Se \underset{\sim}{\underline{M}}=\left[\begin{array}{cc}1 & 2 \\\\ -1 & 0\end{array}\right] e \underset{\sim}{I}=\left[\begin{array}{ll}1 & 0 \\\\ 0 & 1\end{array}\right] é a matriz identidade de ordem 2, calcule \underset{\sim}{\underline{M}^{2}}-2 \underset{\sim}{\underline{M}}+ 3 \underset{\sim}{I}.

Passo 1

O enunciado pede para calcular
\underset{\sim}{\underline{M}}^{2} \;-\; 2\,\underset{\sim}{\underline{M}} \;+\; 3\,\underset{\sim}{I},
onde
\underset{\sim}{\underline{M}} = \begin{pmatrix}1 & 2 \\\\ -1 & 0\end{pmatrix} \quad\text{e}\quad \underset{\sim}{I} = \begin{pmatrix}1 & 0 \\\\ 0 & 1\end{pmatrix}.
Precisaremos usar multiplicação de matrizes, multiplicação por escalar e soma de matrizes.

Passo 2

Calcular \underset{\sim}{\underline{M}}^{2} = \underset{\sim}{\underline{M}}\cdot\underset{\sim}{\underline{M}}:
\underset{\sim}{\underline{M}}^{2} = \begin{pmatrix}1 & 2 \\\\ -1 & 0\end{pmatrix} \begin{pmatrix}1 & 2 \\\\ -1 & 0\end{pmatrix} = \begin{pmatrix} 1\cdot1 + 2\cdot(-1) & 1\cdot2 + 2\cdot0 \\[6pt] -1\cdot1 + 0\cdot(-1) & -1\cdot2 + 0\cdot0 \end{pmatrix} = \begin{pmatrix}-1 & 2 \\\\ -1 & -2\end{pmatrix}.]

Passo 3

Subtrair 2\,\underset{\sim}{\underline{M}} de \underset{\sim}{\underline{M}}^{2}:
\underset{\sim}{\underline{M}}^{2} - 2\,\underset{\sim}{\underline{M}} = \begin{pmatrix}-1 & 2 \\\\ -1 & -2\end{pmatrix} - 2 \begin{pmatrix}1 & 2 \\\\ -1 & 0\end{pmatrix} = \begin{pmatrix}-1-2 & 2-4 \\\\ -1+2 & -2-0\end{pmatrix} = \begin{pmatrix}-3 & -2 \\\\ 1 & -2\end{pmatrix}.

Passo 4

Adicionar 3\,\underset{\sim}{I}:
\underset{\sim}{\underline{M}}^{2} - 2\,\underset{\sim}{\underline{M}} + 3\,\underset{\sim}{I} = \begin{pmatrix}-3 & -2 \\\\ 1 & -2\end{pmatrix} + 3 \begin{pmatrix}1 & 0 \\\\ 0 & 1\end{pmatrix} = \begin{pmatrix}-3+3 & -2+0 \\\\ 1+0 & -2+3\end{pmatrix} = \begin{pmatrix}0 & -2 \\\\ 1 & 1\end{pmatrix}.

Resposta

\begin{pmatrix}0 & -2 \\\\ 1 & 1\end{pmatrix}

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