Dadas as matrizes \boldsymbol{A}=\left[\begin{array}{rrr}\mathbf{2} & \mathbf{3} & \mathbf{8} \\\ \mathbf{4} & \mathbf{- 1} & \mathbf{- 6}\end{array}\right], \boldsymbol{B}=\left[\begin{array}{rrr}\mathbf{5} & \mathbf{- 7} & \mathbf{- 9} \\\ \mathbf{0} & \mathbf{4} & \mathbf{1}\end{array}\right] e \boldsymbol{C}=\left[\begin{array}{lll}\mathbf{0} & \mathbf{9} & \mathbf{8} \\\ \mathbf{1} & \mathbf{4} & \mathbf{6}\end{array}\right] :
Calcular \boldsymbol{X}=4 \boldsymbol{C}+\mathbf{2 A}-\mathbf{6 B}

Passo 1

O problema dá três matrizes de dimensão 2\times3: \boldsymbol{A}=\begin{pmatrix}2 & 3 & 8 \\\\ 4 & -1 & -6\end{pmatrix}, \boldsymbol{B}=\begin{pmatrix}5 & -7 & -9 \\\\ 0 & 4 & 1\end{pmatrix} e \boldsymbol{C}=\begin{pmatrix}0 & 9 & 8 \\\\ 1 & 4 & 6\end{pmatrix}.
Pede-se \boldsymbol{X}=4\boldsymbol{C}+2\boldsymbol{A}-6\boldsymbol{B}.
Vamos usar operações de multiplicação escalar em matrizes e soma/subtração de matrizes de mesmo tamanho.

Passo 2

Calculamos separadamente cada múltiplo escalar:
4\boldsymbol{C} =4\begin{pmatrix}0 & 9 & 8 \\\\ 1 & 4 & 6\end{pmatrix} =\begin{pmatrix}0 & 36 & 32 \\\\ 4 & 16 & 24\end{pmatrix},
2\boldsymbol{A} =2\begin{pmatrix}2 & 3 & 8 \\\\ 4 & -1 & -6\end{pmatrix} =\begin{pmatrix}4 & 6 & 16 \\\\ 8 & -2 & -12\end{pmatrix},
6\boldsymbol{B} =6\begin{pmatrix}5 & -7 & -9 \\\\ 0 & 4 & 1\end{pmatrix} =\begin{pmatrix}30 & -42 & -54 \\\\ 0 & 24 & 6\end{pmatrix}.

Passo 3

Agora somamos e subtraímos as matrizes correspondentes:
\boldsymbol{X} =4\boldsymbol{C}+2\boldsymbol{A}-6\boldsymbol{B} =\begin{pmatrix}0+4-30 & 36+6-(-42) & 32+16-(-54)\\\\ 4+8-0 & 16+(-2)-24 & 24+(-12)-6\end{pmatrix} =\begin{pmatrix}-26 & 84 & 102 \\\\ 12 & -10 & 6\end{pmatrix}.

Resposta

\boldsymbol{X} =\begin{pmatrix}-26 & 84 & 102 \\\\ 12 & -10 & 6\end{pmatrix}.

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