Question
If $\mathrm{A}=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right)$ then $\mathrm{A}^{2005} \mathrm{is}$(a) (a)(b) - A(c) I(d) 0
Step 1
This gives us the matrix $\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)$, which is the negative of the identity matrix, i.e., $A^2 = -I$. Show more…
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If $A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0\end{array}\right]$, then $A+2 A^{\prime}$ equals(a) $A$ (b) $A^{\prime}$ (c) $-\mathrm{A}^{\prime}$ (d) $2 \mathrm{~A}$
$$ A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right], B=\left[\begin{array}{lll} a^{-1} & 0 & 0 \\ 0 & b^{-1} & 0 \\ 0 & 0 & c^{-1} \end{array}\right] \quad(a, b, c \neq 0) $$
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$$ A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & 0 \end{array}\right], B=\left[\begin{array}{lll} a^{-1} & 0 & 0 \\ 0 & b^{-1} & 0 \\ 0 & 0 & 0 \end{array}\right] \quad(a, b \neq 0) $$
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