 2 - 1 & 3 \end{matrix})}^{- 1}$

$= (\begin{matrix} 44 & 88 - 44 & 132 \end{matrix}) ⎛ ⎝ \begin{matrix} \frac{3}{5} & - \frac{2}{5} \frac{1}{5} & \frac{1}{5} \end{matrix} ⎞ ⎠$

$∴ K = (\begin{matrix} 44 & 0 0 & 44 \end{matrix})$

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\vec{a} = 2 \hat{i} - 3 \hat{j} - \hat{k} and \vec{b} = \hat{i} + 4 \hat{j} - 2 \hat{k}, then \vec{a} \times \vec{b}

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