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Negation

If A=[ 0 i...

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If A=\begin{bmatrix}0&i -i&0\end{bmatrix} , then prove that A^{40}= \begin{bmatrix}1&0 0&1\end{bmatrix

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Q. Suppose the vectors

X_{1}

X_{2}

X_{3}

are the solutions of the system of linear equations,

A x = b

when the vector

b

on the right side is equal to

b_{1}

b_{2}

b_{3}

respectively. if

X_{1} = ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦, X_{2} = ⎡ ⎢ ⎣ \begin{matrix} 021 \end{matrix} ⎤ ⎥ ⎦, X_{3} = ⎡ ⎢ ⎣ \begin{matrix} 001 \end{matrix} ⎤ ⎥ ⎦, b_{1} = ⎡ ⎢ ⎣ \begin{matrix} 100 \end{matrix} ⎤ ⎥ ⎦, b_{2} = ⎡ ⎢ ⎣ \begin{matrix} 020 \end{matrix} ⎤ ⎥ ⎦

b_{3} = ⎡ ⎢ ⎣ \begin{matrix} 002 \end{matrix} ⎤ ⎥ ⎦

, then the determinant of

A

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 2 & 1 & 0 3 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦

u_{1}

u_{2}

are column matrices such that

A u_{1} = ⎡ ⎢ ⎣ \begin{matrix} 100 \end{matrix} ⎤ ⎥ ⎦

A u_{2} = ⎡ ⎢ ⎣ \begin{matrix} 010 \end{matrix} ⎤ ⎥ ⎦

u_{1} + u_{2}

A - 2 B = [\begin{matrix} 1 & - 2 3 & 0 \end{matrix}]

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

B =

Q. For a real number

α

, if the system

⎡ ⎢ ⎣ \begin{matrix} 1 & α & α^{2} α & 1 & α α^{2} & α & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 - 1 1 \end{matrix} ⎤ ⎥ ⎦

of linear equations, has infinitely many solutions, then

1 + α + α^{2} =

\frac{a - i b}{a + i b} = \frac{1 + i}{1 - i}

, then prove that

a + b = 0

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