4. \( \left[\begin{array}{ccc}1 & 2 & 2 \\ -1 & 1 & 3\end{array}\right] \cdot\left[\begin{array}{c}2 \\ -1\end{array}\right] \)
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
Extra Insights
To perform the multiplication of the two matrices, we need to follow the rules of matrix multiplication. We're multiplying a \(2 \times 3\) matrix by a \(3 \times 1\) matrix: \[ \left[\begin{array}{ccc}1 & 2 & 2 \\ -1 & 1 & 3\end{array}\right] \cdot\left[\begin{array}{c}2 \\ -1 \\ 0\end{array}\right] \] **Step 1:** Multiply the first row of the first matrix by the column of the second matrix: \(1 \cdot 2 + 2 \cdot (-1) + 2 \cdot 0 = 2 - 2 + 0 = 0\) **Step 2:** Multiply the second row of the first matrix by the column of the second matrix: \(-1 \cdot 2 + 1 \cdot (-1) + 3 \cdot 0 = -2 - 1 + 0 = -3\) So, the result of the multiplication is: \[ \left[\begin{array}{c}0 \\ -3\end{array}\right] \] The final answer is: \[ \left[\begin{array}{c}0 \\ -3\end{array}\right] \]