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Scalar Multiplication of a Matrix

If 2A + B =...

Question

If $2 A + B = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 \begin{matrix} 01 \end{matrix} & \begin{matrix} 1 - 2 \end{matrix} \end{matrix} ⎤ ⎥ ⎦ A - 2 B = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 \begin{matrix} - 2 1 \end{matrix} & \begin{matrix} 0 - 1 \end{matrix} \end{matrix} ⎤ ⎥ ⎦$ find $A$ and $B$

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Solution

$2 A + B = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 0 & 1 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$ and $2 A - 4 B = ⎡ ⎢ ⎣ \begin{matrix} 0 & 2 - 4 & 0 2 & - 2 \end{matrix} ⎤ ⎥ ⎦$

$2 A + B - 2 A + 4 B = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 0 & 1 1 & - 2 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 0 & 2 - 4 & 0 2 & - 2 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow 5 B = ⎡ ⎢ ⎣ \begin{matrix} 1 - 0 & - 1 - 2 0 + 4 & 1 - 0 1 - 2 & - 2 + 2 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow 5 B = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 3 4 & 1 - 1 & 0 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow B = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{5} & \frac{- 3}{5} \frac{4}{5} & \frac{1}{5} \frac{- 1}{5} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

Substitute $B = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{5} & \frac{- 3}{5} \frac{4}{5} & \frac{1}{5} \frac{- 1}{5} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$ in $2 A + B = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 0 & 1 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$

$2 A + ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{5} & \frac{- 3}{5} \frac{4}{5} & \frac{1}{5} \frac{- 1}{5} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 0 & 1 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow 2 A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 0 & 1 1 & - 2 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{5} & \frac{- 3}{5} \frac{4}{5} & \frac{1}{5} \frac{- 1}{5} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$\Rightarrow 2 A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 - \frac{1}{5} & - 1 + \frac{3}{5} 0 - \frac{4}{5} & 1 - \frac{1}{5} 1 + \frac{1}{5} & - 2 - 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$\Rightarrow 2 A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{5 - 1}{5} & \frac{- 5 + 3}{5} - \frac{4}{5} & \frac{5 - 1}{5} \frac{5 + 1}{5} & - 2 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$\Rightarrow 2 A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{4}{5} & \frac{- 2}{5} - \frac{4}{5} & \frac{4}{5} \frac{6}{5} & - 2 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$∴ A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{2}{5} & \frac{- 1}{5} - \frac{2}{5} & \frac{2}{5} \frac{3}{5} & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

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Q = ⎛ ⎜ ⎜ ⎝ \begin{matrix} cos \frac{π}{4} & - sin \frac{π}{4} sin \frac{π}{4} & cos \frac{π}{4} \end{matrix} ⎞ ⎟ ⎟ ⎠

x = ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \begin{matrix} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \end{matrix} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Q^{3} x

{(A B)}^{- 1}

A = [\begin{matrix} 1 & 2 1 & - 2 \end{matrix} \begin{matrix} 3 - 3 \end{matrix}]

B = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 \begin{matrix} 11 \end{matrix} & \begin{matrix} 2 - 2 \end{matrix} \end{matrix} \begin{matrix} \end{matrix} ⎤ ⎥ ⎦

n

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{n}

| ¯ ¯ ¯ a | = 1, | ¯ ¯ b | = 2, | ¯ ¯ ¯ a - ¯ ¯ b |^{2} + | ¯ ¯ ¯ a + 2 ¯ ¯ b |^{2} = 20,

(¯ ¯ ¯ a, ¯ ¯ b) =

A = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 2 & 2 3 & - 1 \end{matrix} ∣ ∣ ∣ ∣

B = ∣ ∣ ∣ ∣ \begin{matrix} 2 & 1 1 & 2 - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣

, then find matrix

C

A + 2 B + C = 0

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