Problema sobre operacions entre matrius

Dades les matrius: $$A = \begin{pmatrix}1 & 0 & -2 \\ 4 & 1 & -3\end{pmatrix}, \quad B = \begin{pmatrix}-3 & 1 & 5 \\ 6 & 2 & 4\end{pmatrix}$$ calculeu: a) $A + B$ b) $2A – 3B$ c) $A^T$, $B^T$

Dades:

$$A = \begin{pmatrix}
1 & 0 & -2 \\
4 & 1 & -3
\end{pmatrix}, \quad
B = \begin{pmatrix}
-3 & 1 & 5 \\
6 & 2 & 4
\end{pmatrix}$$

a) $A + B$

Sumem les matrius element a element:
$$A + B = \begin{pmatrix}
1 + (-3) & 0 + 1 & -2 + 5 \\
4 + 6 & 1 + 2 & -3 + 4
\end{pmatrix} = \begin{pmatrix}
-2 & 1 & 3 \\
10 & 3 & 1
\end{pmatrix}$$

b) $2A – 3B$

• Primer, calculem $2A$:
  $$2A = 2 \cdot \begin{pmatrix}
  1 & 0 & -2 \\
  4 & 1 & -3
  \end{pmatrix} = \begin{pmatrix}
  2 & 0 & -4 \\
  8 & 2 & -6
  \end{pmatrix}$$
• Després, calculem $3B$:
  $$3B = 3 \cdot \begin{pmatrix}
  -3 & 1 & 5 \\
  6 & 2 & 4
  \end{pmatrix} = \begin{pmatrix}
  -9 & 3 & 15 \\
  18 & 6 & 12
  \end{pmatrix}$$
• Finalment, restem:
  $$2A – 3B = \begin{pmatrix}
  2 – (-9) & 0 – 3 & -4 – 15 \\
  8 – 18 & 2 – 6 & -6 – 12
  \end{pmatrix} = \begin{pmatrix}
  11 & -3 & -19 \\
  -10 & -4 & -18
  \end{pmatrix}$$

c) $A^T$, $B^T$

La transposta d’una matriu s’obté intercanviant files per columnes:

• $A^T$:
  $$A^T = \begin{pmatrix}
  1 & 4 \\
  0 & 1 \\
  -2 & -3
  \end{pmatrix}$$
• $B^T$:
  $$B^T = \begin{pmatrix}
  -3 & 6 \\
  1 & 2 \\
  5 & 4
  \end{pmatrix}$$

Resultats finals:

a) $A + B = \begin{pmatrix} -2 & 1 & 3 \\ 10 & 3 & 1 \end{pmatrix}$

b) $2A – 3B = \begin{pmatrix} 11 & -3 & -19 \\ -10 & -4 & -18 \end{pmatrix}$

c) $A^T = \begin{pmatrix} 1 & 4 \\ 0 & 1 \\ -2 & -3 \end{pmatrix}, \quad B^T = \begin{pmatrix} -3 & 6 \\ 1 & 2 \\ 5 & 4 \end{pmatrix}$